天坑。。。
杭电
dls代码：https://ideone.com/Wo55gi
官方题解：http://bestcoder.hdu.edu.cn/blog/
2018 Multi-University Training Contest 1
1001 Maximum Multiple
打表找规律，发现只有当n是3倍数或者4倍数时才有解
而且当n%3 == 0，max(x*y*z) = (n/3) * (n/3) * (n/3) = n^3 / 27，
当n%4 == 0, max(x*y*z) = (n/2) * (n/4) * (n/4) = n^3 / 32。
正解是解不定方程组
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head int main() { int T, n; scanf("%d", &T); while(T--) { scanf("%d", &n); if(n % 3 == 0) printf("%lld\n", 1LL * n * n * n / 27); else if(n % 4 == 0)printf("%lld\n", 1LL * n * n * n / 32); else printf("-1\n"); } return 0; }