hdu2844 Coins（多重背包模板题）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=2844

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16894    Accepted Submission(s): 6703

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

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

AC code：

#include<iostream>#include<stdio.h>#include<algorithm>#include<math.h>#include<string.h>#include<map>#include<set>#include<vector>#define LL long long#define INF 0xfffffff#define PI acos(-1)#define EPS 1e-6using namespace std;int dp[100010],wei[111],val[111],cnt[111];int n,m;void zeroone(int w,int v){int i;for(i=m;i>=w;i--){dp[i]=max(dp[i],dp[i-w]+v);}}void complete(int w,int v){int i;for(i=w;i<=m;i++){dp[i]=max(dp[i],dp[i-w]+v);}}void multi(int w,int v,int num){if(num*w>=m){complete(w,v);}else{int k=1;while(k<=num){zeroone(k*w,k*v);num-=k;k*=2;}zeroone(num*w,num*v);}}int main(){//freopen("in.txt","r",stdin);int i,ans;while(scanf("%d%d",&n,&m)!=EOF){if(n==0&&m==0)break;for(i=0;i<=m;i++){dp[i]=-INF;}dp[0]=0;for(i=1;i<=n;i++){scanf("%d",&wei[i]);val[i]=wei[i];}for(i=1;i<=n;i++){scanf("%d",&cnt[i]);}for(i=1;i<=n;i++){multi(wei[i],val[i],cnt[i]);}ans=0;for(i=1;i<=m;i++){if(dp[i]>0)ans++;}printf("%d\n",ans);}}