生成所有 K 大小子数组的总和除以 N 留下余数 X 的数组

给定三个整数N 、 K和X ，任务是创建一个长度为N的数组，其所有 K 长度子数组的模N 之和为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，为了使任何大小为K模N 的子数组的总和等于X ，该子数组需要有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：

• 从0到N – 1迭代i ，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总和K*N + X 。因此，对于所有这样的子阵列，求和模N是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

Javascript

 
// Javascript implementation of the
// above approach
 
// Funtion prints the required array
function createArray(n, k, x)
{
    for (var i = 0; i < n; i++) {
        // First element of each K
        // length subarrays
        if (i % k == 0) {
            document.write( x + n + " ");
        }
        else {
            document.write( n + " ");
        }
    }
}
 
// Driver Program
var N = 6, K = 3, X = 3;
createArray(N, K, X);
 
// This code is contributed by itsok.

输出：

9 6 6 9 6 6

时间复杂度： O(N)
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live