cf 442B Andrey and Problem

B. Andrey and Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.

Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi(0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.

Output

Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.

Sample test(s)

input

4
0.1 0.2 0.3 0.8

output

0.800000000000

input

2
0.1 0.2

output

0.260000000000

Note

In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.

In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.

 

可以选1个,可以选2个,选i个....

但是选哪i个,似乎还要枚举...这样计算起来很麻烦

但是再一看,很容易发现,选i个概率最大的情况一定是选本身概率最大的i个的情况

那么我们只需要枚举选的个数,取每种情况的最大值就好了.

 1 /*************************************************************************
 2     > File Name: code/2015summer/#3/C.cpp
 3     > Author: 111qqz
 4     > Email: rkz2013@126.com 
 5     > Created Time: 2015年07月28日 星期二 12时27分50秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<iomanip>
10 #include<cstdio>
11 #include<algorithm>
12 #include<cmath>
13 #include<cstring>
14 #include<string>
15 #include<map>
16 #include<set>
17 #include<queue>
18 #include<vector>
19 #include<stack>
20 #define y0 abc111qqz
21 #define y1 hust111qqz
22 #define yn hez111qqz
23 #define j1 cute111qqz
24 #define tm crazy111qqz
25 #define lr dying111qqz
26 using namespace std;
27 #define REP(i, n) for (int i=0;i<int(n);++i)  
28 typedef long long LL;
29 typedef unsigned long long ULL;
30 const int N=1E2+7;
31 double p[N];
32 int n;
33 
34 bool cmp(double a,double b)
35 {
36     if (a>b) return true;
37     return false;
38 }
39 double solve (int x)
40 {
41     double res=0;
42     double poss;
43     for  ( int i = 1 ; i <= x; i++)
44     {
45     poss = 1.0;
46     for ( int j = 1 ; j <= x;  j++ )
47     {
48         if (j==i) poss=poss*p[j-1];
49         else poss = poss * (1-p[j-1]);
50     }
51     res = res + poss;
52     }
53     return res;
54 }
55 int main()
56 {
57     cin>>n;
58     for ( int i = 0 ; i < n ; i++ )
59     {
60     cin>>p[i];
61     }
62     sort(p,p+n,cmp);
63     double ans = 0;
64     for ( int i = 0 ; i < n ; i++ )
65     {
66     ans = max(ans,solve(i+1));
67     }
68     cout<<fixed<<setprecision(12)<<ans<<endl;
69   
70     return 0;
71 }

 

转载于:https://www.cnblogs.com/111qqz/p/4684853.html