查找总和为N的N个不同的整数

给定一个整数N ，任务是找到总和为N的N个不同的整数。如果整数的组合不止一个，请打印其中的任何一个。

例子：

Input: N = 3 
Output: 1, -1, 3 
Explanation: 
On adding the numbers that is 1 + (-1) + 3 the sum is 3.

Input: N = 4 
Output: 1, -1, 0, 4 
Explanation: 
On adding the numbers that is 1 + (-1) + 0 + (4) the sum is 4. 

方法：想法是打印N / 2个对称对，例如(+ x，-x)，这样结果总和将始终为0 。
现在，如果整数N为奇数，则将N与这些整数集一起打印以使所有整数的总和等于N
如果N为偶数，则输出0和N以及这些整数集，以使所有整数之和等于N。

下面是上述方法的实现：

C++

// C++ for the above approach
#include 
using namespace std;
 
// Function to print distinct N
// numbers whose sum is N
void findNumbers(int N)
{
    // To store how many symmetric
    // pairs needs to be calculated
    int half = N / 2;
 
    // For even N we have to print
    // one less symmetric pair
    if (N % 2 == 0) {
        half--;
    }
 
    // Iterate till [1 n/2] and Print
    // all symmetric pairs(i, -i)
    for (int i = 1; i <= half; i++) {
 
        // Print 2 symmetric numbers
        cout << (-1) * i
             << ", " << i << ", ";
    }
 
    // if N is Odd, then print N
    if (N & 1) {
        cout << N << endl;
    }
 
    // Else print(0, N)
  else {
    cout << 0 << ", "
         << N << endl;
   }
}
 
// Driver Code
int main()
{
    // Given Sum
    int N = 5;
 
    // Function Call
    findNumbers(N);
    return 0;
}

Java

// Java for the above approach
class GFG{
     
// Function to print distinct N
// numbers whose sum is N
public static void findNumbers(int N)
{
     
    // To store how many symmetric
    // pairs needs to be calculated
    int half = N / 2;
     
    // For even N we have to print
    // one less symmetric pair
    if (N % 2 == 0)
    {
        half--;
    }
     
    // Iterate till [1 n/2] and Print
    // all symmetric pairs(i, -i)
    for(int i = 1; i <= half; i++)
    {
 
       // Print 2 symmetric numbers
       System.out.print((-1) * i + ", " +
                               i + ", ");
    }
     
    // if N is Odd, then print N
    int check = N & 1;
    if (check != 0)
    {
        System.out.println(N);
    }
     
    // Else print(0, N)
    else
    {
    System.out.println(0 + ", " + N);
    }
}
 
// Driver code
public static void main(String[] args)
{
         
    // Given sum
    int N = 5;
     
    // Function sall
    findNumbers(N);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 code for the above approach
 
# Function to print distinct N
# numbers whose sum is N
def findNumbers(N):
 
    # To store how many symmetric
    # pairs needs to be calculated
    half = int(N / 2)
 
    # For even N we have to print
    # one less symmetric pair
    if (N % 2 == 0):
        half = half - 1
 
    # Iterate till [1 n/2] and Print
    # all symmetric pairs(i, -i)
    for i in range(1, half + 1):
 
        # Print 2 symmetric numbers
        print((-1) * i, end = ', ')
        print(i, end = ', ')
 
    # If N is Odd, then print N
    if (N & 1):
        print(N, end = '\n')
 
    # Else print(0, N)
    else:
        print(0, end = ', ')
        print(N, end = '\n')
 
# Driver Code
N = 5
 
# Function Call
findNumbers(N)
 
# This code is contributed by PratikBasu

C#

// C# for the above approach
using System;
class GFG{
     
// Function to print distinct N
// numbers whose sum is N
public static void findNumbers(int N)
{
     
    // To store how many symmetric
    // pairs needs to be calculated
    int half = N / 2;
     
    // For even N we have to print
    // one less symmetric pair
    if (N % 2 == 0)
    {
        half--;
    }
     
    // Iterate till [1 n/2] and Print
    // all symmetric pairs(i, -i)
    for(int i = 1; i <= half; i++)
    {
 
        // Print 2 symmetric numbers
        Console.Write((-1) * i + ", " +
                             i + ", ");
    }
     
    // if N is Odd, then print N
    int check = N & 1;
    if (check != 0)
    {
        Console.Write(N + "\n");
    }
     
    // Else print(0, N)
    else
    {
    Console.Write(0 + ", " + N + "\n");
    }
}
 
// Driver code
public static void Main(string[] args)
{
         
    // Given sum
    int N = 5;
     
    // Function sall
    findNumbers(N);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// javascript program for the above approach
 
// Function to print distinct N
// numbers whose sum is N
function findNumbers( N)
{
    // To store how many symmetric
    // pairs needs to be calculated
    let half = parseInt(N / 2);
 
    // For even N we have to print
    // one less symmetric pair
    if (N % 2 == 0) {
        half--;
    }
 
    // Iterate till [1 n/2] and Print
    // all symmetric pairs(i, -i)
    for (let i = 1; i <= half; i++) {
 
        // Print 2 symmetric numbers
         document.write( (-1) * i
             + ", " + i + ", ");
    }
 
    // if N is Odd, then print N
    if (N & 1) {
         document.write( N);
    }
 
    // Else print(0, N)
  else {
     document.write(  0 + ", "
         + N +"<br/>");
   }
}
 
// Driver Code
 
    // Given Sum
    let N = 5;
 
    // Function Call
    findNumbers(N);
           
 
    // This code contributed by aashish1995
 

输出：

-1,1,-2,2,5