[POJ1742 Coins]DP

题目链接 http://poj.org/problem?id=1742

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 31612		Accepted: 10756

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

这题和HDU 2844是一模一样的，但是HDU的数据比较弱，用二进制优化后的代码是无法在poj上面通过的，会超时。

HDU 2844 用二进制优化 http://blog.youkuaiyun.com/chenninggo/article/details/48321197

这题还是要用dp，记录该种硬币在和为 i 时用了几枚

具体看代码

/*
用dp记录当前i价值能不能组成
用sum数组记录组成i价值时用了多少枚该种硬币，注意该种硬币的使用次数不能超过该硬币所拥有的数目
当当前i价值无法组成时，进行动归
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n, m;
int dp[100010], sum[100010];
int val[110], c[110];
int main()
{
//	freopen("input.txt", "r", stdin);
	while (scanf("%d%d", &n, &m) != EOF && n && m)
	{
		for (int i = 1; i <= n; i++)
			scanf("%d", &val[i]);
		for (int i = 1; i <= n; i++)
			scanf("%d", &c[i]);
		memset(dp, 0, sizeof(dp));
		dp[0] = 1;
		int ans = 0;
		for (int i = 1; i <= n; i++)
		{
			memset(sum, 0, sizeof(sum));
			for (int l = val[i]; l <= m; l++)
			{
				if (!dp[l] && dp[l - val[i]] && sum[l - val[i]] < c[i])
				{
					dp[l] = 1;
					sum[l] = sum[l - val[i]] + 1;
					ans++;
				}
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}