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: Time Limit Exceeded)

#include <bits/stdc++.h>
#ifdef DBP
#define deb(...) fprintf(stderr,__VA_ARGS__)
#else
#define deb(...) 0
#endif
#define fst first
#define snd second
#define fore(x,a,b) for(int x=(a), qwerty=(b); x<qwerty; x++)
#define pb push_back
#define mset(a,v) memset((a),(v),sizeof(a))
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())
#define FIN ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
using namespace std;
typedef long long ll;

const int N=3000;
bool vis[N][N];
double dp[N][N],ps[N];
int n;

double f(int i, int j){
	if(i+j==n){return i>j?1. : 0.;}
	if(vis[i][j])return dp[i][j];
	double &r=dp[i][j]=0;
	return r=f(i+1,j)*ps[i+j]+f(i,j+1)*(1.-ps[i+j]);
}

int main(){FIN;cout<<fixed<<setprecision(12);
	cin >> n;
	fore(x,0,n)cin>>ps[x];
	cout << f(0,0) << "\n";
}