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 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<stdio.h>
#include<map>
#include<queue>
#include<vector>
using namespace std;
const int maxn=1e5+10;

typedef long long ll;

int a[maxn];

int main(){
	int n,m;
	cin>>n>>m;
	
	int lcm=1;
	for(int i=0;i<n;i++){
		cin>>a[i];
		if(lcm>1e9) continue;
		lcm=lcm*a[i]/2 / __gcd(lcm,a[i]/2);
	
	}
	
	bool flag=0;
	for(int i=0;i<n;i++){
		int t=lcm*2/a[i];
		if(t%2==0){
			flag=1;
			break;
		}
	}
	
	//个数=在此范围内的最小公倍数的奇数倍个数 
	int ans=m/lcm;
	if(ans&1) ans=ans/2+1;
	else ans/=2;
	if(flag) cout<<"0";
	else cout<<ans;
	return 0;
}