POJ 1742 Coins 多重背包

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 24891		Accepted: 8419

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
代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std ;
#define MAX 100010
#define M 110
int  v[M] , c[M] ;
int mun[MAX] ;
bool vi[MAX] ;
int main()
{
	int i , j , n , m ;
	int ans ;
	while( scanf( "%d%d" , &n , &m ) != EOF )
	{
		if( n == 0 || m == 0 ) break ;
		for( i = 1; i <= n ; i++) 
			scanf( "%d" , &v[i] ) ;
		for( i = 1 ; i <= n ;i++)
			scanf( "%d" , &c[i] ) ;
		memset( vi , 0 , sizeof(vi) ) ;
		ans = 0 ;
		vi[0] = 1 ;
		for( i = 1 ; i <= n ;i++)
		{
			for( j = 0 ; j <= m ; j++ ) 
				mun[j] = 0 ;// mun[] 是用当前硬币数量，所以每次都要初始化

			for( j = v[i] ; j <= m ; j++)
			{
				if( !vi[j] && vi[j-v[i]] && mun[j-v[i]] < c[i])
				{// 如果j-v[i]已经能组成而且组成它的数量少于当前硬币的数量
				 // 说明当前 的总量肯定能组成,
					vi[j] = 1 ;
					mun[j] = mun[j-v[i]] + 1 ;
					ans++ ;
				}
			}
		}
		cout << ans << endl ;
	} 
}