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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

double dp[3005][3005];
double p[3005],q[3005];

double def(int N);

int main(void){
  int N,i;
  char str[20000],*P;
  
  fgets(str,sizeof(str),stdin);
  N=atoi(str);
  
  fgets(str,sizeof(str),stdin);
  P=strtok(str," \n");
  p[1]=atof(P);
  q[1]=1.0-p[1];
  for(i=2;i<N+1;i++){
   P=strtok(NULL," \n");
    p[i]=atof(P);
    q[i]=1.0-p[i];
  }
  
  printf("%.10f\n",def(N));
  
  
  return 0;
}


double def(int N){
  int i,j;
  double ans;
  dp[0][0]=1.00;
  for(i=1;i<N+1;i++){
    for(j=0;j<N+1;j++){
      if(j==0){
        dp[i][j]=q[i]*dp[i-1][j];
      }else{
        dp[i][j]=p[i]*dp[i-1][j-1]+q[i]*dp[i-1][j];
      }
    }
  }
  
  for(i=N/2+1,ans=0.0;i<N+1;i++){
    ans+=dp[N][i];
  }
  
  return ans;
  
}