zoukankan      html  css  js  c++  java
  • P

    P - Shopaholic
    Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

    Description

    Lindsay is a shopaholic. Whenever there is a discount of the kind where you can buy three items and only pay for two, she goes completely mad and feels a need to buy all items in the store. You have given up on curing her for this disease, but try to limit its effect on her wallet.

    You have realized that the stores coming with these offers are quite selective when it comes to which items you get for free; it is always the cheapest ones. As an example, when your friend comes to the counter with seven items, costing 400, 350, 300, 250, 200, 150, and 100 dollars, she will have to pay 1500 dollars. In this case she got a discount of 250 dollars. You realize that if she goes to the counter three times, she might get a bigger discount. E.g. if she goes with the items that costs 400, 300 and 250, she will get a discount of 250 the first round. The next round she brings the item that costs 150 giving no extra discount, but the third round she takes the last items that costs 350, 200 and 100 giving a discount of an additional 100 dollars, adding up to a total discount of 350.

    Your job is to find the maximum discount Lindsay can get.

    Input

    The first line of input gives the number of test scenarios, 1 <= t <= 20. Each scenario consists of two lines of input. The first gives the number of items Lindsay is buying, 1 <= n <= 20000. The next line gives the prices of these items, 1 <= pi <= 20000.

    Output

    For each scenario, output one line giving the maximum discount Lindsay can get by selectively choosing which items she brings to the counter at the same time.

    Sample Input

    1
    6
    400 100 200 350 300 250

    Sample Output

    400

    my answer:

    #include<iostream>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    bool cmp(int a,int b)
    {
    	return a>b;
    }
    int main()
    {
    	int T,n;
    	cin>>T;
    	while(T--)
    	{
    		cin>>n;
    		int a[20002];
    		for(int i=0;i!=n;i++){
    			cin>>a[i];
    		}
    		sort(a,a+n,cmp);
    		if(n>=3){
    			int i=2,sum=0;
    			for(;i<n;i+=3){
    				sum+=a[i];
    			}
    			cout<<sum<<endl;
    		}
    		else
    		cout<<0<<endl;	 
    	}
    	return 0;
    }


  • 相关阅读:
    平面几何-9 (海边直播目标2017全国初中数学竞赛班第14周作业题)
    数学奥林匹克问题解答:平面几何-8
    数学奥林匹克问题解答:平面几何-7
    lazyload懒加载插件
    Vue的生命周期
    用Vue来实现音乐播放器(九):歌单数据接口分析
    axios的详细用法以及后端接口代理
    用Vue来实现音乐播放器(八):自动轮播图啊
    Vue实现音乐播放器(七):轮播图组件(二)
    Vue实现音乐播放器(六):jsonp的应用+抓取轮播图数据
  • 原文地址:https://www.cnblogs.com/NYNU-ACM/p/4248807.html
Copyright © 2011-2022 走看看