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  • Codeforces--106C--Buns(背包)

    Buns

    Time Limit: 2000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u

    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 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.

    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;
    }

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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273422.html
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