HDU 2844：Coins

Coins

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16707    Accepted Submission(s): 6636

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

题意：

给定n种物品和一个m的总价格，又给出物品的面值和数量，求可以组成多少个不超过m的数。

解题思路：

首先我们可以判断他是一个多重背包问题。在金额的限制条件下求的最大的价值，如果说在多重背包求得的dp[i]=i.可以认为可以恰好组成这个不差过m的价值。

动态转移方程：dp[i]=max(dp[i],dp[i-w]+w);//i不差过m的金额，w表示面值，dp该金额下的最大价值。

源代码：

#include<cstdio>#include<cstring>const int N=105;//const int NINF=-0x3f3f3f3f;int max(int a, int b){return a>b?a:b;}int V,m;int dp[100005];int w[N],num[N];void MultiplePack(int n,int w){    if(n*w>=V)//完全背包    {        for(int i=w;i<=V;i++)            dp[i]=max(dp[i],dp[i-w]+w);        return ;    }    int k=1;    int nC=n;    while(k<nC)//01背包，做简单的优化    {        for(int i=V;i>=k*w;i--)            dp[i]=max(dp[i],dp[i-k*w]+k*w);        nC-=k;        k*=2;    }    for(int i=V;i>=nC*w;i--)        dp[i]=max(dp[i],dp[i-nC*w]+nC*w);}int main(){    while(scanf("%d%d",&m,&V),m!=0&&V!=0)    {        memset(dp,0,sizeof(dp));        for(int i=0;i<m;i++)            scanf("%d",&w[i]);        for(int i=0;i<m;i++)            scanf("%d",&num[i]);        for(int i=0;i<m;i++)//多重背包        {            MultiplePack(num[i],w[i]);        }        int cont=0;        for(int i=1;i<=V;i++)//判断符合的个数，即最后答案        {            if(dp[i]==i) cont++;        }        printf("%d\n",cont);    }    return 0;}