Coins（多重背包）

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：有n种货币和手表价格为m，单位货币的价值有val[i]，单位货币的数量有vol[i],。求解带哪些 货币val*货币数量vol 可将手表买回且身上的钱总价值最大

题解：注意时间上的优化，为避免超时，将多重背包问题分成完全背包和01背包问题，01背包再用二进制优化

#include<iostream> #include<algorithm>#include<string.h>#define INF 0x3f3f3f3fusing namespace std;//手表价格 int m; //单位货币的价值int val[1005];//单位货币的数量int vol[1005];int dp[100005];void Zero(int cost){for(int i=m;i>=cost;i--)dp[i]=max(dp[i],dp[i-cost]+cost);}//val*vol>>m，硬币可以无限使用，可以使用完全背包 void Com(int cost){for(int i=cost;i<=m;i++)dp[i]=max(dp[i],dp[i-cost]+cost);}//拆分vol,价格系数k*vol组成新硬币 void mul(int val,int vol){if(val*vol>=m)Com(val);else{int k=1;while(k<=vol){Zero(k*val);vol-=k;k*=2;}Zero(vol*val);}}int main(){//货币的种类 int n;while(cin>>n>>m,(n||m)){for(int i=1;i<=n;i++)cin>>val[i];for(int i=1;i<=n;i++)cin>>vol[i];memset(dp,0,sizeof(dp));for(int i=0;i<=m;i++)dp[i]=-INF;dp[0]=0;for(int i=1;i<=n;i++)mul(val[i],vol[i]);int ans=0;for(int i=1;i<=m;i++)if(dp[i]>0)ans++;cout<<ans<<endl;}return 0;}