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

背包問題，一定要多思考，，，參考大神代碼

 

#include
#include
#include
#include
using namespace std;
#define M 120
int val[M],cnt[M];
int dp[100020];// dp[i][j] := 用前i種硬幣湊成j時第i種硬幣最多能剩余多少個
//dp[j] := 在第i次循環時之前表示用前i-1種硬幣湊成j時第i種硬幣最多能剩余多少個
 //（-1表示配不出來），循環之後就表示第i次的狀態
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;
}