dp[i][j]表示到了第i步放了j个减速，造成的伤害。我们用贪心的策略把造成一段伤害的放在最后面，造成持续伤害的与减速放在前i个中这样得到的伤害是最高的。
所以前(i,j)中的伤害为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);
每次造成的伤害就为：dp[i][j]+(n-i)*(j*z+t)*(x+(i-j)*y)。然后取一个最大值，就是伤害的最大值了。
Stupid Tower Defense
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1194 Accepted Submission(s): 350
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
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