poj 1742 coins

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

题目的大意

    有n种硬币 ，求n种硬币加起来不能超过m的方案数

对于dp还是要多进行思考，这个是一个多重背包的问题，详细的解说在代码中

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
    {
        int n,m;
        int a[205],c[205],dp[100005],vis[100005];
        //dp表示该币值的硬币数量有多少个
        //vis表示该币值的硬币是否被用过
        while (cin>>n>>m)
            {
                memset(dp,0,sizeof(dp));
                memset(vis,0,sizeof(vis));
                vis[0]=1;
                int cnt=0;
                if(n==0&&m==0) break;
                for (int i=0;i<n;i++)
                    scanf("%d",&a[i]);
                for (int i=0;i<n;i++)
                    {
                        scanf("%d",&c[i]);
                    }
                for (int i=0;i<n;i++)
                    {
                        memset(dp,0,sizeof(dp));//很好理解啊
                        for (int j=a[i];j<=m;j++)
                            {
                                //dp向前滚动
                                if(!vis[j]&&vis[j-a[i]]&&dp[j-a[i]]<c[i])//0<=dp<=c[i]  0<dp-1<c[i]
                                    {
                                        cnt++;
                                        vis[j]=1;
                                        dp[j]=dp[j-a[i]]+1;
                                    }
                            }
                    }
                printf("%d\n",cnt);
            }
        return 0;
    }