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>
using namespace std;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define sc(n) scanf("%d",&n)
#define scl(n) scanf("%lld",&n)
#define pr(n) printf("%d",n)
#define prl(n) printf("%lld",n)
#define nl printf("\n")
#define fr(i,n) for(i=0;i<n;i++)
#define rep(i,st,en) for(i=st;i<=en;i++)
#define repv(i,en,st) for(i=en;i>=st;i--)
#define fout cout<<fixed<<setprecision(9)
#define bi(n) __builtin_popcount(n)
#define bil(n) __builtin_popcountll(n)


typedef long long ll;
typedef pair<int,int> pii;
const int N = 3010;
ll mod = 1e9+7;
ll fmod(ll b,ll exp){
    ll res =1;
    while(exp){if(exp&1ll)res=(res*b)%mod;
    b =(b*b)%mod;exp/=2ll;
    }
    return res;
}

long double dp[N][N];
long double p[N];

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    int t=1,n,i,j,m;
    cin>>n;
    rep(i,1,n)cin>>p[i];
    dp[0][0]=1;
    rep(i,1,n)
    {
        dp[i][0]= dp[i-1][0]*(1-p[i]);
        for(int j=1;j<i;j++)
        {
            dp[i][j]=(dp[i-1][j]*(1-p[i]) + dp[i-1][j-1]*p[i]);
        }
        dp[i][i]= dp[i-1][i-1]*p[i];
    }
    long double ans = 0;
    for(int i=n/2+1;i<=n;i++)ans+=dp[n][i];
    fout<<ans<<"\n";

    
    return 0;
}