题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5410
Problem Description
Today is CRB‘s birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with M Won(currency unit).
At the shop, there are N kinds of presents.
It costs Wi Won to buy one present of i-th kind. (So it costs k × Wi Won to buy k of them.)
But as the counter of the shop is her friend, the counter will give Ai × x + Bi candies if she buys x(x>0) presents of i-th kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤ T ≤ 20
1 ≤ M ≤ 2000
1 ≤ N ≤ 1000
0 ≤ Ai, Bi ≤ 2000
1 ≤ Wi ≤ 2000
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers M and N.
Then N lines follow, i-th line contains three space separated integers Wi, Ai and Bi.
Output
For each test case, output the maximum candies she can gain.
Sample Input
1
100 2
10 2 1
20 1 1
Sample Output
21
Hint
CRB‘s mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
题目的大概意思是有M元钱，有N件物品，没件物品的花费是W[i]，第i件物品买X件能得到A[i]*X+B[i]块糖，没件物品买的数量没有上限
问这M元钱最多能得到多少块糖。
问题分析：没种物品买第一件能得到A[i]+B[i]块糖，以后每买一件只能多得到A[i]块糖，也就是每种物品买第一件比买后面的得到的糖多，
根据贪心的思想很容易想到要先每种物品买一件再买大于一件，这样就把问题分成了两个阶段，第一阶段每种物品最多只买一件，
0/1背包，第二阶段，没种物品能买无限件，完全背包。
1 #include <iostream>
2 #include <cstring>
3 using namespace std;
4
5 int ma(int a,int b)
6 {
7 if (a>b)
8 return a;
9 else
10 return b;
11 }
12 int main()
13 {
14 int ans[12000],v,a[1100],b[1100],c[1100],n,m,T;
15 cin>>T;
16 while (T--)
17 {
18 cin>>v>>m;
19 for (int i=1;i<=m;i++)
20 cin>>c[i]>>a[i]>>b[i];
21 memset(ans,0,sizeof(ans));
22 for (int i=1;i<=m;i++)//处理第一阶段，0/1背包
23 {
24 for (int j=v;j>=c[i];j--)
25 {
26 ans[j]=ma(ans[j],ans[j-c[i]]+a[i]+b[i]);
27 }
28 }
29 for (int i=1;i<=m;i++)//处理第二阶段，完全背包。
30 {
31 for (int j=c[i];j<=v;j++)
32 {
33 ans[j]=ma(ans[j],ans[j-c[i]]+a[i]);
34 }
35 }
36 cout <<ans[v]<<endl;
37 }
38 return 0;
39 }