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>
///...................................*****.................................................///
///        Author :  Raihan Khan Raka  ( raihankhanraka@gmail.com )                         ///
///                  Department of Computer Science                                         ///
///                  & Engineering                                                          ///
///                  Comilla University , Bangladesh.                                       ///
///...................................*****.................................................///

/*....................................Values................................................*/
#define       p5                   100007
#define       p6                   1000007
#define       PI                   acos(-1)
#define       M                    1000000007
#define       inf                  1LL << 62
#define       white                0
#define       gray                 1
#define       black                2
/*....................................Functions.............................................*/
#define       sqr(x)               x*x
#define       sc                   scanf
#define       pf                   printf
#define       pfn                  printf("\n")
#define       scin(x)              sc("%d",&(x))
#define       scin2(xx,zz)         scanf("%d %d",&xx,&zz)
#define       scln(x)              sc("%lld",&(x))
#define       scln2(xx,zz)         scanf("%lld %lld",&xx,&zz)
#define       min3(a,b,c)          min(a,b<c?b:c)
#define       max3(a,b,c)          max(a,b>c?b:c)
#define       all(v)               v.begin(), v.end()
#define       ok                   cout << "ok" << endl
#define       mem(x,y)             memset(x,y,sizeof(x))
#define       clr(a)               a.clear()
#define       READ(f)              freopen(f, "r", stdin)
#define       WRITE(f)             freopen(f, "w", stdout)

/*...................................Data_Types............................................*/
#define       lli                  long long int
#define       ull                  unsigned long long int
#define       pii                  pair < int, int>
#define       pll                  pair < ll, ll>
#define       veci                 vector<int>
#define       vecl                 vector<long long int>
#define       vecp                 vector<  pair<int,int> >
#define       mapstrint            map< string , int >
#define       mapstrstr            map< string , string >
#define       mapint               map< int, int >
#define       uset                 unordered_set
#define       umap                 unordered_map
#define       pq                   priority_queue

#define       pb                   push_back
#define       mp                   make_pair
#define       ff                   first
#define       ss                   second

/*.....................................Loops...............................................*/
#define       rep( i , a , b )     for( i=a ; i<b ; i++)
#define       rev( i , a , b )     for( i=a ; i>=b ; i--)
#define       repx( i ,a,b, x)     for( i=a ; i<b ; i+=x)

#define       IOS             ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

//int month[]={31,28,31,30,31,30,31,31,30,31,30,31};

///------------------------------- Mudular functions----------------------------------------
/*
inline lli power(lli x, lli y){ lli temp; if( y == 0) return 1; temp = power(x, y/2); if (y%2 == 0) return temp*temp; else return x*temp*temp; }
inline lli add(lli a, lli b) {a += b; return a >= M ? a - M : a;}
inline lli sub(lli a, lli b) {a -= b; return a < 0 ? a + M : a;}
inline lli mul(lli a, lli b) {return  (a * b) % M;}
lli gcd(lli x,lli y)
{
    if(x==0) return y;
    return gcd(y%x,x);
}
lli bigmod(lli n, lli k )
{
    lli ans=1;
    while(k)
    {
        if(k&1)
            ans=(ans*n)%M;
        k=k>>1;
        n=(n*n)%M;
    }

    return ans;
}
*/
///----------------------------------Graph moves----------------------------------------
/*
int dx4[5] = {1, -1, 0,  0};
int dy4[5] = {0,  0, 1, -1};
int dx8[9] = {0,  0, 1, -1, -1, 1, -1,  1};
int dy8[9] = {-1, 1, 0,  0,  1, 1, -1, -1};
int knightx[9] = {-2, -2, -1, -1,  1, 1, 2,  2};
int knighty[9] = {-1,  1, -2,  2, -2, 2, -1, 1};
bool valid( int r , int c , int x , int y ){ if( x >= 1 && x <= r && y >= 1 && y <= c ) return 1 ; return 0 ; }
*/

using namespace std;

double dp[3001][3001];

int main()
{
    int n,i,j;

    cin >> n ;
    double head[n+7],tail[n+7];

    rep(i , 1 , n+1)
    {
        cin >> head[i] ;
        tail[i]=1.0-head[i];
    }

    dp[0][0]=1.0;

    rep(i , 1 , n+1)
    {
        rep(j , 0 , n+1)
        {
            if(j==0)
                dp[i][j]=tail[i]*dp[i-1][j];
            else dp[i][j]= dp[i-1][j-1]*head[i] + dp[i-1][j]*tail[i] ;

            //cout << dp[i][j] << " " ;
        }
       // pfn;
    }

    double ans=0;
    rep(j , (n>>1)+1 , n+1 )
    ans+= dp[n][j];

    cout << fixed << setprecision(10) <<  ans << endl;





#ifdef HOME
    cerr << "Time elapsed: " << clock() / 1000 << " ms" << endl;
#endif
    return 0;
}