Cash Machine_多重背包

Description

A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example, 

N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10 

means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each. 

Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine. 

Notes: 
@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc. 

Input

The program input is from standard input. Each data set in the input stands for a particular transaction and has the format: 

cash N n1 D1 n2 D2 ... nN DN 

where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct. 

Output

For each set of data the program prints the result to the standard output on a separate line as shown in the examples below. 

Sample Input

735 3  4 125  6 5  3 350
633 4  500 30  6 100  1 5  0 1
735 0
0 3  10 100  10 50  10 10

Sample Output

735
630
0
0

 

 

【题意】给出一个sum，还有一个n代表有n个价值和对应个数；

【思路】多重背包，dp数组表示是否存在该状态，不断更新最大价值

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
struct node
{
    int n,v;
} a[20];
int dp[100010];
int main()
{
    int sum,n;
    while(~scanf("%d%d",&sum,&n))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&a[i].n,&a[i].v);
        }
        if(!sum||(!n))
        {
            printf("0\n");
            continue;
        }
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        int maxn=0,tmp;
        for(int i=1; i<=n; i++)
        {
            for(int j=maxn; j>=0; j--)
            {
                if(dp[j])
                {
                    for(int k=1; k<=a[i].n; k++)
                    {
                        tmp=j+k*a[i].v;
                        if(tmp>sum) continue;
                        dp[tmp]=1;
                        if(tmp>maxn) maxn=tmp;
                    }
                }

            }
        }
        printf("%d\n",maxn);
    }
    return 0;
}