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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mset(m,v) memset(m,v,sizeof(m))
#define pr(a,s) printf("Case #%lld: %lld\n",a,s)
#define mod 1000000007
#define mp make_pair
#define all(a)                a.begin(),a.end()
#define pb push_back
#define scan(a) scanf("%d",&a)
#define scanl(a) scanf("%lld",&a)
#define se second
#define fi first
#define pi 3.14159265358979323
#define pii pair<ll,ll>
 
const ll inf = 1e15;
const int MAXM = 3e3+4;
const int MAXN = 1e7+1;

double dp[MAXM][MAXM];
bool visit[MAXM][MAXM];
double arr[MAXM];
int n;

double solve(int idx, int heads) 
{
	if (idx == 0 and heads > n-heads) return 1.0;
	if (idx == 0)	return 0.0;
	if (visit[idx][heads])	return dp[idx][heads];
	double ret = 0.0;
	ret += arr[idx] * solve(idx-1, heads+1);
	ret += (1.0-arr[idx]) * solve(idx-1, heads);
	visit[idx][heads] = true;
	return dp[idx][heads] = ret; 
}

int main()
{
	mset(visit, false);
	cin>>n;
	for (int i=1;i<=n;i++) {
		cin>>arr[i];
	}
	printf("%0.10lf\n", solve(n,0));
	return 0;
}