找出K个连续的整数，使它们的和为N

给定两个整数N和K ，任务是找到K个连续的整数，使得它们的和为N。
注意：如果没有这样的K个整数，则打印-1。
例子：

Input: N = 15, K = 5 
Output: 1 2 3 4 5 
Explanation: 
N can be represented as sum of 5 consecutive integers as follows – 
=> N => 1 + 2 + 3 + 4 + 5 = 15
Input: N = 33, K = 6 
Output: 3 4 5 6 7 8 
Explanation: 
N can be represented as sum of 6 consecutive integers as follows – 
=> N => 3 + 4 + 5 + 6 + 7 + 8 = 33 
 

天真的方法：一个简单的解决方案是运行一个从i = 0到N –(K – 1)的循环，以检查是否从i开始的K个连续整数的总和为N。
高效的方法：想法是使用算术级数来解决此问题，其中具有相同差的K个算术级数项之和可以定义如下：

1. K项总和– $\frac{Number of Terms}{2} * (first Term + Last Term)$
   => $N = \frac{a_1 + a_K}{2} * K$
2. 进一步求解方程以获得可能的第一项
   => $a_1 + a_K = \frac{2*N}{K}$
3. 这里的_K是第K^个项，可以写成₁ + K – 1
   => $2*a_1 + K - 1 = \frac{2*N}{K}$
   => $a_1 = \frac{\frac{2*N}{K} + 1 - K}{2}$
4. 最后，检查计算出的第一个项是整数，如果是，则存在K个连续数，如果N，则其和。

下面是上述方法的实现：

C++

// C++ implementation to check if
// a number can be expressed as
// sum of K consecutive integer
 
#include 
using namespace std;
 
// Function to check if a number can be
// expressed as the sum of k consecutive
void checksum(int n, int k)
{
    // Finding the first
    // term of AP
    float first_term = ((2 * n) / k
                        + (1 - k))
                       / 2.0;
 
    // Checking if first
    // term is an integer
    if (first_term - int(first_term) == 0) {
 
        // Loop to print the K
        // consecutive integers
        for (int i = first_term;
             i <= first_term + k - 1; i++) {
            cout << i << " ";
        }
    }
    else
        cout << "-1";
}
 
// Driver Code
int main()
{
    int n = 33, k = 6;
    checksum(n, k);
    return 0;
}

Java

// Java implementation to check if
// a number can be expressed as
// sum of K consecutive integer
class GFG{
 
// Function to check if a number can be
// expressed as the sum of k consecutive
static void checksum(int n, int k)
{
     
    // Finding the first
    // term of AP
    float first_term = (float) (((2 * n) / k +
                                 (1 - k)) / 2.0);
 
    // Checking if first
    // term is an integer
    if (first_term - (int)(first_term) == 0)
    {
 
        // Loop to print the K
        // consecutive integers
        for(int i = (int)first_term;
            i <= first_term + k - 1; i++)
        {
           System.out.print(i + " ");
        }
    }
    else
        System.out.print("-1");
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 33, k = 6;
     
    checksum(n, k);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to check 
# if a number can be expressed as
# sum of K consecutive integer
 
# Function to check if a number can be
# expressed as the sum of k consecutive
def checksum(n, k):
     
    # Finding the first
    # term of AP
    first_term = ((2 * n) / k + (1 - k)) / 2.0
     
    # Checking if first
    # term is an integer
    if (first_term - int(first_term) == 0):
         
        # Loop to print the K
        # consecutive integers
        for i in range(int(first_term),
                       int(first_term) + k):
            print(i, end = ' ')
    else:
        print('-1')
 
# Driver Code
if __name__=='__main__':
     
    (n, k) = (33, 6)
    checksum(n, k)
 
# This code is contributed by rutvik_56

C#

// C# implementation to check if
// a number can be expressed as
// sum of K consecutive integer
using System;
class GFG{
 
// Function to check if a number can be
// expressed as the sum of k consecutive
static void checksum(int n, int k)
{
     
    // Finding the first
    // term of AP
    float first_term = (float)(((2 * n) / k +
                                (1 - k)) / 2.0);
 
    // Checking if first
    // term is an integer
    if (first_term - (int)(first_term) == 0)
    {
 
        // Loop to print the K
        // consecutive integers
        for(int i = (int)first_term;
                i <= first_term + k - 1; i++)
        {
            Console.Write(i + " ");
        }
    }
    else
        Console.Write("-1");
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 33, k = 6;
     
    checksum(n, k);
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// javascript implementation to check if
// a number can be expressed as
// sum of K consecutive integer   
 
// Function to check if a number can be
// expressed as the sum of k consecutive
    function checksum(n , k)
    {
        // Finding the first
        // term of AP
        var first_term =  (((2 * n) / k + (1 - k)) / 2.0);
 
        // Checking if first
        // term is an integer
        if (first_term - parseInt( (first_term)) == 0) {
 
            // Loop to prvar the K
            // consecutive integers
            for (i = parseInt( first_term); i <= first_term + k - 1; i++) {
                document.write(i + " ");
            }
        } else
            document.write("-1");
    }
 
    // Driver Code
     
        var n = 33, k = 6;
 
        checksum(n, k);
 
// This code contributed by Rajput-Ji
 

输出：

3 4 5 6 7 8