可以使用二分优化掉一层枚举。经由两层枚举，有 d=k-a^3-b^3 。预处理立方数组 c^3，若在数组里能二分找到 d ，就输出 c^3 对应的 c 。

n=10^3 ，时间复杂度为 O(n^2\log n) 。

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define mx 1000
ll k, cube[2 * mx + 10], a, b, c[2 * mx + 10], ci, m, r, cn;
signed main()
{
    scanf("%lld", &k);
    for (ll i = mx; i >= 1; --i, ++cn)
        cube[cn] = -i * i * i, c[cn] = -i;
    for (ll i = 1; i <= mx; ++i, ++cn)
        cube[cn] = i * i * i, c[cn] = i;
    for (a = -mx; a <= mx; ++a)
    {
        for (b = a; b <= mx; ++b)
        {
            if (a == 0 || b == 0)
                continue;
            ll d = k - a * a * a - b * b * b, c3;
            ci = lower_bound(cube, cube + cn, d) - cube;
            c3 = cube[ci];
            d = a * a * a + b * b * b + c3;
            if (d == k)
            {
                printf("%lld %lld %lld", a, b, c[ci]);
                return 0;
            }
        }
    }
    printf("no solution");
    return 0;
}