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 ll long long
#define rep(i,f,e) for(int (i)=(f); (i)<(e); (i)++)
#define repr(i,e,f) for(int (i)=(e-1); (i)>=(f); (i)--)

const ll MOD = 1e9+7;
const int INF = (int)1e9+7;

// vector<tuple<long,long>> items; // (value, weight)
// auto item = items[i];
// long v = get<0>(item);
// long w = get<1>(item);
// long v, w;
//   for( int i = 0; i < N; i++ ){
//     cin >> v >> w;
//     items.push_back( make_tuple(v, w) ); // item : (value, weight)
//   }

double dp[3001][3001];

int main(){
    int N;
    cin >> N;
    double p;
    cin >> p;
    dp[1][0] = 1.0-p;
    dp[1][1] = p;
    rep(i,2,N+1){
        cin >> p;
        dp[i][0] = dp[i-1][0] * (1.0-p);
        rep(j,1,N+1){
            dp[i][j] = dp[i-1][j] * (1.0-p) + dp[i-1][j-1] * p;
        }
    }

    // rep(j, 0, N+1){
    //     cout << j << "/" << N << " : " << dp[N][j] << endl;
    // }

    double ans = 0.0;
    rep(j, (N+1)/2, N+1){
        ans += dp[N][j];
    }
    cout << ans;

    return 0;
}