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 (ll i = 0; i < n; i++)
#define FOR(i, m, n) for(ll i = m;i < n;i++)
#define VSORT(v) sort(v.begin(), v.end());
#define VREV(v) reverse(v.begin(), v.end());
using namespace std;
using ll = long long;
using pll = pair<ll,ll>;
using vll = vector<ll>;
using vdouble = vector<double>;
using vstr = vector<string>;
typedef pair<int,int> P;
int dx[4] = {1, 0, -1, 0},dy[4] = {0, 1, 0, -1};
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
template<class T> inline bool chmax(T& a, T b){if(a<b) {a=b; return 1;} return 0; }
template<class T> inline bool chmin(T& a, T b){if(a>b) {a=b; return 1;} return 0; }
const ll INF = 1LL << 60;
const ll mod = 1000000007;

ll N;
vdouble p(3300);

double dp[3300][3300];//dp[i][j]:=最初のi枚のコインを投げたときに表がj枚になる確率

int main(){
	//乱数srand((unsigned)time(NULL));
	//入力
	cin>>N;
	p.resize(N);
	rep(i,N) cin>>p[i];
	
	dp[0][0]=1;//最初の0枚のコインを投げたときに表が0枚になる確率は1
	//dpテーブル埋め
	
	
	for(int i=0; i<N; i++){
		for(int j=0; j<=i; j++){
			dp[i+1][j+1]+=dp[i][j]*p[i];
			dp[i+1][j]+=dp[i][j]*(1.0-p[i]);
		}
	}
	
	
	double res=0.0;
	for(int i=(N+1)/2; i<=N; i++) res+=dp[N][i];
	cout<<setprecision(10)<<res<<endl;
}