Description
LYD loves codeforces since there are many Russian contests. In an contest lasting for T minutes there are n problems, and for the ith problem you can get ai−di∗ti points, where ai indicates the initial points, di indicates the points decreased per minute (count from the beginning of the contest), and ti stands for the passed minutes when you solved the problem (count from the begining of the contest).
Now you know LYD can solve the ith problem in ci minutes. He can't perform as a multi-core processor, so he can think of only one problem at a moment. Can you help him get as many points as he can?
Input
The first line contains two integers n,T(0≤n≤2000,0≤T≤5000).
The second line contains n integers a1,a2,..,an(0<ai≤6000).
The third line contains n integers d1,d2,..,dn(0<di≤50).
The forth line contains n integers c1,c2,..,cn(0<ci≤400).
Output
Output an integer in a single line, indicating the maximum points LYD can get.
Sample Input
3 10 100 200 250 5 6 7 2 4 10
Sample Output
254
题意：有 n道题目，每一道题都有一个初始分值 ai，每个单位时间这道题的分数便会减少 di，而我们可以在 ci时间内做出这道题而得到分数，求在时间 T 内最多可以获得的分数。
思路：先挑单位时间内分数下降最快（di/ci）的做
看别人的博客，但是不理解01背包为什么要排序，每种物品取或者不取不都是固定的吗？后来一想才明白排序只是要让那些一定会选中的先被选（一直刷01背包的模板水题导致自己思维的固定）！！！
贪心+01背包，主要是推出公式
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
struct node
{
int a;
int d;
int c;
}a[2003];
bool cmp(node x,node y)
{
return (double)(x.d/x.c)>(double)(y.d/y.c);
}
int dp[5005];
int main()
{
int n,t,maxx;
while(cin>>n>>t)
{
for(int i=1;i<=n;i++)
{
cin>>a[i].a;
}
for(int i=1;i<=n;i++)
{
cin>>a[i].d;
}
for(int i=1;i<=n;i++)
{
cin>>a[i].c;
}
memset(dp,0,sizeof(dp));
sort(a+1,a+1+n,cmp);
maxx=0;
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-a[i].d*j);
if(dp[j]>maxx)
maxx=dp[j];
}
}
cout<<maxx<<endl;
}
}