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(void)
{
	int n;
	scanf ("%d",&n);
	double *f;
	f = (double *)malloc(sizeof(double)*n);
	int i=0;
	while (i < n)
	{
		scanf("%lf",&(f[i]));
		i++;
	}
	double **dp;
	dp = (double **)malloc(sizeof(double*)*3000);
	i = 0;
	while (i <= n)
	{
		dp[i] = (double*)malloc(sizeof(double)*3000);
		i++;
	}
	i = 0;
	dp[1][0] = 1-f[0];
	dp[1][1] = f[0];
	i = 2;
	int j = 1;
	//printf("debug00100\n");
	while (i <= n)
	{
		j = 0;
		//printf("debug00150, i;%d, j;%d\n",i,j);
		while (j <= i)
		{
			//printf("debug00151, i;%d, j;%d\n",i,j);
			if (j == 0)
			{
				//printf("debug00152, i;%d, j;%d\n",i,j);
				//printf("dp[i-1][j];%lf\n",dp[i-1][j]);
				//printf("f[i-1];%lf\n",f[i-1]);
				dp[i][j] = dp[i-1][j] * (1-f[i-1]);
				//printf("dp[i][j];%lf\n",dp[i][j]);
			}
			else if (j == i)
			{
				//printf("debug00153, i;%d, j;%d\n",i,j);
				dp[i][j] = dp[i-1][j-1] * (f[i-1]);
			}
			else
			{
				//printf("debug00154, i;%d, j;%d\n",i,j);
				dp[i][j] = dp[i-1][j-1]*f[i-1] + dp[i-1][j]*(1-f[i-1]);
			}
			//printf("debug00158, i;%d, j;%d\n",i,j);
			j++;
		}
		i++;
	}
	i = (n+1)/2;
	//printf("debug00200\n");
	double sum = 0.0;
	while (i <= n)
	{
		sum = sum + dp[n][i];
		i++;
	}
	printf("%.10lf\n",sum);
	return (0);
}