Stupid Tower Defense

Problem Description

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.

 

Input

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)

 

Output

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.

 

Sample Input

1

2 4 3 2 1

 

Sample Output

Case #1: 12

#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;
typedef __int64 LL;
const int ms=1600;
LL dp[ms][ms];
LL max(LL a,LL b)
{
    return a>b?a:b;
}
int main()
{
    LL ans,b,c;//注意 b和c 要定义为LL，因为后面的计算中含有LL形的数。
    int T,p=1;
    int n,x,y,z,t;
    scanf("%d",&T);
    //cin>>T;
    while(T--)
    {
        printf("Case #%d: ",p++);
        //cout<<"Case #"<<p++<<": ";
        scanf("%d%d%d%d%d",&n,&x,&y,&z,&t);
        //cin>>n>>x>>y>>z>>t;
        memset(dp,0,sizeof(dp));
        ans=x*n*t;
        for(b=0;b<=n;b++)
            for(c=0;c+b<=n;c++)
            {
                dp[b+1][c]=max(dp[b+1][c],dp[b][c]+c*y*(t+b*z));
                dp[b][c+1]=max(dp[b][c+1],dp[b][c]+c*y*(t+b*z));
                ans=max(ans,dp[b][c]+(n-b-c)*x*(t+b*z)+(n-b-c)*y*c*(t+b*z));
            }
        printf("%I64d
",ans);
        //cout<<ans<<endl;
    }
    return 0;
}