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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define LOCAL
using namespace std;
template <typename A, typename B>
ostream& operator <<(ostream& out, const pair<A, B>& a) {
out << "(" << a.first << "," << a.second << ")";
return out;
}
template <typename T, size_t N>
ostream& operator <<(ostream& out, const array<T, N>& a) {
out << "["; bool first = true;
for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
return out;
}
template <typename T>
ostream& operator <<(ostream& out, const vector<T>& a) {
out << "["; bool first = true;
for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
return out;
}
template <typename T, class Cmp>
ostream& operator <<(ostream& out, const set<T, Cmp>& a) {
out << "{"; bool first = true;
for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
return out;
}
template <typename U, typename T, class Cmp>
ostream& operator <<(ostream& out, const map<U, T, Cmp>& a) {
out << "{"; bool first = true;
for (auto& p : a) { out << (first ? "" : ", "); out << p.first << ":" << p.second; first = 0;} out << "}";
return out;
}
#ifdef LOCAL
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...) 42
#endif
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << ": " << arg1 << " |";
__f(comma + 1, args...);
}
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
#define int long long
using P = pair<int,int>;
//#########################################
double dp[3001][3001];
signed main(){
    int n;cin >> n;
    vector<double> a(n);
    rep(i,n)cin >> a[i];
    dp[0][0] = 1;
    rep(i,n)rep(j,n+1){
        dp[i+1][j+1] += dp[i][j]*a[i];
        dp[i+1][j] += dp[i][j]*(1-a[i]);
    }
    double ret = 0;
    rep(i,n+1){
        if(i > n-i)ret += dp[n][i];
    }
    printf("%.12lf\n",ret);

    return 0;
}