题解 | #尼科彻斯定理#

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main()
{
    int n;
    int N = 100;
    while(scanf("%d",&n)!=EOF)
    {
        int a[100]={0};
        int num = pow(n, 3);
        int tmp = pow(n, 2);
        int sum =0;
        int t;
        if(tmp%2==0)//根据n的平方是否为偶数分为两种情况
        {
            a[0] = tmp-1;
            a[1] = tmp+1;
        }
        else
        {
            a[0] = tmp;
        }
        int k = 2;
        while(sum!=num)
        {
            sum = 0;
            if(tmp%2==0)
            {
                a[k] = a[0]-(k);
                a[k+1] = a[1]+(k);
            }
            else
            {
                a[k] = a[0]-(k);
                a[k+1] = a[0]+(k);
            }
            k+=2;
            for(int i = 0;i<N;i++)
            {
                sum+=a[i];
            }
        }
        for(int i = 0;i<N;i++)//对找到的结果进项排序输出
        {
           for(int j = i+1;j<N;j++)
           {
               if(a[i]>a[j])
               {
                   t = a[i];
                   a[i] = a[j];
                   a[j] = t;
               }
           }
        }
        for(int i=0;i<N-1;i++)
        {
            if(a[i]==0)
                continue;
            else
            {
                printf("%d",a[i]);
                printf("+");
            }
        }
         printf("%d\n",a[N-1]);
    }
    return 0;
}