题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5501
The Highest Mark
Problem Description
The SDOI in 2045 is far from what it was been 30 years ago. Each competition has t minutes and n problems.
The ith problem with the original mark of Ai(Ai≤106),and it decreases Bi by each minute. It is guaranteed that it does not go to minus when the competition ends. For example someone solves the ith problem after x minutes of the competition beginning. He/She will get Ai−Bi∗x marks.
If someone solves a problem on x minute. He/She will begin to solve the next problem on x+1 minute.
dxy who attend this competition with excellent strength, can measure the time of solving each problem exactly.He will spend Ci(Ci≤t) minutes to solve the ith problem. It is because he is so godlike that he can solve every problem of this competition. But to the limitation of time, it's probable he cannot solve every problem in this competition. He wanted to arrange the order of solving problems to get the highest mark in this competition.
Input
There is an positive integer T(T≤10) in the first line for the number of testcases.(the number of testcases with n>200 is no more than 5)
For each testcase, there are two integers in the first line n(1≤n≤1000) and t(1≤t≤3000) for the number of problems and the time limitation of this competition.
There are n lines followed and three positive integers each line Ai,Bi,Ci. For the original mark,the mark decreasing per minute and the time dxy of solving this problem will spend.
Hint:
First to solve problem 2 and then solve problem 1 he will get 88 marks. Higher than any other order.
Output
For each testcase output a line for an integer, for the highest mark dxy will get in this competition.
Sample Input
1
4 10
110 5 9
30 2 1
80 4 8
50 3 2
Sample Output
88
题目大意：
每道题目有A,B,C三个值，A表示初始分数，B表示每分钟题的分数会减少B，C表示做这道题需要C分钟，数据保证分数不会变为负数。现在给出比赛时长，问安排做题的顺序，求最大得分。
思路：
很清楚的01背包，但是需要注意的是，完成题目的时间不同会产生不同的的得分
现在有A1,B1,C1和A2,B2,C2这两道题，如果先做T1再做T2的得分是A1-B1*C1+A2-B2*(C1+C2)，如果先做T2在做T1的得分是A2-B2*C2+A1-B1*(C1+C2),令先做T1再做T2的得分更高些，那么有A1-B1*C1+A2-B2*(C1+C2) >= A2-B2*C2+A1-B1*(C1+C2),解得B1*C2>=B2*C1所以，只要B1*C2>=B2*C1，那么先做题1再做题2的分数就会更高。
所以，我们需要根据这个来排序，再做01背包
This is the code：
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define LL long long
#define ULL unsigned long long //1844674407370955161
#define INT_INF 0x7f7f7f7f //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。
const int dr[]={0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]={-1, 1, 0, 0, -1, 1, -1, 1};
struct node
{
LL a;//得分
LL b;//
LL c;//时间
bool operator <(const node &tem)const
{
//还要注意这个地方需要用long long 不然乘法会爆
return b*tem.c>tem.b*c;//注意这个排序,可以自己输出一边就能明白了
}
}a[1005];
LL dp[3005];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
LL n,T;
scanf("%lld%lld",&n,&T);
for(int i=1;i<=n;++i)
scanf("%lld%lld%lld",&a[i].a,&a[i].b,&a[i].c);
sort(a+1,a+n+1);//需要排序（贪心）
//排序保证的是时间优先的同时，得分多
//for(int i=1;i<=n;++i)
//printf("%lld %lld %lld\n",a[i].a,a[i].b,a[i].c);
//puts("");
LL ans=-1;
for(int i=1;i<=n;++i)
{
for(int j=T;j>=a[i].c;--j)
{
dp[j]=max(dp[j],dp[j-a[i].c]+a[i].a-j*a[i].b);
ans=max(dp[j],ans);//dp[T]不一定是得分最高
}
}
printf("%lld\n",ans);
}
return 0;
}