devkuma – Algorithm

Introduction to Algorithm Programming

kc@example.com (kc kim) — Wed, 30 Mar 2022 09:27:00 +0900

Algorithm

An algorithm is a collection of operations used to solve a problem by using data stored in data structures.
It means a procedure for solving a problem.
It must satisfy input, output, definiteness, validity, and effectiveness.
An algorithm is logic and mathematics, and it is a basic idea applied to practical development.

Algorithm Courses

Practical Algorithm Course | YouTube
- Materials: https://blog.naver.com/ndb796/221226794899

Algorithm Problem Sites

Korean sites
International sites
- Codility
- LeetCode
- HackerRank
- CodeSignal
- Codewars
- LintCode

Big O Notation | Time Complexity and Space Complexity

kc@example.com (kc kim) — Tue, 06 Jul 2021 09:58:55 +0900

Time Complexity and Space Complexity

Time Complexity

Time complexity is a concept that predicts the execution time of source code logic and shows how efficient the code is. Execution time increases in proportion to operations. It is related to CPU usage.

Space Complexity

Space complexity is a concept that describes how efficiently code uses memory space. It often appears with data structures that reserve space in advance, such as static arrays or hash tables. It is related to RAM usage.

Source code logic must work first, but whether the code is efficient is determined not only by behavior but also by calculating time and space complexity.

What Is Big O Notation?

Big O is a representative way to mathematically express time and space complexity. However, it does not show the actual runtime of code. It is used to logically predict algorithm performance according to the growth rate of input data. Big O notation has the following two rules.

Keep only the highest-order term.

O(n² + 2n) -> O(n²)

Discard coefficients and constants boldly.

O(4n + 5) -> O(n)

Classifying Time Complexity with Big O

O(1) (Constant)

Represents an algorithm that always takes a constant time regardless of input size. Even if data increases, performance is barely affected.

O(log₂ n) (Logarithmic)

An algorithm whose processing time grows logarithmically as input size increases. For example, if the data becomes 10 times larger, processing time doubles. Binary search is representative, and some beneficial recursive cases also apply.

O(n) (Linear)

An algorithm whose processing time increases in proportion to input size. For example, if the data becomes 10 times larger, processing time also becomes 10 times larger. Linear search is representative.

O(n log₂ n) (Linear-Logarithmic)

An algorithm whose processing time increases by a logarithmic factor as data grows. For example, if the data becomes 10 times larger, processing time becomes about 20 times larger. This is the average time complexity of sorting algorithms such as Merge sort and Quick sort.

O(n²) (Quadratic)

An algorithm whose processing time increases rapidly as data grows. For example, if the data becomes 10 times larger, processing time can become up to 100 times larger. Nested loops, or an n² matrix, are representative. However, when m is smaller than n, it is better to write O(nm).

O(2ⁿ) (Exponential)

An algorithm whose processing time increases exponentially as the amount of data grows. The Fibonacci sequence is representative, and recursive algorithms that work inefficiently also fall into this category.

How to Calculate Time Complexity

Write down the number of executions whenever an instruction ends.

Basic Calculation

public static void basic() {
    int[] a = {7, 3, 6, 3, 3, 5, 6, 9};          // 1
    for (int i = 0; i < a.length - 1; i++) {     // n             [i > 0 to n-1 : n times called]
        for (int j = i + 1; j < a.length; j++) { // (n-1) * n     [j > 1 to n   : n times called]
            if (a[i] == a[j]) {                  // (n-1) * (n-1) [if statement eventually runs n-1 times]
                a[j] = 0;
            }
        }
    }

    for (int i : a) {
        System.out.println(i);
    }
}

Adding all instruction counts gives 2n²-2n+2. Omitting constants and considering only the highest-order term, it is written as O(n²).

Calculating Recursive Function Time Complexity

public static int factorial(int n) {
    System.out.println("n:" + n);
    if (n == 1) {
        return 1;
    }
    return n * factorial(n - 1);
}

The recursive call is made n times as shown below, so it is written as O(n).

factorial(n) -> factorial(n -1) .... ->factorial(2) -> factorial(1)

Algorithm | Sum, Average, Maximum, Minimum, and Data Swap

kc@example.com (kc kim) — Sat, 02 Apr 2022 21:55:00 +0900

Sum

Find the sum of all elements in an array.
The method takes out each element one by one and adds it to sum.

package com.devkuma.algorithum.programming.basic;

public class Sum {

    int sum(int[] nums) {
        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        return sum;
    }

    public static void main(String[] args) {
        Sum sum = new Sum();

        int[] nums = {1, 4, 3, 8, 5};
        System.out.println("합계: " + sum.sum(nums));
    }
}

Execution result:

합계: 21

Average

The average is the total value divided by the number of data items.

package com.devkuma.algorithum.programming.basic;

public class Average {
    int average(int[] nums) {
        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        return sum / nums.length;
    }

    public static void main(String[] args) {
        Average average = new Average();

        int[] nums = {1, 4, 3, 8, 5};
        System.out.println("평균:" + average.average(nums));
    }
}

Execution result:

합계: 21

Maximum

package com.devkuma.algorithum.programming.basic;

public class Max {
    int max(int[] nums) {
        int max = 0;
        for (int num : nums) {
            if (max < num) {
                max = num;
            }
        }
        return max;
    }

    public static void main(String[] args) {
        Max max = new Max();

        int[] nums = {1, 4, 3, 8, 5};
        System.out.println("최대값:" + max.max(nums));
    }
}

Execution result:

합계: 21

Minimum

package com.devkuma.algorithum.programming.basic;

public class Min {
    int min(int[] nums) {
        int min = Integer.MAX_VALUE;
        for (int num : nums) {
            if (min > num) {
                min = num;
            }
        }
        return min;
    }

    public static void main(String[] args) {
        Min min = new Min();

        int[] nums = {1, 4, 3, 8, 5};
        System.out.println("최소값:" + min.min(nums));
    }
}

Data Swap

When simply swapping two pieces of data, prepare the variables containing the data and one temporary variable, temp.
The reason is that when there are only two variables, changing one side first overwrites the data that should be placed into the other side.

package com.devkuma.algorithum.programming.basic;

public class Swap {

    public static void main(String[] args) {
        int a = 1;
        int b = 2;

        int temp = a;
        a = b;
        b = temp;

        System.out.println("a:" + a); // 2
        System.out.println("b:" + b); // 1
    }
}

Execution result:

합계: 21

Algorithm | Recursion Function

kc@example.com (kc kim) — Wed, 30 Mar 2022 09:27:00 +0900

Recursion Function

A recursive function is a function that calls itself again during processing.
For example, a function that calculates the sum from 1 to N can be defined with recursion as follows.

public class SumRecursion {

    public int sum(int n) {
        if(n == 0) {
            return 0;
        }
        return n + sum(n - 1);
    }

    public static void main(String[] args) {
        SumRecursion recursion = new SumRecursion();
        int result = recursion.sum(10);
        System.out.println(result); // 55
    }
}

Execution result:

One thing to note is that although sum(10) is called, the first value returned is from sum(0). Finally, the value of sum(10) is returned.

Euclidean Algorithm

The Euclidean algorithm is an algorithm for finding the greatest common divisor(GCD(m, n)) of two integers m and n.

When m is divided by n and the remainder is r, it is defined as follows.

GCD(m, n) = GCD(n, r)

Using a recursive function, you can write the well-known Euclidean algorithm.

public class GcdRecursion {

    public int gcd(int m, int n) {
        if(n == 0) {
            return m;
        }
        return gcd(n,  m % n);
    }

    public static void main(String[] args) {
        GcdRecursion recursion = new GcdRecursion();
        int result = recursion.gcd(30, 7);
        System.out.println(result); // 1
    }
}

Execution result:

Fibonacci Sequence

The Fibonacci sequence algorithm can be made using recursion twice and is defined as follows.

F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2 (n = 2, 3, ...)

public class FibonacciRecursion {
    
    public int fibonacci(int n) {
        if (n == 0) {
            return 0;
        } else if (n == 1) {
            return 1;
        }
        return fibonacci(n - 1) + fibonacci(n - 2);
    }

    public static void main(String[] args) {
        FibonacciRecursion recursion = new FibonacciRecursion();
        int result = recursion.fibonacci(10);
        System.out.println(result); // 55
    }
}

Execution result:

The complexity of the Fibonacci algorithm written above is O((1+√5)^n/2).

Calculating the Sum from a to b

The code below is an example that calculates the sum from a to b.

package com.devkuma.algorithum.programming.recursion;

public class SumRangeRecursion {

    int sumRange(int a, int b) {
        if (a == b) {
            return a;
        }
        return sumRange(a, b - 1) + b;
    }

    public static void main(String[] args) {
        SumRangeRecursion recursion = new SumRangeRecursion();
        int result = recursion.sumRange(1, 3);
        System.out.println(result); // 6
    }
}

Execution result:

In the example above, the calculation proceeds like 1 + 2 + 3 + 4 + 5 = 15.

Written out, it looks like this.

sumRange(1, 5) = sumRange(1, 5 - 1) + 5 > sumRange(1, 4) + 5 > 10 + 5 = 15
sumRange(1, 4) = sumRange(1, 4 - 1) + 4 > sumRange(1, 3) + 4 > 6 + 4 = 10
sumRange(1, 3) = sumRange(1, 3 - 1) + 3 > sumRange(1, 2) + 3 > 3 + 3 = 6
sumRange(1, 2) = sumRange(1, 2 - 1) + 2 > sumRange(1, 1) + 2 > 1 + 2 = 3
sumRange(1, 1) = 1

Determining Whether a Number Is Prime

The code below is an example that determines whether a number is prime.

public class PrimeRecursion {

    boolean hasDivisor(int n, int i) {
        if (i == n) {
            return false;
        }
        if (n % i == 0) {
            return true;
        }
        return hasDivisor(n, i + 1);
    }

    boolean isPrime(int n) {
        if (n == 1) {
            return false;
        } else if (n == 2) {
            return true;
        } else {
            return !hasDivisor(n, 2);
        }
    }

    public static void main(String[] args) {
        PrimeRecursion recursion = new PrimeRecursion();
        boolean result1 = recursion.isPrime(7);
        System.out.println(result1); // true

        boolean result2 = recursion.isPrime(4);
        System.out.println(result2); // false
    }

}

Execution result:

true
false

Mutual Recursion: Determining Odd and Even

The code below is an example that determines odd and even numbers using mutual recursive functions where methods call each other.

package com.devkuma.algorithum.programming.recursion;

public class EvenOddRecursion {

    boolean isEven(int n) {
        if (n == 0) {
            return true;
        }
        return isOdd(n-1);
    }

    boolean isOdd(int n) {
        if (n == 0) {
            return false;
        }
        return isEven(n-1);
    }

    public static void main(String[] args) {
        EvenOddRecursion recursion = new EvenOddRecursion();

        System.out.println(recursion.isEven(8)); // true
        System.out.println(recursion.isEven(9)); // false
        System.out.println(recursion.isOdd(8)); // false
        System.out.println(recursion.isOdd(9)); // true
    }
}

Execution result:

true
false
false
true

Finding the Maximum Value in an Array

The code below is an example that finds the maximum value up to count n in the input array a.

public class MaxArrayRecursion {

    public static int max(int[] a, int n) {
        int x;
        if (n == 1) {
            return a[0];
        } else {
            x = max(a, n - 1);
        }

        if (x > a[n - 1]) {
            return x;
        }
        return a[n - 1];
    }

    public static void main(String[] args) {
        int arr[] = {0, 80, 60, 40, 20, 100};
        System.out.println(max(arr, 3));
    }
}

Execution result:

Up to the third element of the input array a is “0, 80, 60”, and the largest value among them is “80”.

Algorithm | Linear Search

kc@example.com (kc kim) — Sat, 02 Apr 2022 20:39:00 +0900

Linear Search Algorithm

Linear search is an algorithm that searches for a target value sequentially from the beginning of a list to the end.

Advantage: it is the simplest search method, easy to implement, and can be used even with an unsorted list.
Disadvantage: it is inefficient when the list to search is long.

package com.devkuma.algorithum.programming.search.linear;

public class LinearSearch {

    void search(int[] nums, int target) {
        int index = -1;
        for(int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                index = i + 1;
                break;
            }
        }
        if (index > 0) {
            System.out.println(target + "은 " + index + "번째 있습니다.");
        } else {
            System.out.println(target + "은 없습니다.");
        }
    }

    public static void main(String[] args) {
        LinearSearch search = new LinearSearch();
        search.search(new int[]{1, 5, 3, 2, 8, 9}, 8);
        search.search(new int[]{1, 5, 3, 2, 8, 9}, 4);
    }
}

Execution result:

8은 5번째 있습니다.
4은 없습니다.

Finding the Minimum Value with Linear Search

The code below finds the minimum value by updating minValue whenever a value smaller than minValue is found among all elements. The maximum value can be found with the same algorithm.

package com.devkuma.algorithum.programming.search.linear;

public class MinLinearSearch {

    void searchMinimum(int[] nums) {
        int minValue = Integer.MAX_VALUE; // maximum value
        for(int i = 0; i < nums.length; i++) {
            if (nums[i] < minValue) {
                minValue = nums[i];
            }
        }
        System.out.println("최소값은 " + minValue + "입니다.");
    }

    public static void main(String[] args) {
        MinLinearSearch search = new MinLinearSearch();
        search.searchMinimum(new int[]{1, 5, 3, 2, 8, 9});
        search.searchMinimum(new int[]{2, 7, 8, 6, 5, 4});
    }
}

Execution result:

최소값은 1입니다.
최소값은 2입니다.

Solving a Problem with Linear Search

Now let’s solve a problem using the linear search algorithm.

Problem
Given N integers a(0), a(1), ..., a(N-1) and b(0), b(1), ..., b(N-1), choose one value each from the two sequences, a(i), b(j) [i = 0, 1, ..., N-1, j = 0, 1, ..., N-1], and find the minimum value of their sum among sums greater than or equal to integer R.

Solution

package com.devkuma.algorithum.programming.linearsearch;

public class LinearSearchQuiz1 {

    void minSum(int[] aNums, int[] bNums, int r) {
        int minSum = Integer.MAX_VALUE; // maximum value
        for(int i = 0; i < aNums.length; i++) {
            for(int j = 0; j < bNums.length; j++) {
                int sum = aNums[i] + bNums[j];
                if (sum >= r && minSum > sum) {
                    minSum = sum;
                }
            }
        }
        System.out.println("최소값은 " + minSum + "입니다.");
    }

    public static void main(String[] args) {
        LinearSearchQuiz1 search = new LinearSearchQuiz1();

        int[] aNums =   new int[]{9, 5, 3, 7, 8, 9};
        int[] bNums =   new int[]{3, 9, 5, 8, 3, 5};
        int r = 7;
        search.minSum(aNums, bNums, 7);
    }
}

Execution result:

최소값은 8입니다.

The complexity of the example above is O(N^2).

Algorithm | Binary Search

kc@example.com (kc kim) — Sat, 02 Apr 2022 21:24:00 +0900

Binary Search

Binary search is a method for finding a specific value in an ascending sorted list by first choosing the middle value and comparing whether it is larger or smaller than the target value.

Advantage: each repeated search doubles the probability of finding the target value, so it is fast.
Disadvantage: it can only be used on sorted lists.

Method Using a Loop

package com.devkuma.algorithum.programming.search.binary;

public class BinarySearchWhile {

    int search(int[] nums, int target) {
        int startIndex = 0;
        int endIndex = nums.length - 1;

        int middle = 1;

        while (startIndex <= endIndex) {
            middle = (startIndex + endIndex) / 2;
            if (nums[middle] == target)
                return middle + 1;
            else if (nums[middle] > target)
                endIndex = middle - 1;
            else
                startIndex = middle + 1;
        }
        return middle;
    }

    public static void main(String[] args) {
        BinarySearchWhile search = new BinarySearchWhile();

        int[] nums = {15, 28, 65, 67, 88, 92, 100, 150};
        int target = 28;
        int index = search.search(nums, target);

        if (index > 0) {
            System.out.println(target + "은 " + index + "번째 있습니다.");
        } else {
            System.out.println(target + "은 없습니다.");
        }
    }
}

Execution result:

5은 2번째 있습니다.

Method Using Recursion

package com.devkuma.algorithum.programming.search.binary;

public class BinarySearchRecursion {

    int search(int arr[], int target, int startIndex, int endIndex) {
        if (startIndex > endIndex)
            return -1;

        int middle = (startIndex + endIndex) / 2;
        if (arr[middle] == target)
            return middle;
        else if (arr[middle] > target)
            return search(arr, target, startIndex, middle - 1);
        else
            return search(arr, target, middle + 1, endIndex);
    }

    public static void main(String[] args) {
        BinarySearchRecursion search = new BinarySearchRecursion();

        int[] nums = {15, 28, 65, 67, 88, 92, 100, 150};
        int target = 28;
        int startIndex = 0;
        int endIndex = nums.length - 1;

        int index = search.search(nums, target, startIndex, endIndex);

        if (index > 0) {
            System.out.println(target + "은 " + index + "번째 있습니다.");
        } else {
            System.out.println(target + "은 없습니다.");
        }
    }
}

Execution result:

28은 2번째 있습니다.

Binary Search Example Explanation

Using the array from the example above, the binary search for 28 can be written out as follows.

{ 15, 28, 65, 67, 88, 92, 100, 150 }
- Middle value: 88 -> smaller -> { 15, 28, 65, 67 }
- Middle value: 43 -> smaller -> { 15, 28 }
- Middle value: 28 -> end

Baekjoon Algorithm | Problem 1037: Divisors

kc@example.com (kc kim) — Sun, 12 Jun 2022 16:26:00 +0900

Source

https://www.acmicpc.net/problem/1037

Problem

Solve Baekjoon Online Judge problem 1037, Divisors.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

2 4 2

Sample Output 1

Sample Input 2

1 2

Sample Output 2

Sample Input 3

6 3 4 2 12 6 8

Sample Output 3

Sample Input 4

14 14 26456 2 28 13228 3307 7 23149 8 6614 46298 56 4 92596

Sample Output 4

185192

Algorithm Classification

Math
Brute Force
Number Theory

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int n = Integer.parseInt(br.readLine());
            StringTokenizer nums = new StringTokenizer(br.readLine(), " ");

            int max = Integer.MIN_VALUE;
            int min = Integer.MAX_VALUE;
            for(int i = 0; i < n; i++) {
                int num  = Integer.parseInt(nums.nextToken());
                if(num > max) max = num;
                if(num < min) min = num;
            }
            System.out.println(max * min);

        }
    }
}

Baekjoon Algorithm | Problem 1065: Hansu

kc@example.com (kc kim) — Sat, 09 Apr 2022 08:31:00 +0900

Source

https://www.acmicpc.net/problem/1065

Problem

Solve Baekjoon Online Judge problem 1065, Hansu.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

110

Sample Output 1

Sample Input 2

Sample Output 2

Sample Input 3

210

Sample Output 3

105

Sample Input 4

1000

Sample Output 4

144

Sample Input 5

500

Sample Output 5

199

Algorithm Classification

Brute Force

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            int n = Integer.parseInt(br.readLine());

            int count = 0;
            if (n < 100) {
                count = n;
            } else {
                count = 99;

                int loop = n;
                if (n == 1000)
                    loop = 999;

                for (int i = 100; i <= loop; i++) {

                    int a = i % 10;
                    int b = (i / 10) % 10;
                    int c = (i / 100) % 10;

                    if (a - b == b - c) {
                        count++;
                        //System.out.println("i=" + i + ", count=" + count + ", a=" + a + ", b=" + b + ", c=" + c + ", result=" + (a - b == b - c));
                    }

                }
            }

            System.out.println(count);
        }
    }
}

Baekjoon Algorithm | Problem 1110: Addition Cycle

kc@example.com (kc kim) — Thu, 07 Apr 2022 08:53:00 +0900

Source

https://www.acmicpc.net/problem/1110

Problem

Solve Baekjoon Online Judge problem 1110, Addition Cycle.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

Sample Input 2

Sample Output 2

Sample Input 3

Sample Output 3

Sample Input 4

Sample Output 4

Sample Input 5

Sample Output 5

Algorithm Classification

Math
Implementation

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {

        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            final int n = Integer.parseInt(br.readLine());

            int newNum = n;
            int count = 0;
            do {
                int num1 = (newNum / 10);
                int num2 = newNum % 10;

                int sum = num1 + num2;
                int sum2 = sum % 10;

                newNum = (num2 * 10) + sum2;

                count++;
            } while (n != newNum);

            System.out.println(count);
        }
    }
}

Baekjoon Algorithm | Problem 1157: Word Study

kc@example.com (kc kim) — Tue, 19 Apr 2022 09:46:00 +0900

Source

https://www.acmicpc.net/problem/1157

Problem

Solve Baekjoon Online Judge problem 1157, Word Study.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Mississipi

Sample Output 1

Sample Input 2

zZa

Sample Output 2

Sample Input 3

baaa

Sample Output 3

Algorithm Classification

Math
Implementation

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            String input = br.readLine();

            Map<Character, Integer> alphabetCountMap = new HashMap<>();

            Set<Character> maxSet = new HashSet<>();
            int max = Integer.MIN_VALUE;

            for (int i = 0; i < input.length(); i++) {
                Character alphabet = Character.toUpperCase(input.charAt(i));

                int count;
                if (alphabetCountMap.containsKey(alphabet)) {
                    count = alphabetCountMap.get(alphabet) + 1;
                } else {
                    count = 0;
                }
                alphabetCountMap.put(alphabet, count);

                if (max < count) {
                    max = count;
                    maxSet.clear();
                    maxSet.add(alphabet);
                } else if (max == count) {
                    maxSet.add(alphabet);
                }
            }

            if (maxSet.size() > 1) {
                System.out.println("?");
            } else {
                System.out.println(maxSet.iterator().next());
            }
        }
    }
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main2 {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            String input = br.readLine();

            int[] alphabet = new int[26];
            for (int i = 0; i < input.length(); i++) {
                //char c = Character.toUpperCase(input.charAt(i));
                //alphabet[c - 65]++;
                char c = input.charAt(i);
                if (c >= 'a')
                    alphabet[c - 'a']++;
                else
                    alphabet[c - 'A']++;
            }

            char output = 0;
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < alphabet.length; i++) {

                if(max < alphabet[i]) {
                    max = alphabet[i];
                    output = (char) (i + 'A');
                } else if(max == alphabet[i]) {
                    output = '?';
                }
            }

            System.out.println(output);
        }
    }
}

Baekjoon Algorithm | Problem 1316: Group Word Checker

kc@example.com (kc kim) — Fri, 15 Apr 2022 07:13:00 +0900

Source

https://www.acmicpc.net/problem/1316

Problem

Solve Baekjoon Online Judge problem 1316, Group Word Checker.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

3 happy new year

Sample Output 1

Sample Input 2

4 aba abab abcabc a

Sample Output 2

Sample Input 3

5 ab aa aca ba bb

Sample Output 3

Sample Input 4

2 yzyzy zyzyz

Sample Output 4

Sample Input 5

1 z

Sample Output 5

Algorithm Classification

Implementation
String

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            final int n = Integer.parseInt(br.readLine());

            int count = n;
            for (int i = 0; i < n; i++) {
                String line = br.readLine();

                Set set = new HashSet<Character>();
                for (int j = 0; j < line.length(); j++) {

                    if (j != 0 && line.charAt(j) == line.charAt(j - 1)) {
                        continue;
                    }

                    char b = line.charAt(j);
                    if (set.contains(b)) {
                        count--;
                        break;
                    }
                    set.add(b);
                }
            }

            System.out.println(count);
        }
    }
}

Baekjoon Algorithm | Problem 1330: Compare Two Numbers

kc@example.com (kc kim) — Sun, 03 Apr 2022 01:06:00 +0900

Source

https://www.acmicpc.net/problem/1330

Problem

Solve Baekjoon Online Judge problem 1330, Compare Two Numbers.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

1 2

Sample Output 1

Sample Input 2

10 2

Sample Output 2

Sample Input 3

5 5

Sample Output 3

Algorithm Classification

Math
Implementation
Arithmetic

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {

        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            String[] input = br.readLine().split(" ");

            int a = Integer.parseInt(input[0]);
            int b = Integer.parseInt(input[1]);

            String c;
            if (a > b)
                c = ">";
            else if (a < b)
                c = "<";
            else
                c = "==";

            System.out.print(c);
        }
    }
}

Baekjoon Algorithm | Problem 1546: Average

kc@example.com (kc kim) — Sat, 09 Apr 2022 17:29:00 +0900

Source

https://www.acmicpc.net/problem/1546

Problem

Solve Baekjoon Online Judge problem 1546, Average.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

3 40 80 60

Sample Output 1

75.0

Sample Input 2

3 10 20 30

Sample Output 2

66.666667

Sample Input 3

4 1 100 100 100

Sample Output 3

75.25

Sample Input 4

5 1 2 4 8 16

Sample Output 4

38.75

Sample Input 5

2 3 10

Sample Output 5

65.0

Sample Input 6

4 10 20 0 100

Sample Output 6

32.5

Sample Input 7

1 50

Sample Output 7

100.0

Sample Input 8

9 10 20 30 40 50 60 70 80 90

Sample Output 8

55.55555555555556

Algorithm Classification

Math
Arithmetic

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int n = Integer.parseInt(br.readLine());
            StringTokenizer input = new StringTokenizer(br.readLine(), " ");

            double max = Integer.MIN_VALUE;
            double[] scores = new double[n];
            for (int i = 0; i < n; i++) {
                double score = Double.parseDouble(input.nextToken());
                if (score > max) {
                    max = score;
                }
                scores[i] = score;
            }

            double sum = 0;
            for (int i = 0; i < n; i++) {
                sum += scores[i] / max * 100;
            }

            System.out.println(sum/n);
        }
    }
}

Baekjoon Algorithm | Problem 1712: Break-even Point

kc@example.com (kc kim) — Thu, 14 Apr 2022 18:21:00 +0900

Source

https://www.acmicpc.net/problem/1712

Problem

Solve Baekjoon Online Judge problem 1712, Break-even Point.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

1000 70 170

Sample Output 1

Sample Input 2

3 2 1

Sample Output 2

-1

Sample Input 3

2100000000 9 10

Sample Output 3

2100000001

Algorithm Classification

Math
Arithmetic

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            StringTokenizer input = new StringTokenizer(br.readLine(), " ");
            final long a = Long.parseLong(input.nextToken());
            final long b = Long.parseLong(input.nextToken());
            final long c = Long.parseLong(input.nextToken());

            if (b >= c) {
                System.out.println(-1);
                return;
            }

            int n = 0;
            long cost = 0;
            long income = 0;

            do {
                n++;

                cost = a + (b * n);
                income = c * n;

                //System.out.println("n=" + n + ", cost=" + cost + ", income=" + income);
            } while (cost >= income);

            System.out.println(n);
        }
    }
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            StringTokenizer input = new StringTokenizer(br.readLine(), " ");
            final long a = Long.parseLong(input.nextToken());
            final long b = Long.parseLong(input.nextToken());
            final long c = Long.parseLong(input.nextToken());

            int count = -1;
            if (b < c) {
                count = (int) ((a / (c - b)) + 1);
            }

            System.out.println(count);
        }
    }
}

Baekjoon Algorithm | Problem 1759: Making Password

kc@example.com (kc kim) — Sun, 12 Jun 2022 22:26:00 +0900

Source

https://www.acmicpc.net/problem/1759

Problem

Solve Baekjoon Online Judge problem 1759, Making Password.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

4 6 a t c i s w

Sample Output 1

acis acit aciw acst acsw actw aist aisw aitw astw cist cisw citw istw

Algorithm Classification

Math
Brute Force
Combinatorics
Backtracking

Solution

import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

    static int L;
    static int C;
    static char[] chars;
    static char[] keys;

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            StringTokenizer input = new StringTokenizer(br.readLine(), " ");
            L = Integer.parseInt(input.nextToken());
            C = Integer.parseInt(input.nextToken());

            input = new StringTokenizer(br.readLine(), " ");

            chars = new char[C];
            for (int i = 0; i < C; i++) {
                chars[i] = input.nextToken().charAt(0);
            }
        }

        Arrays.sort(chars);

        keys = new char[L];

        cypherText(0, 0);
    }

    private static void cypherText(int start, int index) {
        if (index == L) {
            int mo = 0;
            int ja = 0;

            for (char x : keys) {
                if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u') {
                    mo++;
                } else {
                    ja++;
                }
            }

            if (mo >= 1 && ja >= 2) {
                System.out.println(keys);
            }

            return;
        }

        for (int i = start; i < C; i++) {
            keys[index] = chars[i];
            cypherText(i + 1, index + 1);
        }
    }
}

Baekjoon Algorithm | Problem 1978: Finding Primes

kc@example.com (kc kim) — Wed, 20 Apr 2022 07:15:00 +0900

Source

https://www.acmicpc.net/problem/1978

Problem

Solve Baekjoon Online Judge problem 1978, Finding Primes.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

4 1 3 5 7

Sample Output 1

Algorithm Classification

Math
Number Theory
Primality Testing
Sieve of Eratosthenes

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int n = Integer.parseInt(br.readLine());

            StringTokenizer st = new StringTokenizer(br.readLine());
            int count = 0;
            for (int i = 0; i < n; i++) {
                int a = Integer.parseInt(st.nextToken());
                if (isPrime(a)) {
                    count++;
                }
            }

            System.out.println(count);
        }
    }

    private static boolean isPrime(int num) {
        if (num == 1)
            return false;

        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) throws IOException {

        boolean[] prime = new boolean[1001];
        prime[0] = true;
        prime[1] = true;
        for (int i = 2; i <= Math.sqrt(1000); i++) {
            if (prime[i] == true)
                continue;
            for (int j = i * i; j < 1001; j = j + i)
                prime[j] = true;
        }

        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            int n = Integer.parseInt(br.readLine());
            StringTokenizer st = new StringTokenizer(br.readLine());

            int count = 0;
            for (int i = 0; i < n; i++) {
                int a = Integer.parseInt(st.nextToken());
                if (prime[a] == false) {
                    count++;
                }
            }

            System.out.println(count);
        }
    }
}

Baekjoon Algorithm | Problem 2309: Seven Dwarfs

kc@example.com (kc kim) — Sun, 12 Jun 2022 21:26:00 +0900

Source

https://www.acmicpc.net/problem/2309

Problem

Solve Baekjoon Online Judge problem 2309, Seven Dwarfs.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

20 7 23 19 10 15 25 8 13

Sample Output 1

7 8 10 13 19 20 23

Algorithm Classification

Brute Force
Sorting

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {

    public static void main(String[] args) throws IOException {
        int sum = 0;
        int[] arr = new int[9];
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            for(int i= 0; i < 9;i++){
                arr[i] = Integer.parseInt(br.readLine());
                sum += arr[i];
            }
        }

        Arrays.sort(arr);

        boolean exit = false;
        for (int i = 0; i < arr.length - 1; i++) {
            for (int j = i; j < arr.length; j++) {
                if (sum - arr[i] - arr[j] == 100) {
                    for (int y = 0; y < arr.length; y++) {
                        if(arr[i] == arr[y] || arr[j] == arr[y]) {
                            continue;
                        }
                        System.out.println(arr[y]);
                    }
                    exit =true;
                    break;
                }
            }
            if (exit) {
                break;
            }
        }
    }
}

Baekjoon Algorithm | Problem 2438: Print Stars - 1

kc@example.com (kc kim) — Sun, 03 Apr 2022 01:57:00 +0900

Source

https://www.acmicpc.net/problem/2438

Problem

Solve Baekjoon Online Judge problem 2438, Print Stars - 1.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

\* \** \*** \**** \*****

Algorithm Classification

Implementation

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int input = Integer.parseInt(br.readLine());

            String star = "";
            for (int i = 0; i < input; i++) {
                star = star + "*";
                System.out.println(star);
            }
        }
    }
}

Baekjoon Algorithm | Problem 2439: Print Stars - 2

kc@example.com (kc kim) — Sun, 03 Apr 2022 01:57:00 +0900

Source

https://www.acmicpc.net/problem/2438

Problem

Solve Baekjoon Online Judge problem 2439, Print Stars - 2.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

\* \** \*** \**** \*****

Algorithm Classification

Implementation

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int input = Integer.parseInt(br.readLine());

            String space = "";
            for (int i = 0; i < input - 1; i++) {
                space = space + " ";
            }
            String star = "";
            for (int i = 0; i < input; i++) {
                star = star + "*";
                System.out.println(space.substring(0, space.length() - i) + star);
            }
        }
    }
}

Baekjoon Algorithm | Problem 2557: Hello World

kc@example.com (kc kim) — Sun, 03 Apr 2022 01:57:00 +0900

Source

https://www.acmicpc.net/problem/2557

Problem

Solve Baekjoon Online Judge problem 2557, Hello World.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

Hello World!

Algorithm Classification

Implementation

Solution

public class Main {
    public static void main(String[] args) {
        System.out.println("Hello World!");
    }
}

Baekjoon Algorithm | Problem 2577: Number of Digits

kc@example.com (kc kim) — Thu, 07 Apr 2022 08:37:00 +0900

Source

https://www.acmicpc.net/problem/2577

Problem

Solve Baekjoon Online Judge problem 2577, Number of Digits.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

150 266 427

Sample Output 1

3 1 0 2 0 0 0 2 0 0

Algorithm Classification

Math
Implementation

Solution

import java.io.*;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int a = Integer.parseInt(br.readLine());
            int b = Integer.parseInt(br.readLine());
            int c = Integer.parseInt(br.readLine());

            String m = String.valueOf(a * b * c);

            int[] counts = int[10];
            for (int i = 0; i < m.length(); i++) {
                int num = Integer.parseInt(String.valueOf(m.charAt(i)));
                counts[num]++;
            }

            for(int count : counts) {
                System.out.println(count);
            }
        }
    }
}

Baekjoon Algorithm | Problem 2581: Primes

kc@example.com (kc kim) — Wed, 20 Apr 2022 06:19:00 +0900

Source

https://www.acmicpc.net/problem/2581

Problem

Solve Baekjoon Online Judge problem 2581, Primes.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

60 100

Sample Output 1

620 61

Sample Input 2

64 65

Sample Output 2

-1

Algorithm Classification

Math
Number Theory
Primality Testing

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int m = Integer.parseInt(br.readLine());
            int n = Integer.parseInt(br.readLine());

            int sum = 0;
            int min = Integer.MAX_VALUE;
            for (int i = m; i <= n; i++) {
                if (isPrime(i)) {
                    sum += i;
                    if (min > i) {
                        min = i;
                    }
                }
            }
            if (sum == 0) {
                System.out.println(-1);
            } else {
                System.out.println(sum);
                System.out.println(min);
            }
        }
    }

    private static boolean isPrime(int num) {
        if (num == 1)
            return false;

        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
}

Baekjoon Algorithm | Problem 2675: String Repetition

kc@example.com (kc kim) — Wed, 20 Apr 2022 06:19:00 +0900

Source

https://www.acmicpc.net/problem/2675

Problem

Solve Baekjoon Online Judge problem 2675, String Repetition.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

2 3 ABC 5 /HTP

Sample Output 1

AAABBBCCC /////HHHHHTTTTTPPPPP

Algorithm Classification

Implementation
String

Solution

import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ) {
            int n = Integer.parseInt(br.readLine());

            for (int i = 0; i < n; i++) {
                String[] input = br.readLine().split(" ");
                final int r = Integer.parseInt(input[0]);
                final String s = input[1];

                for (int j = 0; j < s.length(); j++) {
                    for (int x = 0; x < r; x++) {
                        bw.write(s.charAt(j));
                    }
                }
                bw.write("\n");
            }

            bw.flush();
        }
    }
}

Baekjoon Algorithm | Problem 2739: Multiplication Table

kc@example.com (kc kim) — Sun, 03 Apr 2022 01:06:00 +0900

Source

https://www.acmicpc.net/problem/2739

Problem

Solve Baekjoon Online Judge problem 2739, Multiplication Table.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

2 * 1 = 2 2 * 2 = 4 2 * 3 = 6 2 * 4 = 8 2 * 5 = 10 2 * 6 = 12 2 * 7 = 14 2 * 8 = 16 2 * 9 = 18

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int input = Integer.parseInt(br.readLine());

            for (int i = 1; i < 10; i++) {
                System.out.println(input + " * " + i + " = " + (input * i));
            }
        }
    }
}

Baekjoon Algorithm | Problem 2741: Print N

kc@example.com (kc kim) — Sun, 03 Apr 2022 00:42:00 +0900

Source

https://www.acmicpc.net/problem/2741

Problem

Solve Baekjoon Online Judge problem 2741, Print N.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

1 2 3 4 5

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int n = Integer.parseInt(br.readLine());

            for (int i = 1; i <= n; i++) {
                System.out.println(i);
            }
        }
    }
}

Baekjoon Algorithm | Problem 2742: Print N Reversed

kc@example.com (kc kim) — Tue, 05 Apr 2022 09:21:00 +0900

Source

https://www.acmicpc.net/problem/2742

Problem

Solve Baekjoon Online Judge problem 2742, Print N Reversed.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

5 4 3 2 1

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int n = Integer.parseInt(br.readLine());

            for (int i = n; i > 0; i--) {
                System.out.println(i);
            }
        }
    }
}

Baekjoon Algorithm | Problem 2753: Leap Year

kc@example.com (kc kim) — Tue, 05 Apr 2022 07:50:00 +0900

Source

https://www.acmicpc.net/problem/2753

Problem

Solve Baekjoon Online Judge problem 2753, Leap Year.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

2000

Sample Output 1

Sample Input 2

1999

Sample Output 2

Algorithm Classification

Math
Implementation

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int input = Integer.parseInt(br.readLine());

            if (input % 4 == 0 && (input % 100 != 0 || input % 400 == 0)) {
                System.out.println(1);
            } else {
                System.out.println(0);
            }
        }
    }
}

Baekjoon Algorithm | Problem 2869: Climbing Snail

kc@example.com (kc kim) — Sun, 03 Apr 2022 01:57:00 +0900

Source

https://www.acmicpc.net/problem/2869

Problem

Solve Baekjoon Online Judge problem 2869, Climbing Snail.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

2 1 5

Sample Output 1

Sample Input 2

5 1 6

Sample Output 2

Sample Input 3

2 1 5

Sample Output 3

Sample Input 4

100 99 1000000000

Sample Output 4

999999901

Algorithm Classification

Math

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
    
    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            StringTokenizer input = new StringTokenizer(br.readLine(), " ");

            double a = Integer.parseInt(input.nextToken());
            double b = Integer.parseInt(input.nextToken());
            double v = Integer.parseInt(input.nextToken());

            System.out.println((int) Math.ceil((v - a) / (a - b)) + 1);
        }
    }
}

Baekjoon Algorithm | Problem 2884: Alarm Clock

kc@example.com (kc kim) — Tue, 05 Apr 2022 08:04:00 +0900

Source

https://www.acmicpc.net/problem/2884

Problem

Solve Baekjoon Online Judge problem 2884, Alarm Clock.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

10 10

Sample Output 1

9 25

Sample Input 2

0 30

Sample Output 2

23 45

Sample Input 3

23 40

Sample Output 3

22 55

Algorithm Classification

Math
Arithmetic

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            String[] input = br.readLine().split(" ");
            final int inputHour = Integer.parseInt(input[0]);
            final int inputMinute = Integer.parseInt(input[1]);

            int hour;
            int minute;
            if (inputMinute < 45) {
                if (inputHour == 0) {
                    hour = 23;
                } else {
                    hour = (inputHour - 1);
                }
                minute = inputMinute + 60 - 45;
            } else {
                hour = inputHour;
                minute = inputMinute - 45;
            }
            System.out.println(hour + " " + minute);
        }
    }
}

Baekjoon Algorithm | Problem 2908: Constant

kc@example.com (kc kim) — Tue, 19 Apr 2022 09:28:00 +0900

Source

https://www.acmicpc.net/problem/2908

Problem

Solve Baekjoon Online Judge problem 2908, Constant.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

734 893

Sample Output 1

437

Sample Input 2

221 231

Sample Output 2

23 45

Sample Input 3

839 237

Sample Output 3

938

Algorithm Classification

Math
Implementation

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            String[] input = br.readLine().split(" ");

            int a = Integer.parseInt(new StringBuffer(input[0]).reverse().toString());
            int b = Integer.parseInt(new StringBuffer(input[1]).reverse().toString());

            if(a > b) {
                System.out.println(a);
            } else {
                System.out.println(b);
            }
        }
    }
}

Baekjoon Algorithm | Problem 2941: Croatian Alphabet

kc@example.com (kc kim) — Fri, 15 Apr 2022 07:13:00 +0900

Source

https://www.acmicpc.net/problem/2941

Problem

Solve Baekjoon Online Judge problem 2941, Croatian Alphabet.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

ljes=njak

Sample Output 1

Sample Input 2

ddz=z=

Sample Output 2

Sample Input 3

nljj

Sample Output 3

Sample Input 4

c=c=

Sample Output 4

Sample Input 5

dz=ak

Sample Output 5

Algorithm Classification

Implementation
String

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            String input = br.readLine();

            String[] chars = new String[]{"c=", "c-", "dz=", "d-", "lj", "nj", "s=", "z="};

            int count = 0;
            int fromIndex = 0;
            do {
                String searchChar = null;
                for (String ch : chars) {
                    if (input.startsWith(ch, fromIndex)) {
                        searchChar = ch;
                        break;
                    }
                }
                if (searchChar != null) {
                    fromIndex += searchChar.length();
                } else {
                    fromIndex++;
                }
                //System.out.println("searchChar=" + searchChar + ", fromIndex=" + fromIndex + ", input.length=" + input.length());
                count++;
            } while (input.length() > fromIndex);

            System.out.println(count);
        }
    }
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main2 {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            String str = br.readLine();

            String[] chars = new String[]{"c=", "c-", "dz=", "d-", "lj", "nj", "s=", "z="};

            for (String c : chars) {
                str = str.replace(c, "_");
            }

            System.out.println(str.length());
        }
    }
}

Baekjoon Algorithm | Problem 3052: Remainder

kc@example.com (kc kim) — Fri, 08 Apr 2022 07:37:00 +0900

Source

https://www.acmicpc.net/problem/3052

Problem

Solve Baekjoon Online Judge problem 3052, Remainder.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

1 2 3 4 5 6 7 8 9 10

Sample Output 1

Sample Input 2

42 84 252 420 840 126 42 84 420 126

Sample Output 2

Sample Input 3

39 40 41 42 43 44 82 83 84 85

Sample Output 3

Algorithm Classification

Math
Implementation
Geometry

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            Set<Integer> nums = new HashSet();
            for(int i = 0; i < 10; i++) {
                int a = Integer.parseInt(br.readLine());
                nums.add(a % 42);
            }

            System.out.println(nums.size());
        }
    }
}

Baekjoon Algorithm | Problem 4344: Above Average

kc@example.com (kc kim) — Tue, 12 Apr 2022 07:30:00 +0900

Source

https://www.acmicpc.net/problem/4344

Problem

Solve Baekjoon Online Judge problem 4344, Above Average.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

5 5 50 50 70 80 100 7 100 95 90 80 70 60 50 3 70 90 80 3 70 90 81 9 100 99 98 97 96 95 94 93 91

Sample Output 1

40.000% 57.143% 33.333% 66.667% 55.556%

Algorithm Classification

Math
Arithmetic

Solution

import java.io.*;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out))
        ) {

            int c = Integer.parseInt(br.readLine());

            for (int i = 0; i < c; i++) {

                StringTokenizer input = new StringTokenizer(br.readLine(), " ");

                int n = Integer.valueOf(input.nextToken());

                int[] scores = new int[n];

                int sum = 0;
                for (int j = 0; j < n; j++) {
                    int score = Integer.valueOf(input.nextToken());
                    scores[j] = score;
                    sum += score;
                }

                float avg = (float) sum / (float) n;

                int highCount = 0;
                for (int j = 0; j < n; j++) {
                    if (avg < scores[j]) {
                        highCount++;
                    }
                }

                bw.write(String.format("%.3f%%\n", (((float) highCount / n) * 100)));
            }

            bw.flush();
        }
    }
}

Baekjoon Algorithm | Problem 4375: 1

kc@example.com (kc kim) — Sun, 12 Jun 2022 14:24:00 +0900

Source

https://www.acmicpc.net/problem/4375

Problem

Solve Baekjoon Online Judge problem 4375, 1.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

3 7 9901

Sample Output 1

3 6 12

Algorithm Classification

Math
Brute Force
Number Theory

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            String str = null;
            while ((str = br.readLine()) != null) {
                final int n = Integer.parseInt(str);

                int x =1;
                for (int i = 1; ; i++) {
                    x = x % n;
                    x = x * 10 + 1;
                    if (x == 1) {
                        System.out.println(i);
                        break;
                    }
                }
            }
        }
    }
}

Baekjoon Algorithm | Problem 4673: Self Numbers

kc@example.com (kc kim) — Sat, 09 Apr 2022 17:56:00 +0900

Source

https://www.acmicpc.net/problem/4673

Problem

Solve Baekjoon Online Judge problem 4673, Self Numbers.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

1 3 5 7 9 20 31 42 53 64 | | <-- a lot more numbers | 9903 9914 9925 9927 9938 9949 9960 9971 9982 9993

Algorithm Classification

Math
Implementation
Brute Force

Solution

import java.util.HashSet;
import java.util.Set;

public class Main {
    public static void main(String[] args) {

        Set<Integer> notSelfNumber = new HashSet();
        for(int i = 1; i <= 10000; i++) {
            int sum = i;

            if(i < 10){
                sum += i;
            } else {
                int length = String.valueOf(i).length();
                for(int j = 1; j <= length; j++){
                    if(j == 1){
                        sum += i % 10;
                    } else {
                        int y = 1;
                        for(int x = 1; x < j; x++) {
                            y = y * 10;
                        }
                        sum += (i / y) % 10;
                    }
                }
            }
            notSelfNumber.add(sum);
        }

        for(int i = 1; i <= 10000; i++) {
            if (notSelfNumber.contains(i)) {
                continue;
            }
            System.out.println(i);
        }
    }
}

public class Main {
    public static void main(String[] args) {
        boolean[] notSelfNumber = new boolean[10000];
        for (int i = 1; i <= 10000; i++) {

            int sum = i;
            int n = i;
            while (n != 0) {
                sum += (n % 10);
                n /= 10;
            }

            if (sum - 1 < notSelfNumber.length) {
                notSelfNumber[sum - 1] = true;
            }
        }

        for (int i = 0; i < 10000; i++) {
            if(notSelfNumber[i] == true){
                continue;
            }
            System.out.println(i + 1);
        }
    }
}

Baekjoon Algorithm | Problem 5622: Dial

kc@example.com (kc kim) — Fri, 15 Apr 2022 08:58:00 +0900

Source

https://www.acmicpc.net/problem/5622

Problem

Solve Baekjoon Online Judge problem 5622, Dial.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

Sample Input 2

UNUCIC

Sample Output 2

Algorithm Classification

Implementation

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            String input = br.readLine();

            int sum = 0;
            for (int i = 0; i < input.length(); i++) {
                char c = input.charAt(i);

                if ('A' <= c && c <= 'C') { // ABC = 2
                    sum += 3;
                } else if ('D' <= c && c <= 'F') { // DEF = 3
                    sum += 4;
                } else if ('G' <= c && c <= 'I') { // GHI = 4
                    sum += 5;
                } else if ('J' <= c && c <= 'L') { // JKL = 5
                    sum += 6;
                } else if ('M' <= c && c <= 'O') { // MNO = 6
                    sum += 7;
                } else if ('P' <= c && c <= 'S') { // PQRS = 7
                    sum += 8;
                } else if ('T' <= c && c <= 'V') { // TUV = 8
                    sum += 9;
                } else if ('W' <= c && c <= 'Z') { // WXYZ = 9
                    sum += 10;
                }
            }

            System.out.println(sum);
        }
    }
}

Baekjoon Algorithm | Problem 8393: Sum

kc@example.com (kc kim) — Sun, 03 Apr 2022 00:34:00 +0900

Source

https://www.acmicpc.net/problem/8393

Problem

Solve Baekjoon Online Judge problem 8393, Sum.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

Sample Output 1

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int input = Integer.parseInt(br.readLine());

            int sum = 0;
            for (int i = 1; i <= input; i++) {
                sum = sum + i;
            }

            System.out.print(sum);
        }
    }
}

Baekjoon Algorithm | Problem 8958: OX Quiz

kc@example.com (kc kim) — Sat, 09 Apr 2022 19:56:00 +0900

Source

https://www.acmicpc.net/problem/8958

Problem

Solve Baekjoon Online Judge problem 8958, OX Quiz.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

5 OOXXOXXOOO OOXXOOXXOO OXOXOXOXOXOXOX OOOOOOOOOO OOOOXOOOOXOOOOX

Sample Output 1

10 9 7 55 30

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int n = Integer.parseInt(br.readLine());

            int[] sums = new int[n];
            for (int i = 0; i < n; i++) {
                String oxLine = br.readLine();

                int sum = 0;
                int score = 1;
                for(int j=0; j < oxLine.length(); j++) {
                    char ox = oxLine.charAt(j);
                    if (ox == 'O') {
                        sum += score;
                        score++;
                    } else {
                        score = 1;
                    }
                }
                sums[i] = sum;
            }

            for (int sum : sums) {
                System.out.println(sum);
            }
        }
    }
}

Baekjoon Algorithm | Problem 9095: 1, 2, 3 Addition

kc@example.com (kc kim) — Sun, 12 Jun 2022 21:56:00 +0900

Source

https://www.acmicpc.net/problem/9095

Problem

Solve Baekjoon Online Judge problem 9095, 1, 2, 3 Addition.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

3 4 7 10

Sample Output 1

7 44 274

Algorithm Classification

Dynamic Programming

Solution

import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ) {
            int T = Integer.parseInt(br.readLine());

            for (int t = 0; t < T; t++) {
                int n = Integer.parseInt(br.readLine());
                int[] dp = new int[11];

                dp[1] = 1;
                dp[2] = 2;
                dp[3] = 4;

                for (int i = 4; i <= n; i++) {
                    dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
                }

                bw.write(dp[n] + "\n");
            }
            bw.flush();
        }
    }
}

Baekjoon Algorithm | Problem 9498: Exam Score

kc@example.com (kc kim) — Sun, 03 Apr 2022 01:44:00 +0900

Source

https://www.acmicpc.net/problem/9498

Problem

Solve Baekjoon Online Judge problem 9498, Exam Score.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

100

Sample Output 1

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            int input = Integer.parseInt(br.readLine());

            String result;
            if (input >= 90)
                result = "A";
            else if (input >= 80)
                result = "B";
            else if (input >= 70)
                result = "C";
            else if (input >= 60)
                result = "D";
            else
                result = "F";

            System.out.print(result);
        }
    }
}

Baekjoon Algorithm | Problem 10430: Remainder

kc@example.com (kc kim) — Sun, 12 Jun 2022 14:11:00 +0900

Source

https://www.acmicpc.net/problem/10430

Problem

Solve Baekjoon Online Judge problem 10430, Remainder.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

5 8 4

Sample Output 1

1 1 0 0

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

            StringTokenizer input = new StringTokenizer(br.readLine(), " ");
            int a = Integer.parseInt(input.nextToken());
            int b = Integer.parseInt(input.nextToken());
            int c = Integer.parseInt(input.nextToken());

            System.out.println((a + b) % c);
            System.out.println(((a % c) + (b % c)) % c);
            System.out.println((a * b) % c);
            System.out.println(((a % c) * (b % c)) % c);
        }
    }
}

Baekjoon Algorithm | Problem 10757: Big Number A+B

kc@example.com (kc kim) — Wed, 20 Apr 2022 07:51:00 +0900

Source

https://www.acmicpc.net/problem/10757

Problem

Solve Baekjoon Online Judge problem 10757, Big Number A+B.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

9223372036854775807 9223372036854775808

Sample Output 1

18446744073709551615

Algorithm Classification

Math
Implementation
Arithmetic
Arbitrary Precision / Big Integers

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            StringTokenizer st = new StringTokenizer(br.readLine(), " ");

            BigInteger a = new BigInteger(st.nextToken());
            BigInteger b = new BigInteger(st.nextToken());

            System.out.println(a.add(b));
        }
    }
}

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine(), " ");

        String str_A = st.nextToken();
        String str_B = st.nextToken();

        int max_length = Math.max(str_A.length(), str_B.length());

        int[] A = new int[max_length + 1];
        int[] B = new int[max_length + 1];

        for (int i = str_A.length() - 1, idx = 0; i >= 0; i--, idx++) {
            A[idx] = str_A.charAt(i) - '0';
        }

        for (int i = str_B.length() - 1, idx = 0; i >= 0; i--, idx++) {
            B[idx] = str_B.charAt(i) - '0';
        }

        for (int i = 0; i < max_length; i++) {
            int value = A[i] + B[i];
            A[i] = value % 10;
            A[i + 1] += (value / 10);
        }

        StringBuilder sb = new StringBuilder();

        if (A[max_length] != 0) {
            sb.append(A[max_length]);
        }

        for (int i = max_length - 1; i >= 0; i--) {
            sb.append(A[i]);
        }

        System.out.println(sb);
    }
}

Baekjoon Algorithm | Problem 10809: Find Alphabet

kc@example.com (kc kim) — Tue, 12 Apr 2022 10:00:00 +0900

Source

https://www.acmicpc.net/problem/10809

Problem

Solve Baekjoon Online Judge problem 10809, Find Alphabet.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

baekjoon

Sample Output 1

1 0 -1 -1 2 -1 -1 -1 -1 4 3 -1 -1 7 5 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1

Algorithm Classification

Implementation
String

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {

        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            final String s = br.readLine();

            int[] alphabet = new int[26];
            for (int i = 0; i < alphabet.length; i++) {
                alphabet[i] = -1;
            }

            for (int i = 0; i < s.length(); i++) {
                int index = s.charAt(i) - 97;
                if (alphabet[index] == -1) {
                    alphabet[index] = i;
                }
            }

            for (int i = 0; i < alphabet.length; i++) {
                System.out.print(alphabet[i] + " ");
            }
            System.out.println();
        }
    }
}

Baekjoon Algorithm | Problem 10818: Minimum, Maximum

kc@example.com (kc kim) — Thu, 07 Apr 2022 08:19:00 +0900

Source

https://www.acmicpc.net/problem/10818

Problem

Solve Baekjoon Online Judge problem 10818, Minimum, Maximum.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

5 20 10 35 30 7

Sample Output 1

7 35

Algorithm Classification

Math
Implementation

Solution

import java.io.*;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int n = Integer.parseInt(br.readLine());
            StringTokenizer nums = new StringTokenizer(br.readLine(), " ");

            int a = Integer.parseInt(nums.nextToken());

            int min = a;
            int max = a;
            for (int i = 1; i < n; i++) {
                a = Integer.parseInt(nums.nextToken());

                if (a > max) {
                    max = a;
                } else if (a < min) {
                    min = a;
                }
            }

            System.out.println(min + " " + max);
        }
    }
}

Baekjoon Algorithm | Problem 10871: Less Than X

kc@example.com (kc kim) — Sun, 03 Apr 2022 00:42:00 +0900

Source

https://www.acmicpc.net/problem/10871

Problem

Solve Baekjoon Online Judge problem 10871, Less Than X.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

10 5 1 10 4 9 2 3 8 5 7 6

Sample Output 1

1 4 2 3

Algorithm Classification

Math
Implementation

Solution

import java.io.*;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ) {
            StringTokenizer input1 = new StringTokenizer(br.readLine(), " ");
            final int n = Integer.parseInt(input1.nextToken());
            final int x = Integer.parseInt(input1.nextToken());

            StringTokenizer input2 = new StringTokenizer(br.readLine(), " ");
            for (int i = 0; i < n; i++) {
                int a = Integer.parseInt(input2.nextToken());
                if (a < x) {
                    bw.write(a + " ");
                }
            }

            bw.flush();
        }
    }
}

Baekjoon Algorithm | Problem 10950: A+B - 3

kc@example.com (kc kim) — Sun, 03 Apr 2022 00:42:00 +0900

Source

https://www.acmicpc.net/problem/10950

Problem

Solve Baekjoon Online Judge problem 10950, A+B - 3.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

5 1 1 2 3 3 4 9 8 5 2

Sample Output 1

2 5 7 17 7

Algorithm Classification

Math
Implementation
Arithmetic

Solution

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int input = sc.nextInt();

        for (int i = 0; i < input; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            System.out.println(a + b);
        }
    }
}

Baekjoon Algorithm | Problem 10951: A+B - 4

kc@example.com (kc kim) — Thu, 07 Apr 2022 02:45:00 +0900

Source

https://www.acmicpc.net/problem/10951

Problem

Solve Baekjoon Online Judge problem 10951, A+B - 4.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

1 1 2 3 3 4 9 8 5 2

Sample Output 1

2 5 7 17 7

Algorithm Classification

Math
Implementation
Arithmetic

Solution

import java.io.*;

public class Main {

    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ) {
            String line;
            while ((line = br.readLine()) != null) {
                String[] input = line.split(" ");
                final int a = Integer.parseInt(input[0]);
                final int b = Integer.parseInt(input[1]);

                bw.write((a + b) + "\n");
            }

            bw.flush();
        }
    }
}

Baekjoon Algorithm | Problem 10952: A+B - 5

kc@example.com (kc kim) — Thu, 07 Apr 2022 02:44:00 +0900

Source

https://www.acmicpc.net/problem/10952

Problem

Solve Baekjoon Online Judge problem 10952, A+B - 5.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

1 1 2 3 3 4 9 8 5 2 0 0

Sample Output 1

2 5 7 17 7

Algorithm Classification

Math
Implementation
Arithmetic

Solution

import java.io.*;

public class Main {

    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ) {
            while (true) {
                String[] input = br.readLine().split(" ");
                final int a = Integer.parseInt(input[0]);
                final int b = Integer.parseInt(input[1]);

                if(a == 0 && b == 0) {
                    break;
                }
                bw.write((a + b) + "\n");
            }

            bw.flush();
        }
    }
}

Baekjoon Algorithm | Problem 11021: A+B - 7

kc@example.com (kc kim) — Tue, 05 Apr 2022 08:38:00 +0900

Source

https://www.acmicpc.net/problem/11021

Problem

Solve Baekjoon Online Judge problem 11021, A+B - 7.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

5 1 1 2 3 3 4 9 8 5 2

Sample Output 1

Case #1: 2 Case #2: 5 Case #3: 7 Case #4: 17 Case #5: 7

Algorithm Classification

Math
Implementation
Arithmetic

Solution

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;

public class Main {
    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ) {
            int n = Integer.parseInt(br.readLine());

            for (int i = 1; i <= n; i++) {
                String[] input = br.readLine().split(" ");
                final int a = Integer.parseInt(input[0]);
                final int b = Integer.parseInt(input[1]);
                bw.write("Case #" + i + ": " + (a + b) + "\n");
            }

            bw.flush();
        }
    }
}

Baekjoon Algorithm | Problem 11022: A+B - 8

kc@example.com (kc kim) — Tue, 05 Apr 2022 08:38:00 +0900

Source

https://www.acmicpc.net/problem/11022

Problem

Solve Baekjoon Online Judge problem 11022, A+B - 8.

Input

Follow the input format and constraints given in the original problem statement.

Output

Print the answer required by the problem.

Sample Input 1

5 1 1 2 3 3 4 9 8 5 2

Sample Output 1

Case #1: 1 + 1 = 2 Case #2: 2 + 3 = 5 Case #3: 3 + 4 = 7 Case #4: 9 + 8 = 17 Case #5: 5 + 2 = 7

Algorithm Classification

Math
Implementation
Arithmetic

Solution

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;

public class Main {
    public static void main(String[] args) throws IOException {
        try (
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ) {
            int n = Integer.parseInt(br.readLine());

            for (int i = 1; i <= n; i++) {
                String[] input = br.readLine().split(" ");
                final int a = Integer.parseInt(input[0]);
                final int b = Integer.parseInt(input[1]);
                bw.write("Case #" + i + ": " + a + " + " + b + " = " + (a + b) + "\n");
            }

            bw.flush();
        }
    }
}