Codeforces--106C--Buns（背包）

Buns

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

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 a_i grams left of the i-th stuffing. It takes exactly b_i grams of stuffing i and c_i grams of dough to cook a bun with the i-th stuffing. Such bun can be sold for d_i 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 a_i, b_i, c_i and d_i (1 ≤ a_i, b_i, c_i, d_i ≤ 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

241

Input

100 1 25 50
15 5 20 10

Output

200

Sample Output

Hint

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.

n克面粉，m种佐料，n克的面粉每c0克可以生成d0价值的产品，同时也可以与任意一种b克佐料混合生成d价值的产品，问这些东西最多制成多少价值的产品

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

#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;
	int a,b,c,d;
	for(int i=0;i<m;i++)
	{
		cin>>a>>b>>c>>d;
		for(int j=1;j<=a/b;j++)//使用i最多生成a/b个合成品 
		{
			for(int k=n;k>=c;k--)
			dp[k]=max(dp[k-c]+d,dp[k]);
		}
	}
	cout<<dp[n]<<endl;
	return 0;
}