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 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int main(void)
{
  int SIZE = 3000;

  int n;
  scanf("%d", &n);

  double *p = (double *)calloc(SIZE, sizeof(double));
  int i, j;
  for (i = 0; i < n; i++)
  {
    scanf("%lf", &p[i]);
  }

  double **dp = (double **)calloc(SIZE, sizeof(double *));
  for (i = 0; i <= n; i++)
  {
    dp[i] = (double *)calloc(SIZE, sizeof(double));
  }

  dp[0][0] = 1;
  for (i = 0; i < n; i++)
  {
    dp[i + 1][0] = dp[i][0] * (1 - p[i]);
  }

  for (i = 0; i < n; i++)
  {
    for (j = 1; j <= n; j++)
    {
      dp[i + 1][j] = dp[i][j - 1] * p[i] + dp[i][j] * (1 - p[i]);
    }
  }

  double ans = 0;
  for (j = n / 2 + 1; j <= n; j++)
  {
    ans += dp[n][j];
  }

  printf("%lf", ans);

  for (i = 0; i <= n; i++)
  {
    free(dp[i]);
  }
  free(dp);
  free(p);
  return 0;
}