HDU 5999—The Third Cup is Free

Panda and his friends were hiking in the forest. They came across a coffee bar inside a giant tree trunk. 
Panda decided to treat everyone a cup of coffee and have some rest. Mr. Buck, the bartender greeted Panda and his animal friends with his antler. He proudly told them that his coffee is the best in the forest and this bar is a Michelin-starred bar, thats why the bar is called Starred Bucks. 
There was a campaign running at the coffee bar: for every 3 cups of coffee, the cheapest one is FREE. After asking all his friends for their flavors, Panda wondered how much he need to pay. 

Input

The first line of the input gives the number of test cases, T. 
T test cases follow. Each test case consists of two lines. The first line contains one integer N, the number of cups to be bought. 
The second line contains N integers p1,p2,....,pN representing the prices of each cup of coffee. 

Output

For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the least amount of money Panda need to pay. 

limits

∙∙ 1 ≤ T ≤ 100. 
∙∙ 1 ≤ N ≤ 105105. 
∙∙ 1 ≤ pi ≤ 1000. 

Sample Input

231 2 3510 20 30 20 20

Sample Output

Case #1: 5Case #2: 80

题目大意：一个熊猫打算请他很多的朋友在咖啡馆里喝咖啡，咖啡馆有个优惠：买两杯咖啡，第三杯会免费。已知人数和每个所选的咖啡的价格，求熊猫最少只需花多少钱。

分析：一道明显的贪心题目，若要花钱最少，则只需免费的那些咖啡原价越贵越好，因此将所有咖啡的价格装进一个数组后从大到小排序，忽略第3及3的倍数杯咖啡的价格，然后统计总价即可。

AC代码：

#include <cstdio>#include <string>#include <cmath>#include <algorithm>#include <iostream>#include <vector>#include <queue>using namespace std;int p[100005];bool compare(int x,int y){    return x>y;}int main(){#ifdef LOCALfreopen("data.in","r",stdin);freopen("data.out","w",stdout);#endif    int t;    int sum;    int kase=0;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%d",&p[i]);        sort(p,p+n,compare);        sum=0;        for(int i=0;i<n;i++)        {            if((i+1)%3==0)                p[i]=0;            sum+=p[i];        }        printf("Case #%d: %d\n",++kase,sum);    }return 0;}