poj_1742_Coins

#include<iostream>
#include<sstream>
#include<string>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#pragma warning(disable:4996)
#include<cstdio>
#include<cstdlib>
using std::cin;
using std::cout;
using std::endl;
using std::stringstream;
using std::string;
using std::vector;
using std::list;
using std::pair;
using std::set;
using std::multiset;
using std::map;
using std::multimap;
using std::stack;
using std::queue;
using std::priority_queue;
int main()
{
	//freopen("input.txt", "r", stdin);  
	//freopen("output.txt", "w", stdout);  
	int n, m;
	//n表示硬币种类，m表示背包容量
	while (scanf("%d%d", &n, &m))
	{
		if (!n&&!m)
		{
			break;
		}
		//.first表示硬币价值，.second表示硬币个数
		vector<pair<int, int> >coin(n);
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &coin[i].first);
		}
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &coin[i].second);
		}
		vector<bool>dp(m + 1); dp[0] = true;
		int count = 0;
		for (int i = 0; i < coin.size(); i++)
		{
			//这是这道题的关键所在，也可以说是多重背包可行性的关键所在
			//用一个向量表示当金额为i的时候，已经用了多少当前的硬币
			vector<int>used(m + 1);
			for (int j = coin[i].first; j <=m; j++)
			{
				if (!dp[j]&& dp[j - coin[i].first]&&used[j - coin[i].first]+1<=coin[i].second)
				{
					dp[j] = true;
					used[j] = used[j - coin[i].first] + 1;
					count++;
				}
			}
		}
		printf("%d\n", count);
	}
	return 0;
}