Problem:
Score : 100 points

Problem Statement

Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins.
Find the probability of having more heads than tails.

Constraints

N is an odd number.

1 \leq N \leq 2999

p_i is a real number and has two decimal places.

0 < p_i < 1

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 \ldots p_N

Output

Print the probability of having more heads than tails.
The output is considered correct when the absolute error is not greater than 10^{-9}.

Sample Input 1

3
0.30 0.60 0.80

Sample Output 1

0.612

The probability of each case where we have more heads than tails is as follows:

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 × 0.6 × 0.8 = 0.144;

The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 × 0.6 × 0.8 = 0.336;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 × 0.4 × 0.8 = 0.096;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 × 0.6 × 0.2 = 0.036.

Thus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096 + 0.036 = 0.612.

Sample Input 2

1
0.50

Sample Output 2

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

Sample Input 3

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3

0.3821815872

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

/*
___mennat khoday ra azz o va jall ke taatash mojebe ghorbat ast o be shokr andarash mazide nemat.
___har nafasi ke foro miravad momedde hayatast o chon bar miayad mofarrahe zat.
___pas dar har nafasi 2 nemat mojod ast o bar har nematy shokry vajeb.

___az dast o zabane ke barayad ?
___kaz ohdeye shokrash be dar ayad !


___man be to dami gharar natvanam kard
___va ehsane to ra shomar natvanam kard
####################
___gar bar tane man zaban shavad har moyee
___yek shokre to az hezar natvanam kard

*/

#include <bits/stdc++.h>

#define pb push_back
#define SALI main
#define left(x) ((x)*2)
#define right(x) (((x)*2)+1)
#define fuck(x) cout << #x << " : " << x << "   ";
#define findLowVec(v,x) (lower_bound(v.begin(), v.end(), x) - v.begin())
#define findUpVec(v,x) (upper_bound(v.begin(), v.end(), x) - v.begin())
#define findLowArr(a,n,x) (lower_bound(a, a+n, x) - a)
#define findUpArr(a,n,x) (upper_bound(a, a+n, x) - a)
#define int long long

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef long double ld;

template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p)
{return out << "(" << p.first << ", " << p.second << ")";}

template<class A> ostream& operator <<(ostream& out, const vector<A> &v)
{out << "[";for(int i = 0; i < v.size(); i++) {if(i) out << ", ";out << v[i];}return out << "]";}

template<class A,class B> ostream& operator <<(ostream& out, const set<A,B> &s)
{out << "[";for(auto i = s.begin(); i != s.end(); i++) {out << ", ";out << *i;}return out << "]";}

template<class A> ostream& operator <<(ostream& out, const set<A> &s)
{out << "[";for(auto i = s.begin(); i != s.end(); i++) {out << ", ";out << *i;}return out << "]";}

const int mod = 1e9 + 7;

int32_t SALI()
{
    int n; cin >> n;
    ld p[n+1]; for(int i = 1; i <= n; i++) cin >> p[i];

    ld dp[n+1][n+1]; // dp[i][j] = i taye avval daghighan j tashon head biad head

    for(int i = 0; i <= n; i++) for(int j = 0; j <= n; j++) dp[i][j] = 0.00000000000;

    dp[1][0] = 1 - p[1];
    dp[1][1] = p[1];

    for(int i = 2; i <= n; i++){
        dp[i][0] = dp[i-1][0] * (1-p[i]);
        for(int j = 1; j <= i; j++){
            dp[i][j] = dp[i-1][j] * (1-p[i]) + dp[i-1][j-1] * p[i];
        }
    }
    ld ans = 0.0000000;
    for(int i = (n/2) + 1; i <= n; i++) ans += dp[n][i];
    cout << fixed << setprecision(10) << ans;
}


/**< WRITEN BY ALI ADELKHAH */