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.
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).
Print the only number — the maximum number of tugriks Lavrenty can earn.
10 2 2 1
7 3 2 100
12 3 1 10
241
100 1 25 50
15 5 20 10
200
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.
题目其实不难理解，第一行会给出你现有的面团值，而你每做一个馅饼都会消耗面团，馅饼可以分为两类，有馅的和没有陷的，每种馅饼都给出了相应的价值
这道题就是模板类的多重背包，只是在dp初始化的时候需要做出改变,具体的看代码就了解了~
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define maxn 1000+10 int main() { int n,m,c1,d1; while(scanf("%d%d%d%d",&n,&m,&c1,&d1)!=EOF) { int cost[1500],wealth[1500],dp[maxn]; int a,b,x,y,cnt=0; memset(dp,0,sizeof(dp)); for(int i=c1;i<=n;i++) dp[i]=i/c1*d1; //cost[1]=c1,wealth[1]=d1; for(int i=1;i<=m;i++) { scanf("%d%d%d%d",&a,&b,&x,&y); for(int i=1;i<=a/b;i++) { for(int j=n;j>=x;j--) dp[j]=max(dp[j],dp[j-x]+y); } } printf("%d\n",dp[n]); } return 0; }
好的，就这样了~
原文：https://www.cnblogs.com/tombraider-shadow/p/10486201.html