南邮 OJ 1529 Shopaholic

Shopaholic

时间限制(普通/Java) :  1000 MS/ 3000 MS          运行内存限制 : 65536 KByte
总提交 : 28            测试通过 : 19 

比赛描述

Lindsay is a shopaholic. Whenever there isa 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.

输入

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.

输出

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.

样例输入

1
6
400 100 200 350 300 250

样例输出

400

提示

undefined

题目来源

Nordic Collegiate Programming Contest 2007

#include<stdio.h>
#include<algorithm>
int a[20000];
int main(){
	int t,n,i,m,sum;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		m = n%3;
		sum = 0;
		for(i=0; i<n; i++){
			scanf("%d",a+i);
		}
		std::sort(a,a+n);
		for(i=0;i<n;i++){
			if(i%3==m){
				sum += a[i];
			}
		}
		printf("%d\n",sum);
	}
}