因为瞬间伤害的塔一定是放在终点端的，所以枚举这种塔的个数

然后就能把n^3变成n^2了

#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
const int N = 1500 + 2;
ll d[N][N]; // j = 3;
int Tt = 0, n, x, y, z, t;

inline void up(ll& a, ll v) {
    if (v > a)
        a = v;
}

void work() {
    ll ans = 0, T, D;
    scanf("%d%d%d%d%d", &n, &x, &y, &z, &t);
    for (int i = 0; i <= n; ++i)
        for (int j = 0; j <= i; ++j)
            d[i][j] = 0;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j <= i; ++j) {
            T = t + (ll)z * j;
            D = T * (n - i) * x + T * (n - i) * y * (i - j) + d[i][j];
            up(ans, D);
            // put a 2
            up(d[i + 1][j], d[i][j] + (ll)(i - j) * y * T);
            up(d[i + 1][j + 1], d[i][j] + (ll)(i - j) * y * T);
        }
    for (int i = 0; i <= n; ++i)
        up(ans, d[n][i]);
    printf("Case #%d: %I64d\n", ++Tt, ans);
}

int main() {
    int cas;
    scanf("%d", &cas);
    while (cas -- > 0) 
        work();
    return 0;
}