将N分成(X，2X，…，KX)形式的K个部分以获得X的某个值

给定正整数N和K ，任务是将N分成K个部分，使得第一部分的值为X ，第二部分的值为2X ，以此类推，对于X的某个值。如果无法进行这种划分，则打印-1 。

例子：

Input: N = 10, K = 4
Output: 1 2 3 4
Explanation:
If we take 1 as first number second number will be 2 third number will be 3 times first which is 3 and the last number will be 4 times third number so last number is 4. We can note that sum=1+2+3+4=10 which is the required sum.

Input N = 10, K = 3
Output: -1
Explanation:
Distributing N in 3 parts with given constraint is not possible.

方法：为了解决上述问题，让我们了解它的数学实现。让划分为X _{1，X 2，X 3}到X _K，其中所述第二整数为X ₁ * 2，第三个是X ₁ * 3和第K个一个是X ₁ * K.

We know that,
=> $X_{1} + X_{2} + X_{3} + ... + X_{K} = N$
=> $X_{1} + 2*X_{1} + 3*X_{1} + ... + K*X_{1} = N$
=> $X_{1}( 1 + 2 + 3 + ... + K) = N$
=> $X_{1} * \frac{K*(K+1))}{2} = N$, where ( 1 + 2 + 3 + … + K) = $\frac{K*(K+1))}{2}$
=> $X_{1} = \frac{2*N}{K*(K+1)}$

因此，要解决该问题，我们必须遵循以下步骤：

• 计算K *(K +1)的值，然后将2 * N除以K *(K +1) ，以获得X _1的值。
• 如果在上述步骤中X ₁不是整数，则打印-1，因为不可能进行这种除法。
• 要获得X ₂的值，我们将X ₁乘以2。类似地，要获得X _K将X ₁乘以K。
• 找到所有值后，将它们打印出来。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
typedef long long int ll;
  
// Function to find the division
void solve(int n, int k)
{
    int x1, d;
  
    // Calculating value of x1
    d = k * (k + 1);
  
    // Print -1 if division
    // is not possible
    if ((2 * n) % d != 0) {
        cout << "-1";
        return;
    }
  
    x1 = 2 * n / d;
  
    // Get the first number ie x1
    // then successively multiply
    // it by x1 k times by index number
    // to get the required answer
    for (int i = 1; i <= k; i++) {
        cout << x1 * i << " ";
    }
    cout << endl;
}
  
// Driver Code
int main()
{
    // Given N and K
    int n = 10, k = 4;
  
    // Function Call
    solve(n, k);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the division
static void solve(int n, int k)
{
    int x1, d;
  
    // Calculating value of x1
    d = k * (k + 1);
  
    // Print -1 if division
    // is not possible
    if ((2 * n) % d != 0) 
    {
        System.out.print("-1");
        return;
    }
  
    x1 = 2 * n / d;
  
    // Get the first number ie x1
    // then successively multiply
    // it by x1 k times by index number
    // to get the required answer
    for (int i = 1; i <= k; i++) 
    {
        System.out.print(x1 * i+ " ");
    }
    System.out.println();
}
  
// Driver Code
public static void main(String[] args)
{
    // Given N and K
    int n = 10, k = 4;
  
    // Function Call
    solve(n, k);
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
  
# Function to find the division
def solve(n, k):
  
    # Calculating value of x1
    d = k * (k + 1);
  
    # Print -1 if division
    # is not possible
    if ((2 * n) % d != 0):
        print("-1");
        return;
      
  
    x1 = 2 * n // d;
  
    # Get the first number ie x1
    # then successively multiply
    # it by x1 k times by index number
    # to get the required answer
    for i in range(1, k + 1):
        print(x1 * i, end = " ");
  
# Driver Code
  
# Given N and K
n = 10; k = 4;
  
# Function Call
solve(n, k);
  
# This code is contributed by Code_Mech

C#

// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the division
static void solve(int n, int k)
{
    int x1, d;
  
    // Calculating value of x1
    d = k * (k + 1);
  
    // Print -1 if division
    // is not possible
    if ((2 * n) % d != 0) 
    {
        System.out.print("-1");
        return;
    }
  
    x1 = 2 * n / d;
  
    // Get the first number ie x1
    // then successively multiply
    // it by x1 k times by index number
    // to get the required answer
    for (int i = 1; i <= k; i++) 
    {
        System.out.print(x1 * i+ " ");
    }
    System.out.println();
}
  
// Driver Code
public static void main(String[] args)
{
    // Given N and K
    int n = 10, k = 4;
  
    // Function Call
    solve(n, k);
}
}
  
// This code is contributed by 29AjayKumar

输出：

1 2 3 4

时间复杂度： O(K)
辅助空间： O(1)