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.

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.

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.

4
0.1 0.2 0.3 0.8

0.800000000000

2
0.1 0.2

0.260000000000

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.

解題報(bào)告

好久不做cf，大晚上想來(lái)水水。。。

看來(lái)自己還是太嫩了。

Andrey想讓朋友出題，他知道每一個(gè)朋友出題的概率，他僅僅想要一道題，而且想要實(shí)現(xiàn)的概率最大。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
double num[1000];
int cmp(double a,double b)
{
    return a>b;
}
int main()
{
    int n;
    while(cin>>n)
    {
        double sum=0,cnt=0,maxx=0;
        for(int i=0; i<n; i  )
            cin>>num[i];
        sort(num,num n,cmp);
        for(int k=1; k<=n; k  )
        {
            sum=0;
            for(int i=0; i<k; i  )
            {
                cnt=1;
                for(int j=0; j<k; j  )
                {
                    if(i!=j)
                    {
                        cnt*=(double)(1-num[j]);
                    }
                    else cnt*=num[j];
                }
                sum =cnt;
            }
            maxx=max(sum,maxx);
        }
        printf("%.12lf\n",maxx);
    }
    return 0;
}