codeforces 735 B Urbanization
題目描述:
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 n
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 a
Please, help authorities find the optimal way to pick residents for two cities.
Input
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.
Output
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 .
Example
Input
2 1 1
1 5
Output
6.00000000
Input
4 2 1
1 4 2 3
Output
6.50000000
Note
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (a3 + a4) / 2 + a2 = (3 + 2) / 2 + 4 = 6.5
題目分析:排2次序就ok啦
AC程式碼:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int n,n1,n2;
int v[200005];
int main()
{
while(~scanf("%d%d%d",&n,&n1,&n2))
{
if(n1>n2)
{
int t=n1;
n1=n2;
n2=t;
}
for(int i=0;i<n;i++)
scanf("%d",&v[i]);
sort(v,v+n);
double sum1=0,sum2=0;
double p;
for(int i=n-1;i>=n-n1;i--)
sum1+=v[i];
for(int i=n-n1-1;i>=n-n1-n2;i--)
sum2+=v[i];
p=sum1/n1+sum2/n2;
//cout<<p<<endl;
printf("%lf\n",p);
}
return 0;
}