题目链接
http://acm.split.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
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.
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.
Author
KUT（DPRK）
Source
Recommend
wange2014
题意：输入M，N，分别表示总的钱数和物品种数，接下来输入N行，每行3个数，单价、买一件送的糖数 、买一次送的糖数 求最多能得到多少糖？
思路：01背包，dp[i]表示i钱下能得到最多的糖数，vis[i][j]表示i钱下得到最多糖时，是否买j物品，状态转移方程dp[i]=dp[i-kind[j][0]]+kind[j][1]+(vis[i-kind[j][0]][j]==0) * kind[j][2]；
代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#define eps 1e-8
#define maxn 105
#define inf 0x3f3f3f3f3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;
int dp[];
int vis[][];
int kind[][]; int main()
{
int T;
int M,N;
cin>>T;
while(T--)
{
scanf("%d%d",&M,&N);
for(int i=;i<N;i++)
scanf("%d%d%d",&kind[i][],&kind[i][],&kind[i][]);
memset(dp,,sizeof(dp));
memset(vis,,sizeof(vis)); for(int i=;i<=M;i++)
{
int flag=-;
for(int j=;j<N;j++)
{
if(i<kind[j][]) continue;
int s=dp[i-kind[j][]]+kind[j][];
if(!vis[i-kind[j][]][j]) s+=kind[j][];
if(dp[i]<s)
{
dp[i]=s;
flag=j;
}
}
if(flag>=)
{
for(int j=;j<N;j++)
{
vis[i][j]=vis[i-kind[flag][]][j];
}
vis[i][flag]++;
}
}
int tmp=;
for(int i=;i<=M;i++)
tmp=max(tmp,dp[i]);
printf("%d\n",tmp);
}
return ;
}