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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

#define MAXN 2999

double S[MAXN+1][MAXN+1];
double A[MAXN+1];

double Coins(int n)
{
	int i,j;
	double sum = 0;


	S[0][0] = 1;
	for ( i = 1; i <= n; i++ ) S[i][0] = S[i-1][0]*(1-A[i]);
	for ( i = 1; i <= n; i++ ) S[i][i] = S[i-1][i-1]*A[i];

	for ( i = 1; i <= n; i++ ) {
		for ( j = 1; j < i; j++ ) {
			S[i][j] = S[i-1][j-1]*A[i]+S[i-1][j]*(1-A[i]);
		}
	}

	for ( i = n/2+1; i <= n; i++ ) sum += S[n][i];

	return sum;
}

int main(void)
{
	int i, n;

	scanf("%d", &n);
	for ( i = 1; i <= n; i++ ) scanf("%lf", &A[i]);


	printf("%.10f\n", Coins(n));

	return 0;
}