Buns 多重背包

最新推荐文章于 2019-07-22 10:00:48 发布

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本文介绍了一个关于多重背包问题的编程挑战，通过一个具体的场景——Lavrenty制作和销售肉包来阐述。Lavrenty拥有一定数量的面团和不同种类的馅料，每种馅料有不同的存量、需求量、成本和售价。文章详细解释了如何通过两重循环的算法，计算出在有限的材料下，Lavrenty能获得的最大收益。

Description

Lavrenty, a baker, is going to make several buns with stuffings and sell them.

Lavrenty has n grams of dough as well as m different stuffing types. The stuffing types are numerated from 1 to m. Lavrenty knows that he has ai grams left of the i-th stuffing. It takes exactly bi grams of stuffing i and ci grams of dough to cook a bun with the i-th stuffing. Such bun can be sold for di tugriks.

Also he can make buns without stuffings. Each of such buns requires c0 grams of dough and it can be sold for d0 tugriks. So Lavrenty can cook any number of buns with different stuffings or without it unless he runs out of dough and the stuffings. Lavrenty throws away all excess material left after baking.

Find the maximum number of tugriks Lavrenty can earn.

Input

The first line contains 4 integers n, m, c0 and d0 (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10, 1 ≤ c0, d0 ≤ 100). Each of the following m lines contains 4 integers. The i-th line contains numbers ai, bi, ci and di (1 ≤ ai, bi, ci, di ≤ 100).

Output

Print the only number — the maximum number of tugriks Lavrenty can earn.

Sample Input

Input

10 2 2 1
7 3 2 100
12 3 1 10

Output

Input

100 1 25 50
15 5 20 10

Output

Sample Output

Hint

//多重背包

初始化dp值为只用自身所得价值

第一重循环是可以组成包子个数

第二重是面团n...c多重背包

方程是dp[k]=max{dp[k-c]+d,dp[k]}

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[10000+10];
int main()
{
    int n,m,c0,d0;
    cin>>n>>m>>c0>>d0;
    memset(dp,0,sizeof(dp));
    for(int i=c0; i<=n; i++)
        dp[i]=i/c0*d0;  //初始化只用面做馒头的价值;
    int a,b,c,d;
    for(int i=0; i<m; i++)
    {
        cin>>a>>b>>c>>d;//a代表的是它拥有的面团个数,b代表的是馅料个数,c代表的是面团个数
        for(int j=1; j<=a/b; j++) //使用i最多生成a/b个合成品
        {
            //第一层循环是合成品个数
            for(int k=n; k>=c; k--) //第二重循环是从n...c
                dp[k]=max(dp[k-c]+d,dp[k]);  //包与不包两种情况;
        }
    }
    cout<<dp[n]<<endl;
    return 0;
}