Solution

• If $a$ is divisible by $b$, we need not perform any operations.
• Otherwise, we can increase $a$ by $b-(a\%b)$ and make $a$ divisible by $b$.
• Time Complexity: $O(1)$

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Implementation in C++

#include <iostream>

using namespace std;

int main() {
    int t, a, b;
    cin>>t;
    
    while (t--) {
        cin>>a>>b;
        
        // If a is a multiple of b, we need not perform
        // any operations. Otherwise, we can increase a
        // by b-(a%b)
        if (a%b == 0)
            cout<<0<<"\n";
        else
            cout<<b-(a%b)<<"\n";
    }
    
    return 0;
}

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Implementation in Java

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

public class Solution {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        
        int t, a, b;
        t = scan.nextInt();

        while (t-- > 0) {
            a = scan.nextInt();
            b = scan.nextInt();

            // If a is a multiple of b, we need not perform
            // any operations. Otherwise, we can increase a
            // by b-(a%b)
            if (a%b == 0)
                System.out.println(0);
            else
                System.out.println(b-(a%b));
        }
    }
}

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Implementation in Python

t = int(input())
for _ in range(t):
    a, b = map(int, input().split())
    
    # If a is divisible by b, we need not perform
    # any operations. Otherwise, we can increase
    # a by b-(a%b)
    if a%b == 0:
        print(0)
    else:
        print(b-(a%b))

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Contest Material

• Codeforces - Divisibility Problem