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

最优情况下，红塔必然都在最后，dp[i][j]代表前i个有j个蓝塔，枚举最后红塔的个数和前面蓝塔的个数进行dp即可
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define w(x) while(x)
#define ll __int64

ll n,x,y,z,t,dp[1505][1505],ss,cas=1,i,j,k,r,ans;

int main()
{
    scanf("%I64d",&ss);
    w(ss--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
        mem(dp,0);
        ans=n*t*x;//都是红塔
        up(i,1,n)
        up(j,0,i)
        {
            if(!j)
                dp[i][j]=dp[i-1][j]+t*(i-j-1)*y;
            else
                dp[i][j]=max(dp[i-1][j]+(j*z+t)*max(0LL,(i-1-j))*y,dp[i-1][j-1]+((j-1)*z+t)*(i-j)*y);
            ans=max(ans,dp[i][j]+(n-i)*(j*z+t)*(x+(i-j)*y));
        }
        printf("Case #%I64d: %I64d\n",cas++,ans);
    }
    return 0;
}