zoukankan      html  css  js  c++  java
  • [ACM] hdu 2844 Coins (多重背包)

    Coins

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 6347    Accepted Submission(s): 2589


    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
     


     

    Source

    解题思路:

    给出一些价值的硬币,每种硬币有一定的数量,给出一个数m,问用这些硬币可以组合成价值1到m中的几个数。

    代码:

    #include <iostream>
    #include <algorithm>
    #include <string.h>
    using namespace std;
    const int maxn=100005;//有多少种物品
    const int inf=0x7fffffff;
    int c[maxn];//每种的价值
    int num[maxn];//每种的数量
    int f[maxn];
    int v;//最大容量
    
    void ZeroOnePack(int cost,int value)
    {
        int i;
        for(i=v;i>=cost;i--)
            f[i]=max(f[i],f[i-cost]+value);
    }
    
    void CompletePack(int cost ,int value)
    {
        int i;
        for(i=cost;i<=v;i++)
            f[i]=max(f[i],f[i-cost]+value);
    }
    
    void MultiPack(int cost ,int value,int amount)
    {
        if(v<=cost*amount)
        {
            CompletePack(cost,value);
            return;
        }
        else
        {
            int k=1;
            while(k<amount)
            {
                ZeroOnePack(k*cost,k*value);
                amount-=k;
                k*=2;
            }
            ZeroOnePack(amount*cost,amount*value);
        }
    }
    int main()
    {
        int n;
        while(cin>>n>>v&&(n||v))
        {
            for(int i=0;i<n;i++) cin>>c[i];
            for(int i=0;i<n;i++) cin>>num[i];
            for(int i=0;i<=v;i++)
                f[i]=-1*inf;
            f[0]=0;
            for(int i=0;i<n;i++)
                MultiPack(c[i],c[i],num[i]);
            int sum=0;
            for(int i=1;i<=v;i++)
            {
                if(f[i]>0)
                    sum++;
            }
            cout<<sum<<endl;
        }
        return 0;
    }
    


     

  • 相关阅读:
    CentOS5.6下SVN的安装
    在servlet中的init方法中使用getInitParameter方法空指针错误
    Linux iostat监测IO状态【转】
    自己实现一个list比较器 实现Comparator()接口
    一些常用的随机实现
    java里null强转为某个类会报错吗?
    java游戏服务器简单工厂模式
    起个头!准备写一个 设计模式系列
    HashMap根据value值排序
    java游戏服务器 策略+简单工厂
  • 原文地址:https://www.cnblogs.com/vivider/p/3697506.html
Copyright © 2011-2022 走看看