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
 

一道简单的模板题，也就不多说什么了

#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)//如果总容量比这个物品的容量要小，那么这个物品可以直到取完，相当于完全背包
    {
        CompletePack(cost,wei);
        return ;
    }
    else//否则就将多重背包转化为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;
}

补充：软件开发 , C++ ,

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