hdu-2844-Coins (多重背包+二进制优化)

Coins

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 9 Accepted Submission(s) : 5
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：tony 想要买一个手表 他后n个硬币面值为 v[i],每种硬币的数量为 num[i],要买一个价值不超过 m的东西 （价钱小于等于m即可），
输入：一个 n 和一个 m 以 0 0结束；第二行是 2n个数 分别是 v[i]和num[i].
输出：1-m之间有多少种价格tony 可以支付。

思路： 多重背包，在dp初始赋值是 负成 负数 作为标记即可。

code

#include<cstdio>#include<iostream>#include<cstring>#define inf 0x3f3f3f3fusing namespace std;int dp[101000],v[110],num[110];int n,m;//m背包的总容量、v物品的体积、w物品的价值void OneZeroPack(int m,int v,int w)  //0-1背包{    for(int i=m;i>=v;i--)        dp[i]=max(dp[i],dp[i-v]+w);}//m背包的总容量、v物品的体积、w物品的价值void CompletePack(int m,int v,int w)  //完全背包{    for(int i=v;i<=m;i++)        dp[i]=max(dp[i],dp[i-v]+w);}//m背包的总容量、v物品的体积、w物品的价值、num物品的数量void MultiplePack(int m,int v,int w,int num)//多重背包{    if(v*num>=m)    {        CompletePack(m,v,w);        return ;    }    int k=1;    for(k=1;k<=num;k<<=1)    {        OneZeroPack(m,k*v,k*w);        num=num-k;    }    if(num)        OneZeroPack(m,num*v,num*w);}int main(){    while(scanf("%d%d",&n,&m),n||m){        for(int i=0;i<n;i++)            scanf("%d",&v[i]);        for(int i=0;i<n;i++)            scanf("%d",&num[i]);        for(int i=0;i<=m;i++)            dp[i]=-inf;        dp[0]=0;        for(int i=0;i<n;i++){            MultiplePack(m,v[i],v[i],num[i]);        }        int ans=0;        for(int i=1;i<=m;i++)            if(dp[i]>0) ans++;        printf("%d\n",ans);    }    return 0;}