poj 1742 Coins（多重背包）

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 31369		Accepted: 10679

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

Source

LouTiancheng@POJ

背包问题，一定要多思考，，，参考大神代码

#include
#include
#include
#include
using namespace std;
#define M 120
int val[M],cnt[M];
int dp[100020];// dp[i][j] := 用前i种硬币凑成j时第i种硬币最多能剩余多少个
//dp[j] := 在第i次循环时之前表示用前i-1种硬币凑成j时第i种硬币最多能剩余多少个
 //（-1表示配不出来），循环之后就表示第i次的状态
int n,m;
int main(){
	int i,j;
	while(scanf("%d%d",&n,&m),n+m){
		memset(dp,-1,sizeof(dp));
		for(i=0;i=0)
					dp[j]=cnt[i];
				else if(j=0) ans++;
		printf("%d\n",ans);
	}
	return 0;
}