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 9.2.1) (judge verdict: Accepted)

# include <stdio.h>
# include <stdlib.h>
int main(){
	int i,n,j,k;
	double *arr,**ans,res=0;
	scanf ("%d",&n);
	arr=(double*)malloc(n*sizeof(double));
	for(i=0;i<n;i++) scanf ("%lf",&arr[i]);
	ans=(double**)malloc(n*sizeof(double*));
	for(i=0;i<n;i++) ans[i]=(double*)malloc((n+2)*sizeof(double));
	for(i=0;i<n;i++){
		for(j=0;j<n+2;j++) ans[i][j]=0;
	}
	ans[0][0]=1-arr[0];
	ans[0][1]=arr[0];
	for(i=1;i<n;i++){
    	for (j=0;j<i+2;j++){
        	if (j==0) ans[i][j]=ans[i-1][j]*(1-arr[i]);
        	else ans[i][j]=ans[i-1][j]*(1-arr[i])+ans[i-1][j-1]*arr[i];
        }
    }
    for(i=n;i>n/2;i--){
    	res+=ans[n-1][i];
	}
	printf ("%.10lf\n",res);
    return 0;	
}