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>

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;
typedef vector<pi> vpi;
typedef vector<bool> vbool;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define forall(it,s) for(auto it = s.begin(); it != s.end(); ++it)
#define F0(i,n) for(ll i = 0; i < n; i++)
#define F1(i,n) for(ll i = 1; i <= n; i++)
#define rep(i,a,b) for(int i = a; i <= b; i++)
#define per(i,a,b) for(int i = b-1; i >= a; i--)
#define FIN ios::sync_with_stdio(false);cin.tie(0);
#define db(x) cout<<#x<<" = "<<x<<"\n"
#define all(v) v.begin(),v.end()
#define modulo(a,b) ((a%b)+b)%b
#define F first
#define S second
#define INF 1000000010
#define eps 0.000000000001
#define mod 1000000007
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

//=======================//
const int maxn = 3000;
double dp[maxn][maxn];
int n;
double p[maxn];

int main () {
	//FIN
	cin >> n;
	F1(i,n) cin >> p[i];
	F0(i,n) dp[1][i] = 0.0;
	dp[1][0] = 1-p[1];
	dp[1][1] = p[1];
	rep(i,2,n) rep(j, 0, i) {
		dp[i][j] = 0.0;
		if (j < i) dp[i][j] += dp[i-1][j] * (1-p[i]);
		if (j > 0) dp[i][j] += dp[i-1][j-1] * p[i];
	}
	double res = 0.0;
	for (int i = (n+1)/2; i<=n; i++) {
		res += dp[n][i];
	}
	cout << fixed << setprecision(12) << res << "\n";


	return 0;
}