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>
#include <stdlib.h>

int main(){
	int N, i, j;
	scanf("%d", &N);
	double *p = (double *)malloc(sizeof(double) * (N + 1));
	for(i = 1; i <= N; i++){
		scanf("%lf", &p[i]);
	}
	double **dp = (double **)malloc(sizeof(double *) * (N + 1));
	for(i = 0; i <= N; i++){
		dp[i] = (double *)malloc(sizeof(double) * (N + 1));
		for(j = 0; j <= N; j++){
			dp[i][j] = 0;
		}
	}
	dp[0][0] = 1;
	for(i = 1; i <= N; i++){
		for(j = 0; j <= N; j++){
			if(j == 0){
				dp[i][j] = (1 - p[i]) * dp[i - 1][j];
			}
			else{
				dp[i][j] = p[i] * dp[i - 1][j - 1] + (1 - p[i]) * dp[i - 1][j];
			}
		}
	}
	double ans = 0;
	for(i = 0; i <= N; i++){
		if(i > N - i){
			ans += dp[N][i];
		}
	}
	printf("%.20lf\n", ans);
	return 0;
}