求 N 个不同的零和整数
给定一个整数 N ，我们的任务是打印 N 个不同的数字，使得它们的和为 0。 例:
输入: N = 3 输出: 1，-1，0 说明: 将 1 + (-1) + 0 的数字相加，和为 0。 输入: N = 4 输出: 1，-1，2，-2 说明: 在加 1 + (-1) + 2 + (-2)的数时，和为 0。
方法:解决上述问题的主要思路是将 对称对 像(+x，-x)一样打印出来，这样和永远为 0。问题的边缘情况是观察到如果整数 N 为奇数，则随数字一起打印一个 0，这样总和不受影响。 以下是上述方法的实施:
C++
// C++ implementation to Print N distinct
// numbers such that their sum is 0
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// 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);
</script>
Output:
1, -1, 2, -2, 3, -3, 4, -4, 5, -5,
时间复杂度:O(N)T4】
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