hdu2844

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8398    Accepted Submission(s): 3410

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

3 101 2 4 2 1 12 51 4 2 10 0

 

Sample Output

84

 

/*
题意：这是一题经典的多重背包问题，直接根据背包九讲里面的公式就可以得出，在main函数中用dp[i]==i
来判别第i个是否能由给出的这些硬币得出；
  */

#include<iostream>#include <cstdio>#include <cstring>#define N 100010using namespace std;int dp[N];int a[N],c[N];//a代表价值，c代表个数int V;//背包总容量int max(int a,int b){return a>b?a:b;}void ComeletePack(int value,int weight){int i;for(i=value;i<=V;i++)dp[i]=max(dp[i],dp[i-value]+weight);}void ZeroOnePack(int cost,int weight){int j;for(j=V;j>=cost;j--){dp[j]=max(dp[j],dp[j-cost]+weight);}}void MultiplePack(int value,int weight,int amount){int i;if(weight*amount>=V){ComeletePack(value,weight);return ;}else{int k=1;while(k<=amount){ZeroOnePack(k*value,k*weight);amount=amount-k;k*=2;}ZeroOnePack(amount*value,amount*weight);}}int main(){int i,n;while(scanf("%d%d",&n,&V)!=EOF&&V+n){memset(a,0,sizeof(a));memset(c,0,sizeof(c));memset(dp,0,sizeof(dp));for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<n;i++)scanf("%d",&c[i]);for(i=0;i<n;i++){MultiplePack(a[i],a[i],c[i]);}int sum=0;for(i=1;i<=V;i++){if(dp[i]==i)sum++;}printf("%d\n",sum);}        return 0;}