生成其所有K大小子数组的总和除以N个叶子余数X的数组

给出三个整N，K和X，该任务是创建长度N SUACH的阵列，其所有的K-长度子阵列的总和模N是X。

例子：

Input: N = 6, K = 3, X = 3
Output: 9 6 6 9 6 6
Explanation:
All subarrays of length 3 and their respective sum%N values are as follows:
[9, 6, 6] Sum = 21 % 6 = 3
[6, 6, 9] sum = 21 % 6 = 3
[6, 9, 6] sum = 21 % 6 = 3
[9, 6, 6] sum = 21 % 6 = 3
Since all its subarrays have sum % N = X (=3), the generated array is valid.

Input: N = 4, K = 2, X = 2
Output: 6 4 6 4

方法：
我们可以观察到，为了使大小为K模N的子数组的总和等于X ，子数组需要具有K – 1个元素等于N和1个元素等于N +X 。

插图：

If N = 6, K = 3, X = 3
Here a K length subarray needs to be a permutation of {9, 6, 6} where 2 (K – 1) elements are divisible by 6 and 1 element has modulo N equal to X( 9%6 = 3)
Sum of subarray % N = (21 % 6) = 3 (same as X)
Hence, each K length

因此，请按照以下步骤解决问题：

• 将i从0迭代到N – 1 ，以打印所需子数组的第i^个元素。
• 如果i％K等于0 ，则打印N +X 。否则，对于i的所有其他值，打印N。
• 这确保了每个可能的K长度子数组都有一个和K * N + X。因此，对于所有这样的子阵列，求和模N为X。

下面是上述方法的实现。

C++

// C++ implementation of the
// above approach
#include 
using namespace std;
  
// Funtion prints the required array
void createArray(int n, int k, int x)
{
    for (int i = 0; i < n; i++) {
        // First element of each K
        // length subarrays
        if (i % k == 0) {
            cout << x + n << " ";
        }
        else {
            cout << n << " ";
        }
    }
}
  
// Driver Program
int main()
{
  
    int N = 6, K = 3, X = 3;
    createArray(N, K, X);
}

Java

// Java implementation of the above approach
import java.util.*;
class GFG{
  
// Function prints the required array
static void createArray(int n, int k, int x)
{
    for(int i = 0; i < n; i++)
    {
          
       // First element of each K
       // length subarrays
       if (i % k == 0) 
       {
           System.out.print((x + n) + " ");
       }
       else
       {
           System.out.print(n + " ");
       }
    }
}
  
// Driver Code
public static void main(String args[])
{
    int N = 6, K = 3, X = 3;
      
    createArray(N, K, X);
}
}
  
// This code is contributed by Code_Mech

Python3

# Python3 implementation of the 
# above approach 
  
# Funtion prints the required array 
def createArray(n, k, x):
      
    for i in range(n):
          
        # First element of each K
        # length subarrays
        if (i % k == 0):
            print(x + n, end = " ")
        else :
            print(n, end = " ")
  
# Driver code
N = 6
K = 3
X = 3
  
createArray(N, K, X)
  
# This code is contributed by Vishal Maurya.

C#

// C# implementation of the above approach
using System;
class GFG{
  
// Function prints the required array
static void createArray(int n, int k, int x)
{
    for(int i = 0; i < n; i++)
    {
         
       // First element of each K
       // length subarrays
       if (i % k == 0) 
       {
           Console.Write((x + n) + " ");
       }
       else
       {
           Console.Write(n + " ");
       }
    }
}
  
// Driver Code
public static void Main()
{
    int N = 6, K = 3, X = 3;
      
    createArray(N, K, X);
}
}
  
// This code is contributed by Code_Mech

输出：

9 6 6 9 6 6

时间复杂度： O(N)
辅助空间： O(1)