查找N个具有零和的不同整数

给定整数N ，我们的任务是打印N个不同的数字，以使它们的总和为0。
例子：

Input: N = 3 
Output: 1, -1, 0 
Explanation: 
On adding the numbers that is 1 + (-1) + 0 the sum is 0.
Input: N = 4 
Output: 1, -1, 2, -2 
Explanation: 
On adding the numbers that is 1 + (-1) + 2 + (-2) the sum is 0. 
 

方法：解决上述问题的主要思想是打印对称对，如(+ x，-x)，这样总和将始终为0。问题的边缘情况是观察到，如果整数N为奇数，则打印一个0和数字，以使总和不受影响。
下面是上述方法的实现：

C++

// C++ implementation to Print N distinct
// numbers such that their sum is 0
 
#include 
using namespace std;
 
// Function to print distinct n
// numbers such that their sum is 0
void findNumbers(int N)
{
    for (int i = 1; i <= N / 2; i++) {
 
        // Print 2 symmetric numbers
        cout << i << ", " << -i << ", ";
    }
 
    // print a extra 0 if N is odd
    if (N % 2 == 1)
        cout << 0;
}
 
// Driver code
int main()
{
    int N = 10;
 
    findNumbers(N);
}

Java

// Java implementation to Print N distinct
// numbers such that their sum is 0
 
class GFG{
 
// Function to print distinct n
// numbers such that their sum is 0
static void findNumbers(int N)
{
    for (int i = 1; i <= N / 2; i++)
    {
        // Print 2 symmetric numbers
        System.out.print(i + ", " + -i + ", ");
    }
 
    // Print a extra 0 if N is odd
    if (N % 2 == 1)
        System.out.print(0);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 10;
    findNumbers(N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to print N distinct
# numbers such that their sum is 0
 
# Function to print distinct n
# numbers such that their sum is 0
def findNumbers(N):
     
    for i in range(1, N // 2 + 1):
         
        # Print 2 symmetric numbers
        print(i, end = ', ')
        print(-i, end = ', ')
         
    # Print a extra 0 if N is odd
    if N % 2 == 1:
        print(0, end = '')
     
# Driver code
if __name__=='__main__':
     
    N = 10
    findNumbers(N)
 
# This code is contributed by rutvik_56

C#

// C# implementation to print N distinct
// numbers such that their sum is 0
using System;
 
class GFG {
 
// Function to print distinct n
// numbers such that their sum is 0
static void findNumbers(int N)
{
    for(int i = 1; i <= (N / 2); i++)
    {
 
       // Print 2 symmetric numbers
       Console.Write(i + ", " + -i + ", ");
    }
     
    // Print a extra 0 if N is odd
    if (N % 2 == 1)
        Console.Write(0);
}
 
// Driver code
static void Main()
{
    int N = 10;
     
    findNumbers(N);
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

 
// Javascript implementation to Print N distinct
// numbers such that their sum is 0
 
// Function to print distinct n
// numbers such that their sum is 0
function findNumbers(N)
{
    for (var i = 1; i <= N / 2; i++) {
 
        // Print 2 symmetric numbers
        document.write( i + ", " + -i + ", ");
    }
 
    // print a extra 0 if N is odd
    if (N % 2 == 1)
        document.write( 0);
}
 
// Driver code
var N = 10;
findNumbers(N);
 

输出：

1, -1, 2, -2, 3, -3, 4, -4, 5, -5,

时间复杂度： O(log N)