hdu 2844 Coins 多重背包（模板） *

Coins

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

 

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
typedef long long ll;
#define INF 0x3f3f3f3f
using namespace std;
#define maxn 100005
int a[maxn],c[maxn],dp[maxn];
int main(){
   int n,m;
   while(cin >> n >> m){
           if(!n && !m){
               break;
        }
        for(int i=;i<n;i++){
            cin >> a[i];//a[i]既是物体的体积，又是物体的价值
        }
        for(int i=;i<n;i++){
            cin >> c[i]; //c[i]是物体的数量
        }
        memset(dp,,sizeof(dp));
        for(int i=;i<n;i++){
            if(a[i]*c[i]>=m){//体积乘以数量大于总体积，说明不能完全装完，相当于有无穷件，用完全背包
                for(int j=a[i];j<=m;j++){
                    dp[j] = max(dp[j],dp[j-a[i]]+a[i]);
                }
            }
            else{//可以装完，用01背包
                int k = ;
                while(k<c[i]){//二进制优化
                    for(int j=m;j>=a[i]*k;j--){
                        dp[j] = max(dp[j],dp[j-a[i]*k]+a[i]*k);
                    }
                    c[i] -= k;
                    k *= ;
                }
                for(int j=m;j>=a[i]*c[i];j--){
                    dp[j] = max(dp[j],dp[j-a[i]*c[i]]+a[i]*c[i]);
                }
            }
        }
        int count = ;//计数
        for(int i=;i<=m;i++){
            if(dp[i] == i){
                count++;//可以组合且不用找钱
            }
        }
        cout << count << endl;
   }
   return ;
}