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 fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define ll long long
#define ld long double
#define pb push_back
#define fe first
#define se second
#define nl "\n"
#define pp pair < ll , ll > 
#define sz(x) (ll)x.size()
#define st(x) sort(x.begin(),x.end())
#define rst(x) sort(x.rbegin(), x.rend())
#define all(x) x.begin(),x.end()
long double pi = 3.14159265358979323;
 
const double EPS = 1e-12;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
 
 
 
 
int main()
{
    fast;
    ll n;
    cin >> n ;
    ld p[n+1];
    for(int i = 1 ; i <= n ; i++ ){
        cin >> p[i];
    }
    ld dp[n+1][n+1] = {};
    dp[0][0] = 1;
    for(int i = 1 ; i <= n ;i++ ){
        for(int heads = 0 ; heads <= i ; heads++ ){
            dp[i][heads] += dp[i-1][heads]*(1-p[i]);
            if(heads){
                //cout <<  dp[i][heads] << " ";
                //cout << dp[i-1][heads-1] << " ";
                /*cout <<  dp[i][heads] << " ";*/
                dp[i][heads] += (dp[i-1][heads-1])*p[i];
            }
           //cout << dp[i][heads] << " ";
        }
        //cout << "\n";
    }

    ld ans = 0 ;
    for(int i = n/2 + n%2 ; i <= n ; i++ ){
        ans += dp[n][i];
    }    
    cout << fixed << setprecision(6) << ans;





















    
    return 0;
    
}