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 Go (1.6) (judge verdict: Accepted)

package main

import (
  "fmt"
)

func main() {
  var N int
  fmt.Scanf("%d\n", &N)
  p := make([]float64, N)
  for i:=0;i<N;i++ {
    fmt.Scanf("%f", &p[i])
  }
  fmt.Println(solve(N, p))
}

func solve(n int, p []float64) float64 {
  dp := make([][]float64, n+1)
  for i:=range dp {
    dp[i] = make([]float64, n+1)
  }
  dp[0][0] = 1
  //
  for i:=1;i<=n;i++ {
    for j:=0;j<=i;j++ {
      if j == 0 {
        dp[i][j] = dp[i-1][j]*(1-p[i-1])
        continue
      }
      dp[i][j] = dp[i-1][j]*(1-p[i-1])+dp[i-1][j-1]*p[i-1]
    }
  }
  //
  var ret float64
  for j:=n/2+1;j<=n;j++ {
    ret += dp[n][j]
  }
  return ret
}