创建一个总和为S的大小为N的数组，使得不存在总和为S或SK的子数组

给定一个数字N和一个整数S ，任务是创建一个N个整数的数组，以使所有元素的总和等于S并打印一个元素K ，其中0≤K≤S，从而不存在总和等于的子数组K或(S – K) 。
如果没有这样的数组，则打印“ -1” 。

注意： K可以有多个值。您可以打印其中任何一个。

例子：

Input: N = 1, S = 4
Output: {4}
K = 2
Explanation:
There exists an array {4} whose sum is 4.
From all possible value of K i.e., 0 ≤ K ≤ 4, K = 1, 2, and 3 satisfy the given conditions.
For K = 2, there is no subarray whose sum is 2 or S – K i.e., 4 – 2 = 2.

Input: N = 3, S = 8
Output: {2, 2, 4}
K = 1
Explanation:
There exists an array {2, 2, 4} and there exists K as 1 such that there is no subarray whose sum is 1 and S – K i.e., 8 – 1 = 7.

方法：要解决上述问题，我们必须注意：

1. 如果2 * N> S，则不可能有数组。
   例如：

   For N = 3 and S = 4, then the possible arrays are {1, 2, 1}, {1, 1, 2}, {2, 1, 1}.
   The possible values for K are 0, 1, 2, 3 (0 < = k < = S).
   But there is no value for K which satisfy the condition.
   So the solution to this is not possible.

2. 仅当2 * N <= S且数组可以使用元素(N-1)乘以2且最后一个元素为S –(2 *(N – 1))和K始终为1时，才可以创建数组。

下面是上述方法的实现：

C++

// C++ for the above approach
#include
using namespace std;
      
// Function to create an array with
// N elements with sum as S such that
// the given conditions satisfy
void createArray(int n, int s)
{
      
    // Check if the solution exists
    if (2 * n <= s)
    {
  
        // Print the array as
        // print (n-1) elments
        // of array as 2
        for(int i = 0; i < n - 1; i++) 
        {
           cout << "2" << " ";
           s -= 2;
        }
  
        // Print the last element
        // of the array
        cout << s << endl;
  
        // Print the value of k
        cout << "1" << endl;
    }
    else
      
        // If solution doesnot exists
        cout << "-1" << endl;
}
  
// Driver Code
int main()
{
      
    // Given N and sum S
    int N = 1;
    int S = 4;
  
    // Function call
    createArray(N, S);
}
  
// This code is contributed by Ritik Bansal

Java

// Java for the above approach
class GFG{
      
// Function to create an array with
// N elements with sum as S such that
// the given conditions satisfy
static void createArray(int n, int s)
{
  
    // Check if the solution exists
    if (2 * n <= s)
    {
  
        // Print the array as
        // print (n-1) elments
        // of array as 2
        for (int i = 0; i < n - 1; i++) 
        {
            System.out.print(2 + " ");
            s -= 2;
        }
  
        // Print the last element
        // of the array
        System.out.println(s);
  
        // Print the value of k
        System.out.println(1);
    }
    else
      
        // If solution doesnot exists
        System.out.print("-1");
}
  
// Driver Code
public static void main(String[] args) 
{
  
    // Given N and sum S
    int N = 1;
    int S = 4;
  
    // Function call
    createArray(N, S);
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 for the above approach
  
# Function to create an array with
# N elements with sum as S such that
# the given conditions satisfy
def createArray(n, s):
   
    # Check if the solution exists
    if (2 * n<= s):             
          
        # Print the array as
        # print (n-1) elments 
        # of array as 2
        for i in range(n-1):
            print(2, end =" ")
            s-= 2
              
        # Print the last element
        # of the array 
        print(s) 
          
        # Print the value of k 
        print(1)
    else:
        # If solution doesnot exists
        print('-1')
  
  
# Driver Code 
  
# Given N and sum S  
N = 1
S = 4
  
# Function call
createArray(N, S)

C#

// C# program for the above approach
using System;
class GFG{
      
// Function to create an array with
// N elements with sum as S such that
// the given conditions satisfy
static void createArray(int n, int s)
{
      
    // Check if the solution exists
    if (2 * n <= s)
    {
  
        // Print the array as
        // print (n-1) elments
        // of array as 2
        for(int i = 0; i < n - 1; i++) 
        {
           Console.Write(2 + " ");
           s -= 2;
        }
  
        // Print the last element
        // of the array
        Console.WriteLine(s);
  
        // Print the value of k
        Console.WriteLine(1);
    }
    else
      
        // If solution doesnot exists
        Console.Write("-1");
}
  
// Driver Code
public static void Main() 
{
  
    // Given N and sum S
    int N = 1;
    int S = 4;
  
    // Function call
    createArray(N, S);
}
}
  
// This code is contributed by Code_Mech

输出：

4
1

时间复杂度： O(N)
辅助空间： O(1)