HDU2844:Coins(多重背包)

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 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output
8
4
 

一道簡單的模板題，也就不多說什麼了

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

const int MAX=100000;
int dp[MAX];
int c[MAX],w[MAX];
int v;

void ZeroOnePack(int cost,int wei)//01
{
    int i;
    for(i = v;i>=cost;i--)
    {
        dp[i] = max(dp[i],dp[i-cost]+wei);
    }
}

void CompletePack(int cost,int wei)//完全
{
    int i;
    for(i = cost;i<=v;i++)
    {
        dp[i] = max(dp[i],dp[i-cost]+wei);
    }
}

void MultiplePack(int cost,int wei,int cnt)//多重
{
    if(v<=cnt*cost)//如果總容量比這個物品的容量要小，那麼這個物品可以直到取完，相當於完全背包
    {
        CompletePack(cost,wei);
        return ;
    }
    else//否則就將多重背包轉化為01背包
    {
        int k = 1;
        while(k<=cnt)
        {
            ZeroOnePack(k*cost,k*wei);
            cnt = cnt-k;
            k = 2*k;
        }
        ZeroOnePack(cnt*cost,cnt*wei);
    }
}

int main()
{
    int n;
    while(~scanf("%d%d",&n,&v),n+v)
    {
        int i;
        for(i = 0;i<n;i++)
        scanf("%d",&c[i]);
        for(i = 0;i<n;i++)
        scanf("%d",&w[i]);
        memset(dp,0,sizeof(dp));
        for(i = 0;i<n;i++)
        {
            MultiplePack(c[i],c[i],w[i]);
        }
        int sum = 0;
        for(i = 1;i<=v;i++)
        {
            if(dp[i]==i)
            {
                sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}