Given two numbers N and K, the task is to split this number into K positive integers such that their sum is equal to N and none of these K integers is a multiple of K.
Note: N>=2
Examples:
Input: N = 10, K = 3
Output: 1, 1, 8
Input: N = 18, K = 3
Output:1, 1, 16
Approach:
To split N into K numbers, we need to approach the problem by the following steps:
- Check if N is divisible by K or not.
- If N is divisible by K, then N – K + 1 is not divisible by K. Hence, we can split N into N – K + 1 as one part and all the remaining K – 1 parts as 1.
- If N is not divisible by K:
- If K is 2, then it is not possible to split.
- Otherwise, split N into K-2 parts as all 1 and 2 and N – K as the remaining two parts.
Below code is the implementation of the above approach:
C++
// C++ program to split a number// as sum of K numbers which are// not divisible by K#include <iostream>using namespace std;// Function to split into K parts// and print themvoid printKParts(int N, int K){ if (N % K == 0) { // Print 1 K - 1 times for (int i = 1; i < K; i++) cout << "1, "; // Print N - K + 1 cout << N - (K - 1) << endl; } else { if (K == 2) { cout << "Not Possible" << endl; return; } // Print 1 K-2 times for (int i = 1; i < K - 1; i++) cout << 1 << ", "; // Print 2 and N - K cout << 2 << ", " << N - K << endl; }}// Driver codeint main(){ int N = 18, K = 5; printKParts(N, K); return 0;} |
Java
// Java program to split a number// as sum of K numbers which are// not divisible by Kclass GFG{// Function to split into K parts// and print themstatic void printKParts(int N, int K){ if (N % K == 0) { // Print 1 K - 1 times for(int i = 1; i < K; i++) System.out.print("1, "); // Print N - K + 1 System.out.print(N - (K - 1) + "\n"); } else { if (K == 2) { System.out.print("Not Possible" + "\n"); return; } // Print 1 K-2 times for(int i = 1; i < K - 1; i++) System.out.print(1 + ", "); // Print 2 and N - K System.out.print(2 + ", " + (N - K) + "\n"); }}// Driver codepublic static void main(String[] args){ int N = 18, K = 5; printKParts(N, K);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to split a number# as sum of K numbers which are# not divisible by K# Function to split into K parts# and print themdef printKParts(N, K): if (N % K == 0): # Print 1 K - 1 times for i in range(1, K): print("1, "); # Print N - K + 1 print(N - (K - 1), end = ""); else: if (K == 2): print("Not Possible", end = ""); return; # Print 1 K-2 times for i in range(1, K - 1): print(1, end = ", "); # Print 2 and N - K print(2, ", ", (N - K), end = "");# Driver codeif __name__ == '__main__': N = 18; K = 5; printKParts(N, K);# This code is contributed by Rohit_ranjan |
C#
// C# program to split a number// as sum of K numbers which are// not divisible by Kusing System;class GFG{// Function to split into K parts// and print themstatic void printKParts(int N, int K){ if (N % K == 0) { // Print 1 K - 1 times for(int i = 1; i < K; i++) Console.Write("1, "); // Print N - K + 1 Console.Write(N - (K - 1) + "\n"); } else { if (K == 2) { Console.Write("Not Possible" + "\n"); return; } // Print 1 K-2 times for(int i = 1; i < K - 1; i++) Console.Write(1 + ", "); // Print 2 and N - K Console.Write(2 + ", " + (N - K) + "\n"); }}// Driver codepublic static void Main(String[] args){ int N = 18, K = 5; printKParts(N, K);}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// Javascript program to split a number// as sum of K numbers which are// not divisible by K// Function to split into K parts// and print themfunction printKParts(N, K){ if (N % K == 0) { // Print 1 K - 1 times for (var i = 1; i < K; i++) document.write( "1, "); // Print N - K + 1 document.write( (N - (K - 1)) + "<br>"); } else { if (K == 2) { document.write( "Not Possible" + "<br>"); return; } // Print 1 K-2 times for (var i = 1; i < K - 1; i++) document.write( 1 + ", "); // Print 2 and N - K document.write( 2 + ", " + (N - K) + "<br>"); }}// Driver codevar N = 18, K = 5;printKParts(N, K);</script> |
Output:
1, 1, 1, 2, 13
Time Complexity: O(K)
Auxiliary Space: O(1)
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