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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize ("O2")
#include <bits/stdc++.h>

#define int long long
#define double long double 

using namespace std;

const int INF = 1e+18;
const int MAXN = 100000;
const int LOGN = 20;
const int mod = 1e9+7;

signed main() {
    ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int n;
    cin >> n;
    double a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    double dp[n][n + 1];

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

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

    for (int i = 1; i < n; i++) {
        dp[i][0] = dp[i - 1][0]*(1 - a[i]);
        for (int j = 1; j <= n; j++) {
            dp[i][j] += dp[i - 1][j] * (1 - a[i]);
            dp[i][j] += dp[i - 1][j - 1] * a[i];
        }
    }

    double ans = 0.0;
    for (int i = (n + 1) / 2; i <= n; i++) {
        ans += dp[n - 1][i];
    }

    cout << ans;
}