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)

// /*
// power code taken from geeks for geeks
// */
// cout << setprecision (2) << fixed << 1.2;

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;

const ll INF = 1e18 + 5;
const ll MAX_N = 1e5 + 5;
const ll MODD = 1e9 + 7;

ll max(ll a,ll b){return a>b?a:b;}
ll min(ll a,ll b){return a<b?a:b;}
double max(double a,double b){return a>b?a:b;}
double min(double a,double b){return a<b?a:b;}

ll modd(ll a){if(a<0LL) return a*-1LL;return a;}
double modd(double a){if(a<0.0) return a*-1.0;return a;}

ll modulo(ll a){return a % MODD; }

ll power(ll k, ll p){
    if (p == 0)
        return 1LL;
 
    ll P = power(k, p/2LL);
    P = (P * P) % MODD;
 
    if (p % 2 == 1)
        P = (P * k) % MODD;
 
    return P;
}


struct p_str{
    ll a, b;
};

void solve(){
    ll n;
    cin >> n;
    ld p[n + 1];
    for(ll i = 1; i <= n; i++) {
        cin >> p[i];
    }

    ld dp[n+1][n+1]; // dp[j][i] => prob. of 1st i-coins having j-heads...
    for(ll i = 0; i <= n; i++) {
        for(ll j = 0; j <= n; j++) {
            dp[i][j] = 0.0;
        }
    }

    dp[0][1] = 1.0 - p[1];
    dp[1][1] = p[1];
    for(ll i = 2; i <= n; i++) {
        ld h = p[i] * 1.0;
        ld t = 1.0 - p[i] * 1.0;
        for(ll j = 0; j <= i; j++) {
            dp[j][i] = dp[j][i - 1] * t;
            if(j > 0) {
                dp[j][i] += dp[j - 1][i - 1] * h;
            }
        }
    }

    // cout << "dp \n";
    // for(ll i = 0; i <= n; i++) {
    //     for(ll j = 1; j <= n; j++) {
    //         cout << dp[i][j] << " ";
    //     }
    //     cout << endl;
    // }
    // cout << endl;

    ld ans = 0.0;
    for(ll i = (n / 2LL) + 1; i <= n; i++) {
        ans += dp[i][n] * 1.0;
    }

    cout << setprecision (9) << fixed << ans << endl;
}



int main(){


    #ifndef ONLINE_JUDGE 

    // For getting input from input.txt file 
    freopen("input.txt", "r", stdin); 

    // Printing the Output to output.txt file 
    freopen("output.txt", "w", stdout); 
    #endif 


    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    



    ll t=1;


    // cin>>t;
    


    while(t--){
        solve();
    }
    return 0;
}