FSF is addicted to a stupid tower defense game. The goal of tower defense games is to try to stop enemies from crossing a map by building traps to slow them down and towers which shoot at them as they pass.

The map is a line, which has n unit length. We can build only one tower on each unit length. The enemy takes t seconds on each unit length. And there are 3 kinds of tower in this game: The red tower, the green tower and the blue tower.

The red tower damage on the enemy x points per second when he passes through the tower.

The green tower damage on the enemy y points per second after he passes through the tower.

The blue tower let the enemy go slower than before (that is, the enemy takes more z second to pass an unit length, also, after he passes through the tower.)

Of course, if you are already pass through m green towers, you should have got m*y damage per second. The same, if you are already pass through k blue towers, the enemy should have took t + k*z seconds every unit length.

FSF now wants to know the maximum damage the enemy can get.

There are multiply test cases.

The first line contains an integer T (T<=100), indicates the number of cases.

Each test only contain 5 integers n, x, y, z, t (2<=n<=1500,0<=x, y, z<=60000,1<=t<=3)

For each case, you should output "Case #C: " first, where C indicates the case number and counts from 1. Then output the answer. For each test only one line which have one integer, the answer to this question.

1
2 4 3 2 1

Case #1: 12

Hint
For the first sample, the first tower is blue tower, and the second is red tower. So, the total damage is 4*(1+2)=12 damage points.
 

UESTC

题意：

给你一段长为n的路，每一个单位长度可以放一种塔，这里有三种塔。

1）对正在经过这座塔的敌人进行 x 每秒伤害的攻击

2）对于已经经过这塔的敌人进行y每秒的伤害攻击

3）对已经经过这个塔的敌人放慢速度，使得原先为 经过一个单位时间为  t的速度变为  t+k

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=1617;
typedef __int64 LL;
LL dp[maxn][maxn];
LL n,x,y,z,t;
int main()
{
    int T;
    int cas = 0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
        memset(dp,0,sizeof(dp));
        LL ans = n*t*x;//全部放红塔
        
        for(LL i = 1; i <= n; i++)
            for(LL j = 0; j <= i; j++)
            {
                if(j == 0)//没有减缓塔（蓝塔）
                    dp[i][j]=dp[i-1][j]+t*(i-j-1)*y;
                else
                    dp[i][j]=max(dp[i-1][j-1]+(i-j)*y*(t+(j-1)*z),dp[i-1][j]+(i-j-1)*y*(t+j*z));
                ans = max(ans,dp[i][j]+(n-i)*(j*z+t)*(x+(i-j)*y));
            }
        printf("Case #%d: %I64d\n",++cas,ans);
    }
    return 0;
}