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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
int main(){
    int n;
    scanf("%d", &n);
    double te;
    double dp[2][3000];
    memset(dp[0], 0, sizeof(dp[0])); 
    memset(dp[1], 0, sizeof(dp[0])); 
    dp[0][0] = 1;
    for (int i = 1; i <= n; ++i){
        scanf("%lf", &te);
        dp[i&1][0] = dp[i-1&1][0]*(1-te);
        for (int j = 1; j <= i; ++j){
            dp[i&1][j] = dp[i-1&1][j-1]*te + dp[i-1&1][j]*(1-te);
        }
    }
    te = 0;
    for (int i = (n>>1)+1; i <= n; ++i)
        te += dp[n&1][i];
    printf("%0.10lf", te);
    return 0;
}