需要添加到 M 以使其可被 N 整除的最小数

给定两个正整数M和N ，任务是计算需要添加到M的最小数字，以使其可被N整除。

例子：

Input: M = 6, N = 7
Output: 1
Explanation: 1 is the smallest number that can be added to 6 to make it divisible by 7.

Input: M = 100, N = 28
Output: 12

方法：这个想法是找到大于或等于M的最小数字，即可以被N整除，然后从中减去M。要获得N ≥ M的最小倍数，请将M + N除以N 。如果余数为0 ，则值为M 。否则，值为M + N – 余数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the smallest
// number greater than or equal
// to N, that is divisible by k
int findNum(int N, int K)
{
    int rem = (N + K) % K;
 
    if (rem == 0)
        return N;
    else
        return N + K - rem;
}
 
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
int findSmallest(int M, int N)
{
    // Stores the smallest multiple
    // of N, greater than or equal to M
    int x = findNum(M, N);
 
    // Return the result
    return x - M;
}
 
// Driver Code
int main()
{
    // Given Input
    int M = 100, N = 28;
 
    // Function Call
    cout << findSmallest(M, N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to find the smallest
// number greater than or equal
// to N, that is divisible by k
public static int findNum(int N, int K)
{
    int rem = (N + K) % K;
 
    if (rem == 0)
        return N;
    else
        return N + K - rem;
}
 
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
public static int findSmallest(int M, int N)
{
     
    // Stores the smallest multiple
    // of N, greater than or equal to M
    int x = findNum(M, N);
 
    // Return the result
    return x - M;
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given Input
    int M = 100, N = 28;
 
    // Function Call
    System.out.println(findSmallest(M, N));
}}
 
// This code is contributed by SoumikMondal

Python3

# Python3 program for the above approach
 
# Function to find the smallest
# number greater than or equal
# to N, that is divisible by k
def findNum(N, K):
     
    rem = (N + K) % K
 
    if (rem == 0):
        return N
    else:
        return N + K - rem
 
# Function to find the smallest
# number required to be added to
# to M to make it divisible by N
def findSmallest(M, N):
     
    # Stores the smallest multiple
    # of N, greater than or equal to M
    x = findNum(M, N)
 
    # Return the result
    return x - M
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    M = 100
    N = 28
 
    # Function Call
    print(findSmallest(M, N))
     
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
         
class GFG{
     
// Function to find the smallest
// number greater than or equal
// to N, that is divisible by k
public static int findNum(int N, int K)
{
    int rem = (N + K) % K;
  
    if (rem == 0)
        return N;
    else
        return N + K - rem;
}
  
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
public static int findSmallest(int M, int N)
{
     
    // Stores the smallest multiple
    // of N, greater than or equal to M
    int x = findNum(M, N);
  
    // Return the result
    return x - M;
}
     
// Driver Code
public static void Main()
{
     
    // Given Input
    int M = 100, N = 28;
  
    // Function Call
    Console.WriteLine(findSmallest(M, N));
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript

 
// Javascript program for the above approach
 
// Function to find the smallest
// number greater than or equal
// to N, that is divisible by k
function findNum(N, K)
{
    var rem = (N + K) % K;
 
    if (rem == 0)
        return N;
    else
        return N + K - rem;
}
 
// Function to find the smallest
// number required to be added to
// to M to make it divisible by N
function findSmallest(M, N)
{
     
    // Stores the smallest multiple
    // of N, greater than or equal to M
    var x = findNum(M, N);
 
    // Return the result
    return x - M;
}
 
// Driver Code
 
// Given Input
var M = 100, N = 28;
 
// Function Call
document.write(findSmallest(M, N));
 
// This code is contributed by itsok
 

输出：

12

时间复杂度： O(1)
辅助空间： O(1)