Problem:
Score : 400 points

Problem Statement

Given are a sequence A= {a_1,a_2,......a_N} of N positive even numbers, and an integer M.

Let a semi-common multiple of A be a positive integer X that satisfies the following condition for every k (1 \leq k \leq N):

There exists a non-negative integer p such that X= a_k \times (p+0.5).

Find the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^9

2 \leq a_i \leq 10^9

a_i is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 a_2 ... a_N

Output

Print the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Sample Input 1

2 50
6 10

Sample Output 1

2

15 = 6 \times 2.5

15 = 10 \times 1.5

45 = 6 \times 7.5

45 = 10 \times 4.5

Thus, 15 and 45 are semi-common multiples of A. There are no other semi-common multiples of A between 1 and 50, so the answer is 2.

Sample Input 2

3 100
14 22 40

Sample Output 2

0

The answer can be 0.

Sample Input 3

5 1000000000
6 6 2 6 2

Sample Output 3

166666667

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

long GCD(long a, long b){
  long dividend, divisor;
  if (a >= b) {
    dividend = a;
    divisor = b;
  } else {
    dividend = b;
    divisor = a;
  }
  long remainder;
  while (1) {
    if (divisor == 1) {
      return 1;
    }
    remainder = dividend % divisor;
    if (remainder == 0) {
      break;
    } else {
      dividend = divisor;
      divisor = remainder;
    }
  }
  return divisor;
}

int main(void) {

  long n,m;
  scanf("%ld %ld", &n, &m);
  long a[n];
  for (long i = 0; i < n; i++) {
    scanf("%ld", &a[i]);
  }
  long lcm_all = a[0];
  for (long i = 1; i < n; i++) {
    lcm_all = (lcm_all/GCD(lcm_all,a[i]))*a[i];
    if (lcm_all > 3000000000) {
      lcm_all = 3000000000;
    }
  }
  long half = lcm_all/2;
  for (long i = 0; i < n; i++) {
    if (half%a[i] != a[i]/2) {
      printf("0\n");
      return 0;
    }
  }
  long mod = m%lcm_all;
  long ans = m/lcm_all;
  if (mod >= half) {
    ans++;
  }
  printf("%ld\n", ans);

  return 0;
}