Problem

E - Problem

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit  Status

Description

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 numbersp_i(0.0 ≤ p_i ≤ 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 Input

Input

4
0.1 0.2 0.3 0.8

Output

0.800000000000

Input

2
0.1 0.2

Output

0.260000000000

Hint

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.

题意：n个人，每个人提出一个问题的概率为p，Andrey 可以选择请求一个人，两个人，，，，n个人，来得到一个问题，最后要求得到一个问题的最大概率

将n个事件成功的概率从大到小排序，

然后，

只选择前1件事去做成功的概率就是“从中选择1件事情去做，最大的实现概率”

只选择前2件事去做成功的概率就是“从中选择2件事情去做，最大的实现概率”

以此类推。

证明： 选两种的情况：设a为其中一个的概率，b为另一个的概率，则得到一个问题的概率为a*(1-b)+b*(1-a)=a+b-2*a*b;因为a,b是小于11的，a*b的增长，远远小于a+b的增长，同理，三种，四种情况也是这样证明的，所以选择的时候，先从概率从大到小排序，选几个就选前几个

AC代码如下：

#include<stdio.h>
#include<stdlib.h>
int cmp(const void *a,const void *b)
{
    return *(double*)a>*(double*)b?-1:1;
}
int main()
{
    int n,i,j,k;
    double f[110],m,m1;
    while(scanf("%d",&n)!=EOF)
    {
        double sum[110]={0.0};
        for(i=0;i<n;i++)
        {
            scanf("%lf",&f[i]);
        }
        qsort(f,n,sizeof(f[0]),cmp);
        sum[0]=f[0];
        for(i=1;i<n;i++)//n-1种选法
        {
m1=0.0;
            for(j=0;j<i+1;j++)//选i+1个
            {
                m=f[j];
                for(k=0;k<i+1;k++)
                {
if(k!=j)
                    m*=(1-f[k]);
                }
m1+=m;
            }
            sum[i]+=m1;
        }
        qsort(sum,n,sizeof(sum[0]),cmp);
        printf("%.12lf\n",sum[0]);
    }
    return 0;
}