题解 | #尼科彻斯定理#

#include <stdio.h>
#include<math.h>
int main() {
    int n = 0;
    scanf("%d", &n);
    int sum = 0;
    int answer = pow(n, 3);
    int j = 1;

    while (sum < answer) {
        int i = 0;
        sum = 0;
        for (i = 1; i <= n; i++) {
            int m = j;
            sum += m;
            j += 2;
        }
        if (answer == sum) {
            j = j - (2 * n) ;
            for (i = 1; i < n; i++) {
                printf("%d+", j);
                j += 2;
            }
            printf("%d", j);
        }
        j = j - (2 * n) + 2;

    }
    return 0;
}