zoukankan      html  css  js  c++  java
  • hdu 2844 Coins 多重背包问题

    Coins

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
    Total Submission(s) : 5   Accepted Submission(s) : 3

    Font: Times New Roman | Verdana | Georgia

    Font Size:  

    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

    对于多重背包问题可以这样解:
     
    MultiplePack(cost,weight,amount)
    {
        if (cost*weight>=V) CompletePack(cost,weight)
        else
        {
            k=1;
            while(k<amount)
            {
                Zero_OnePack(k*cost,k*weight);
                amount-=k;
                k*=2;
            }
            Zero_OnePack(amount*cost,amount*weight)
        }
    }
    

      

    这道题目的cost 和weight 是同一种量.

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    using namespace std;
    
    int a[105],c[105];
    int dp[100005];
    
    void CompletePack(int cost ,int weight,int n)
    {
    	int i;
    	for(i=cost;i<=n;i++)						//从cost 到n
    		dp[i]=max(dp[i],dp[i-cost]+weight);
    	
    }
    
    void Zero_OnePack(int cost,int weight,int n)
    {
    	int i;
    	for(i=n;i>=cost;i--)					//从n到cost
    		dp[i]=max(dp[i],dp[i-cost]+weight);
    }
    
    int main()
    {
    	int n,m;
    	int i,j,k;
    	while(scanf("%d%d",&n,&m) && n!=0 && m!=0)
    	{
    		for(i=0;i<n;i++)
    			scanf("%d",&a[i]);
    		for(i=0;i<n;i++)
    			scanf("%d",&c[i]);
    		memset(dp,0,sizeof(dp));
    		for(i=0;i<n;i++)
    		{
    			if(a[i]*c[i]>=m) CompletePack(a[i],a[i],m);
    			else
    			{
    				k=1;
    				while(k<c[i])
    				{
    					Zero_OnePack(k*a[i],k*a[i],m);
    					c[i]-=k;
    					k*=2;
    				}
    				Zero_OnePack(c[i]*a[i],c[i]*a[i],m);
    			}
    		}
    		int ans=0;
    		for(i=1;i<=m;i++)
    		{
    			if(dp[i]==i) ans++;
    		}
    		cout<<ans<<endl;
    	}
    	return 0;
    }
    

      

  • 相关阅读:
    Jmeter关联-获取token值
    Jmeter接口测试
    Jmeter关联,正则表达式提取器使用2
    Jmeter关联,正则表达式提取器使用1
    Jmeter-CSV Data Set Config参数化
    Fiddler基本用法:手机抓包1
    接口测试该怎样入手?
    Fiddler简介与Web抓包,远程抓包(IE、360、谷歌、火狐)
    PHP 插件资源
    利用PHP递归 获取所有的上级栏目
  • 原文地址:https://www.cnblogs.com/shenshuyang/p/2644277.html
Copyright © 2011-2022 走看看