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: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define rrep(i,n) for(int i=((int)(n)-1);i>=0;(i)--)
#define rrep1(i,n) for(int i=((int)(n));i>0;(i)--)
#define sz(c) ((int)(c).size())
#define all(x) (x).begin(),(x).end()
#define mp make_pair
#define pb push_back

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int, int> P;
typedef pair<int, pair<int, int> > PP;
const int MAX=3e3+10;
const int MOD=1e9+7;

int N;
double p[MAX];
void input(){
    cin>>N;
    rep1(i,N)cin>>p[i];
}

double dp[MAX][MAX];
void solve(){
    dp[0][0]=1;
    rep1(i,N)rep(up, i+1){
        if(up==0) dp[i][up]=dp[i-1][up]*(1-p[i]);
        else dp[i][up]=dp[i-1][up]*(1-p[i])+dp[i-1][up-1]*p[i];
    }
}

double ans;
void output(){
  	rep(up,N+1)if(up>N-up){
        ans+=dp[N][up];
    }
    cout<<fixed;
    cout<<setprecision(12)<<ans<<endl;
}

int main(){
    input();
    solve();
    output();
}