Número más pequeño mayor o igual a N divisible por K

Dado un número N y un número K, la tarea es encontrar el número más pequeño mayor o igual que N que sea divisible por K.
Ejemplos: 
 

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

Enfoque: La idea es dividir N+K entre K. Si el resto es 0, imprima N , de lo contrario, imprima N + K – resto .
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
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 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 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# 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
}

<?php
// PHP 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)
{
    $rem = ($N + $K) % $K;
 
    if ($rem == 0)
        return $N;
    else
        return $N + $K - $rem;
}
 
// Driver code
$N = 45; $K = 6;
 
echo "Smallest number greater than " .
                   "or equal to ", $N;
echo "\nthat is divisible by " , $K ,
            " is " , findNum($N, $K);
 
// This code is contributed by anuj_67
?>

<script>
 
// 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
</script>

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

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)