Find N distinct integers with sum N

Given an integer N, the task is to find N distinct integers whose sum is N. If there is more than one combination of the integers, print any one of them.

Examples:

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.

Approach: The idea is to print N/2 Symmetric Pairs like (+x, -x) so that the resultant sum will always be 0.



Now if integer N is odd, then print N along with these set of integers to make sum of all integers equals to N
If N is even, print 0 and N along with these set of integers to make sum of all integers equals to N.

Below is the implementation of the above approach:

C++

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// C++ for the above approach
#include <bits/stdc++.h>
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 & 2) {
        cout << N << endl;
    }
  
    // Else print(0, N)
    cout << 0 << ", "
         << N << endl;
}
  
// Driver Code
int main()
{
    // Given Sum
    int N = 5;
  
    // Function Call
    findNumbers(N);
    return 0;
}

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Java

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// 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 & 2;
    if (check != 0)
    
        System.out.println(N);
    
      
    // Else print(0, N)
    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        

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Python3

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# 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 & 2): 
        print(N, end = '\n')
  
    # Else print(0, N) 
    print(0, end = ', ')
    print(N, end = '\n'
  
# Driver Code 
N = 5
  
# Function Call 
findNumbers(N)
  
# This code is contributed by PratikBasu    

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C#

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// 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 & 2;
    if (check != 0)
    
        Console.Write(N + "\n");
    
      
    // Else print(0, N)
    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

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Output:

-1, 1, -2, 2, 0, 5

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