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 JavaScript (node.js v5.12) (judge verdict: Time Limit Exceeded)

'use strict'

function solve(N, P) {
  var permutation = [['H', 1, P[0]], ['T', -1, 1 - P[0]]]
  var pos = [['H', 1], ['T', -1]]
  var res = 0
  for (var i = 1; i < N; i++) {
    var newPermutation = []
    for (var k = 0; k < pos.length; k++) {
      for (var j = 0; j < permutation.length; j++) {
        var p = []
        p[0] = permutation[j][0] + pos[k][0]
        p[1] = permutation[j][1] + pos[k][1]
        p[2] = pos[k][0] === 'H' ? permutation[j][2]*P[i] : permutation[j][2]*(1-P[i])
        if (p[1] < 0 && Math.abs(p[1]) >= (N -1 - i)) {
          continue
        }
        newPermutation.push(p)
      }
    }
    permutation = newPermutation
  }
  permutation = permutation.filter(p => p[1] > 0)
  for (var i = 0; i < permutation.length; i++) {
    res = res + permutation[i][2]
  }
  return res
}

function main(input) {
  var lines = input.split("\n")
  var N = lines[0]
  var P = lines[1].split(' ').map(l => parseFloat(l))
  var ans = solve(N, P)
  console.log(ans)
}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));