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>
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
  int N;
  cin >> N;

  vector<double> p(N + 1, 0.0);
  for (int i = 1; i <= N; i++) {
    cin >> p.at(i);
  }

  vector<vector<double>> dp(N + 1, vector<double>(N + 1, 0.0));
  dp.at(1).at(0) = 1 - p.at(1);
  dp.at(1).at(1) = p.at(1);

  for (int i = 2; i <= N; i++) {
    dp.at(i).at(0) = dp.at(i - 1).at(0) * (1 - p.at(i));
    for (int j = 1; j <= N; j++) {
      dp.at(i).at(j) = dp.at(i - 1).at(j - 1) * p.at(i) + dp.at(i - 1).at(j) * (1 - p.at(i));
    }
  }

  double ans = 0.0;
  for (int i = N/2 + 1; i <= N; i++) {
    ans += dp.at(N).at(i);
  }

  cout << fixed << setprecision(10) << ans << endl;

  return 0;
}