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.
这是一道多重背包问题，可以转化为0-1背包求解，朴素的转化复杂度高，但这题数据范围不是很大可以过。
dp[i][j]表示用i去装前j件物品的最大值，转化后跟平时做的0-1背包是一样的；这边c0的dough可以有d0的tugriks，其实意思就是
dp是有初值的，就是不取物品也是有值的。
AC代码：
# 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; }
