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>

#define DEBUG 1
#define NINF -1
#define PINF 1E9+1

int n;
double *p;
double *dp; // x:throw coins y:surface

int pos(int x,int y)
{
	return n*y+x;
}

int main(void)
{
	int i,j;
	double ans;
	scanf("%d",&n);
	p = (double *)malloc(sizeof(double) * n);
	for (i=0;i<n;i++) scanf("%lf",&p[i]);

	dp = (double *)malloc(sizeof(double) * n*(n+1));
	if (dp==NULL) {
		perror("malloc");
		exit(1);
	}
	for (j=0;j<n+1;j++) {
		for (i=0;i<n;i++) {
			dp[pos(i,j)] = 0;
		}
	}
	dp[pos(0,1)] = p[0];
	dp[pos(0,0)] = 1.0 - p[0];

	for (j=0;j<n+1;j++) {
		for (i=1;i<n;i++) {
			dp[pos(i,j)] += dp[pos(i-1,j)] * (1.0-p[i]);
			if (j==0) continue;
			dp[pos(i,j)] += dp[pos(i-1,j-1)] * p[i];
		}
	}

	ans = 0.0;
	for (j=(n+1)/2;j<n+1;j++) {
		ans += dp[pos(n-1,j)];
	}
	printf("%.10f\n",ans);

	free(p);
	free(dp);
	return 0;
}