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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define dbg(x) cout << #x << '=' << x << '\n';
#define ll long long
#define x first
#define y second
#define pi pair <int, int>
#define vi vector <int>
const ll mod = 998244353;
#define int ll
int n;
double dp[3001];

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(); cout.tie();
    cin>>n;
    dp[0]=1;
    for(int i=1; i<=n; i++){
        double x;
        cin>>x;
        for(int cnt=i; cnt>=1; cnt--){
            dp[cnt]=dp[cnt-1]*x+dp[cnt]*(1-x);
        }
        dp[0]*=(1-x);
    }
    double ans=0;
    //for(int i=0; i<=n; i++) cout<<dp[i]<<" ";
    for(int i=n/2+1; i<=n; i++) ans+=dp[i];
    cout<<ans;
}