B. Urbanization
题目链接
http://codeforces.com/contest/735/problem/B
题面
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are n people who plan to move to the cities. The wealth of the i of them is equal to ai. Authorities plan to build two cities, first for n1 people and second for n2 people. Of course, each of n candidates can settle in only one of the cities. Thus, first some subset of candidates of size n1 settle in the first city and then some subset of size n2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth ai among all its residents divided by the number of them (n1 or n2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
输入
The first line of the input contains three integers n, n1 and n2 (1 ≤ n, n1, n2 ≤ 100 000, n1 + n2 ≤ n) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000), the i-th of them is equal to the wealth of the i-th candidate.
输出
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .
样例输入
2 1 1
1 5
样例输出
6.00000000
题意
一共有n个数，第i个数是a[i]，现在你需要选出n1个数和n2个数，使得那n1个数的和除以n1加上n2个数的和除以n2的值最大。
题解
贪心，如果n1>n2，那么交换
然后选择最大的n1个数为n1集合，然后次大的n2个数为n2集合。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int n,n1,n2,nn1,nn2;
double a[maxn];
int main()
{
scanf("%d%d%d",&n,&n1,&n2);
if(n1>n2)swap(n1,n2);
nn1=n1,nn2=n2;
for(int i=0;i<n;i++)
scanf("%lf",&a[i]);
sort(a,a+n);
reverse(a,a+n);
double sum1=0,sum2=0;
for(int i=0;i<n;i++){
if(n1){
sum1+=a[i];
n1--;
}else if(n2){
sum2+=a[i];
n2--;
}
}
double ans = (sum1/nn1)+(sum2/nn2);
printf("%.12f
",ans);
}