将一个数字拆分为不能被 K 整除的 K 个数字的总和

给定两个数字N和K ，任务是将这个数字分成K 个正整数，使得它们的和等于N并且这些K 个整数中没有一个是K的倍数。
注： N>=2
例子：

Input: N = 10, K = 3 
Output: 1, 1, 8
Input: N = 18, K = 3 
Output:1, 1, 16 
 

方法：
要将N拆分为K 个数字，我们需要通过以下步骤来解决问题：

• 检查N是否可以被K整除。
• 如果N可被K整除，则N – K + 1不可被K整除。因此，我们可以将 N 拆分为N – K + 1部分，并将所有剩余的K – 1部分拆分为1 。
• 如果N不能被K整除：
  • 如果K为 2，则无法拆分。
  • 否则，将N分成K-2部分作为所有1和2 ，将N – K作为剩余的两部分。

下面的代码是上述方法的实现：

C++

// C++ program to split a number
// as sum of K numbers which are
// not divisible by K
 
#include 
using namespace std;
 
// Function to split into K parts
// and print them
void 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 code
int 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 K
class GFG{
 
// Function to split into K parts
// and print them
static 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 code
public 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 them
def 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 code
if __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 K
using System;
 
class GFG{
 
// Function to split into K parts
// and print them
static 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 code
public static void Main(String[] args)
{
    int N = 18, K = 5;
     
    printKParts(N, K);
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
 
// 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 them
function 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 code
var N = 18, K = 5;
printKParts(N, K);
 
 

输出：

1, 1, 1, 2, 13

时间复杂度： O(K)