CQUOJ 2731 - Coins

Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse 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

/*
2016年5月18日01:03:42

题意 ：给你一些不同价值和一定数量的硬币，求用这些硬币可以组合成价值在[1 , m]之间的有多少。

多重背包
学习: http://blog.csdn.net/lyhvoyage/article/details/8545852
*/

# include <iostream>
# include <algorithm>
# include <cstdio>
# include <cstring>
# include <queue>
# include <vector>
# include <cmath>
# include <map>
# include <set>
# define pi acos(-1)
# define LL long long
# define INF 0x3f3f3f3f
using namespace std;
const int N = 1e5 + 5;

int dp[N], n, m;
int val[105], cnt[105];
void _01Pack(int w, int v)    //01背包 
{
    for (int i = m; i >= w; i--)
        dp[i] = max(dp[i], dp[i-w]+v);
}
void comPack(int w, int v)  //完全背包 
{
    for (int i = w; i <= m; i++)
        dp[i] = max(dp[i], dp[i-w]+v);
}
void multiPack(int w, int v, int c) // 多重背包 
{
    if (w*c >= m)
        comPack(w, v);
    else {
        int k = 1;
        while (k <= c){
            _01Pack(k*w, k*v);
            c -= k;
            k *= 2;
        }
        _01Pack(c*w, c*v);
    }
}
int main(void)
{
    int i, ans; 
    while (~scanf("%d %d", &n, &m), n, m){
        for (i = 1; i <= n; i++)
            scanf("%d", &val[i]);
        for (i = 1; i <= n; i++)
            scanf("%d", &cnt[i]);
        ans = 0;
        memset(dp, 0, sizeof(dp));
        for (i = 1; i <= n; i++)
            multiPack(val[i], val[i], cnt[i]);
        for (i = 1; i <= m; i++)
            if (dp[i] == i) ans++;
        cout<<ans<<endl;
    } 

    return 0;
}