Codeforces 106 Buns【多重背包】

参看资料：

https://blog.youkuaiyun.com/qq_37748451/article/details/86486389

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题目：

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 4integers. 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.

Examples

Input

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

Output

241

Input

100 1 25 50
15 5 20 10

Output

200

Note

To get the maximum number of tugriks in the first sample, you need to cook 2 buns with stuffing 1, 4 buns with stuffing 2 and a bun without any stuffing.

In the second sample Lavrenty should cook 4 buns without stuffings.

题目大意：

       面团为n，有M种馅饼的配案，给定C0，D0，表示只用面粉C0蒸馒头可以卖D0元。

       其后M行，第 i 行 Ai 表示有多少 i 馅的馅料，Bi 表示 做 第 i 种馅饼需要多少馅，Ci 表示该饼需要多少面，Di表示这种饼能卖多少钱。求总价值 最大为多少。

解题思路：

       裸的多重DP，（等补）。

       dp[ i ][ j ][ k ],表示i种佐料使用j*c克混合k克面粉生成的价值，第i种佐料我们可以使用for循环控制，j*b克的i面粉跟k克的佐料是对应的，也就是说dp的值是由最后一维控制的，所以j*b也可以由循环表示，所以最后的dp是一维的

实现代码：

   

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<set>
#define ll long long
using namespace std;
int main()
{
    int n,m,c0,d0,a,b,c,d;
    int dp[10010];
    memset(dp,0,sizeof(dp));
    cin>>n>>m>>c0>>d0;
    for(int i=c0;i<=n;i++)
        dp[i]=i/c0*d0;
   
    for(int i=0;i<m;i++)    //m种馅饼
    {
        cin>>a>>b>>c>>d;    
        for(int j=1;j<=a/b;j++)    //第 i 种馅饼 的个数
           for(int k=n;k>=c;k--)   //用了多少面粉
             dp[k]=max(dp[k-c]+d,dp[k]);
    }
    cout<<dp[n]<<endl;
    return 0;
}