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>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ld long double
#define ull unsigned long long int
#define ll long long int
#define SPEED ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod 1000000007ll
#define rep(i,L,R) for(ll i=L;i<R;i++)
#define f(n) rep(i,0,n)
#define pll pair<ll,ll>
#define DEBUG
#ifdef DEBUG
#define deb2(args...) {dbg,args; cerr<<endl;}
#else
#define deb2(args...)
#endif
struct debb{template<typename T> debb& operator,(const T &v){cerr<<v<<" ";return *this;}}dbg;
#define deb1(args...) do {cerr << #args << " : "; dbg,args; cerr << "\n";} while(0)
#define printa(a,L,R) rep(i,L,R) cout << a[i] <<' ';cout<<"\n";
typedef tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
template<typename T>T gcd(T a,T b){if(b==0)return a;else return gcd(b,a%b);}
template<typename T>T lcm(T a,T b){return a/gcd(a,b)*b;}
ll power(ll x,ll y,ll p){ll res=1;x=x%p;while(y>0){if(y&1)res=((res%p)*(x%p))%p;y=y>>1;x=(x*x)%p;}return res;}
ll pow1(ll x,ll y){ll res=1;while(y>0){if(y&1)res=(res*x);y=y>>1;x=x*x;}return res;}
#define sq(X) ((X)*(X))
#define PI 3.141592653589
const ll INF = 1e18l+10,N = 1e5+10,MAX = 2e5+10;

int main()
{
    SPEED
    int n,i,j;
    cin>>n;
    double p[n];
    double dp[n+1][n+1];
    memset(dp, 0, sizeof(dp));
    dp[0][0] =1.0;
    f(n) {
        cin>>p[i];
        dp[0][i+1] = (1-p[i])*dp[0][i];
        for(j=1;j<=i+1;j++)
        dp[j][i+1] = p[i]*dp[j-1][i] + (1-p[i])*dp[j][i];
    }

    double ans = 0;
    for(i=n/2+1;i<=n;i++)ans+=dp[i][n];
    cout<<fixed<<setprecision(10)<<ans<<'\n';

}