大于或等于N且可被K整除的最小数字

给定数字N和数字K，任务是找到大于或等于N且可被K整除的最小数字。
例子：

Input: N = 45, K = 6
Output: 48
48 is the smallest number greater than or equal to 45
which is divisible by 6.

Input: N = 11, K = 3
Output: 12

方法：想法是将N + K除以K。如果余数为0，则打印N，否则打印N + K –余数。
下面是上述方法的实现：

C++

// C++ implementation of 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;
}
 
// Driver code
int main()
{
    int N = 45, K = 6;
 
    cout << "Smallest number greater than or equal to " << N
         << "\nthat is divisible by " << K << " is " << findNum(N, K);
 
    return 0;
}

Java

// Java implementation of the above approach
 
public class GFG{
 
    // Function to find the smallest number
    // greater than or equal to N
    // that is divisible by k
    static int findNum(int N, int K)
    {
        int rem = (N + K) % K;
     
        if (rem == 0)
            return N;
        else
            return N + K - rem;
    }
 
 
     // Driver Code
     public static void main(String []args){
          
        int N = 45, K = 6;
 
    System.out.println("Smallest number greater than or equal to " + N
          +"\nthat is divisible by " + K + " is " + findNum(N, K));
 
     }
     // This code is contributed by ANKITRAI1
}

Python

# Python 3 implementation of 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)
 
# Driver Code
N = 45
K = 6
print('Smallest number greater than',
                   'or equal to' , N,
           'that is divisible by', K,
               'is' , findNum(45, 6))
 
# This code is contributed by Arnab Kundu

C#

// C# implementation of the above approach
 
public class GFG{
 
    // Function to find the smallest number
    // greater than or equal to N
    // that is divisible by k
    static int findNum(int N, int K)
    {
        int rem = (N + K) % K;
     
        if (rem == 0)
            return N;
        else
            return N + K - rem;
    }
 
 
    // Driver Code
    static void Main(){
         
        int N = 45, K = 6;
 
    System.Console.WriteLine("Smallest number greater than or equal to " + N
        +"\nthat is divisible by " + K + " is " + findNum(N, K));
 
    }
    // This code is contributed by mits
}

Javascript

 
// javascript implementation of 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;
    }   
// Driver Code
 
          
var N = 45, K = 6;
 
document.write("Smallest number greater than or equal to " + N
  +"<br>that is divisible by " + K + " is " + findNum(N, K));
 
 
// This code contributed by shikhasingrajput

输出：

Smallest number greater than or equal to 45
that is divisible by 6 is 48