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: Accepted)

#include <bits/stdc++.h>
#define debug(x) cerr << #x << " = " << x << ", ";
#define ll long long
#define pb push_back
using namespace std;
#define FAST_IO                       \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
const int MOD = 2 * (1e9 + 7);
long double dp[3003][3003] = {};
int main()
{
    FAST_IO
    int n;
    cin >> n;
    vector<long double> a(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];

    dp[0][0] = 1;
    // dp[0][1] = 1;
    // dp[0][2] = 1;
    // dp[0][3] = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= i; j++)
        {
            dp[i][j] = dp[i - 1][j] * (1.00000 - a[i]);
            if (j > 0)
            dp[i][j] += dp[i - 1][j - 1] * a[i];
        }
    }
    // for (int i = 0; i <= n; i++)
    // {
    //     for (int j = 0; j <= n; j++)
    //     {
    //         // if (j > n - j)
    //         cout << fixed << setprecision(5) << dp[i][j] << " ";
    //     }
    //     cout << endl;
    // }
    long double ans = 0;
    for (int i = 0; i <= n; i++)
    {
        if (i > n - i)
        {
            ans += dp[n][i];
        }
    }
    cout << fixed << setprecision(15) << ans << '\n';
}