AtCoder：AtCoder Group Contest（思维）

A - AtCoder Group Contest

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

There are 3N participants in AtCoder Group Contest. The strength of the i-th participant is represented by an integer ai. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.

The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.

Find the maximum possible sum of the strengths of N teams.

Constraints

• 1≤N≤105
• 1≤ai≤109
• ai are integers.

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Input

Input is given from Standard Input in the following format:

N
a1 a2 … a3N

Output

Print the answer.

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Sample Input 1

Copy

2
5 2 8 5 1 5

Sample Output 1

Copy

10

The following is one formation of teams that maximizes the sum of the strengths of teams:

• Team 1: consists of the first, fourth and fifth participants.
• Team 2: consists of the second, third and sixth participants.

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Sample Input 2

Copy

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 2

Copy

10000000000

The sum of the strengths can be quite large.

题意：给出3*n个数，要求将他们分成n组，每组对答案贡献值为它们的中位数，问怎样分配使得中位数的和最大，输出最大值。

思路：先对数由小到大排序，显然最优解选择的中位数是3n-1，3n-3，3n-5......加起来就行。

# include <iostream>
# include <cstdio>
# include <algorithm>
using namespace std;

int a[300003];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<3*n; ++i)
        scanf("%d",&a[i]);
    sort(a, a+3*n, greater<int>());
    long long sum = 0;
    for(int i=1, j=0; j<n; i+=2)
        sum += a[i], ++j;
    printf("%lld\n",sum);
    return 0;
}