HDU 2844 Coins[【经典题】【模板题】

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15971    Accepted Submission(s): 6340

Problem Description

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

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

多重背包

#include<iostream>  
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
#define pi 3.14159265358979
#define mod 1000000007
#define maxn 100010
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;
int n, m, c[maxn], v[maxn], d[maxn];
void zeroonepack(int cost, int weight)
{
    for (int i = m; i >= cost; i--)
    {
        d[i] = max(d[i], d[i - cost] + weight);
    }
}
void completepack(int cost, int weight)
{
    for (int i = cost; i <= m; i++)
    {
        d[i] = max(d[i], d[i - cost] + weight);
    }
}
void multipack(int cost, int weight, int amount)
{
    if (cost*amount >= m)
        completepack(cost, weight);
    else
    {
        int k = 1;
        while (k <= amount)
        {
            zeroonepack(k*cost, k*weight);
            amount -= k;
            k *= 2;
        }
        zeroonepack(amount*cost, amount*weight);
    }
}
int main()
{
    while (cin >> n >> m)
    {
        if (!n && !m) break;
        ms(d, 0);
        for (int i = 0; i < n; i++)
        {
            cin >> v[i];
        }
        for (int i = 0; i < n; i++)
        {
            cin >> c[i];
        }
        for (int i = 0; i < n; i++)
        {
            multipack(v[i], v[i], c[i]);
        }
        int sum = 0;
        for (int i = 1; i <= m; i++)
        {
            if (d[i] == i)
                sum++;
        }
        cout << sum << endl;
    }

}