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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int n;
double prob[3001];
double dp[3001][3001];
bool vis[3001][3001];

double f(int c, int h){
    if(c==0 and h>0) return 0;
    if(c==0 and h==0) return 1.0;
    if(c<0) return 0.0;
    
    if(vis[c][h]) return dp[c][h];
    
    dp[c][h] = (double)prob[c]*f(c-1,h-1)+(double)(1-prob[c])*f(c-1,h);
    vis[c][h] = true;
    
    return dp[c][h];
}


int main(){
    memset(dp,-1,sizeof(dp));
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>prob[i];
    }
    double ans = 0.0;
    for(int i=n/2+1;i<=n;i++){
        ans = ans + f(n,i); 
    }
    
    cout<< fixed <<setprecision(20)<<ans;
}