题意：

给定n种面值的硬币面值分别为$W_i$个数为$C_i$，问用这些硬币可以组成1~m之间的多少面值。

题目：

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

分析：

1.关于是否装满问题，其实之前的博客讨论过这个问题，但只需考虑装满问题时，就可以考虑情况，“标记当前出现状况”的形式来计算装满问题。
2.定义一个sum数组。每次填dp[j]时直接由dp[j-a[i]]推出，前提是sum[j-a[i]]<b[i]。sum每填一行都要清零，sum[j]表示当前物品填充j大小的包需要至少使用多少个。复杂度O(n*m)。

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int M=1e5+10;
int n,m,ans;
int a[M],b[M],num[M],dp[M];
int main(){while(~scanf("%d%d",&n,&m)){if(n==0&&m==0)break;ans=0;memset(dp,0,sizeof(dp));dp[0]=1;for(int i=0;i<n;i++)scanf("%d",&a[i]);for(int i=0;i<n;i++)scanf("%d",&b[i]);for(int i=0;i<n;i++){memset(num,0,sizeof(num));for(int j=a[i];j<=m;j++){if(!dp[j]&&dp[j-a[i]]&&num[j-a[i]]<b[i]){dp[j]=1;ans++;num[j]=num[j-a[i]]+1;}}}printf("%d\n",ans);}return 0;
}