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.
這是一道多重揹包問題，可以轉化爲0-1揹包求解，樸素的轉化複雜度高，但這題數據範圍不是很大可以過。
dp[i][j]表示用i去裝前j件物品的最大值，轉化後跟平時做的0-1揹包是一樣的；這邊c0的dough可以有d0的tugriks，其實意思就是
dp是有初值的，就是不取物品也是有值的。
AC代碼：
# include <stdio.h>
# include <iostream>
# include <string.h>
# include <algorithm>
# include <queue>
using namespace std;
typedef long long int ll;
struct stuff{
int c, w, v;
};
stuff s[100010];
int dp[1010][10010];
int main(){
int i, j, k, m, n, c0, d0;
int a, b, c, d;
int cnt=1;
scanf("%d%d%d%d", &n, &m, &c0, &d0);
for(i=1; i<=m; i++){
scanf("%d%d%d%d", &a, &b, &c, &d);
int temp=a/b;
for(j=1; j<=temp; j++){
s[cnt].c=c;
s[cnt].v=d;
s[cnt].w=b;
cnt++;
}
}
for(i=1; i<=n; i++){
for(j=0; j<=cnt-1; j++){
dp[i][j]=i/c0*d0;
}
}
for(i=1; i<=n; i++){
for(j=1; j<=cnt-1; j++){
dp[i][j]=max(dp[i][j], dp[i][j-1]);
if(i>=s[j].c){
dp[i][j]=max(dp[i][j], dp[i-s[j].c][j-1]+s[j].v);
}
}
}
printf("%d", dp[n][cnt-1]);
return 0;
}