Number Puzzle

Time Limit: 2 Seconds Memory Limit: 65536 KB

Given a list of integers (A, A, ..., A), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given list.

Input

The input contains several test cases.

For each test case, there are two lines. The first line contains N (1 <= N <= 10) and M (1 <= M <= 200000000), and the second line contains A, A, ..., A(1 <= A <= 10, for i = 1, 2, ..., N).

Output

For each test case in the input, output the result in a single line.

Sample Input

3 2
2 3 7
3 6
2 3 7

Sample Output

1
4

分析：1~m中至少能被列表里其中一个数整除的数的个数；

　　　容斥即可，注意列表中的数不一定互质，所以要求lcm；

代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <bitset>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define sys system("pause")

const int maxn=1e5+;

using namespace std;

inline ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

inline ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

inline void umax(ll &p,ll q){if(p<q)p=q;}

inline void umin(ll &p,ll q){if(p>q)p=q;}

inline ll read()

{

    ll x=;int f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,m,k,t,fac[];

int main()

{

    int i,j;

    while(~scanf("%d%d",&n,&m))

    {

        rep(i,,n-)scanf("%d",&fac[i]);

        int ret=;

        rep(i,,(<<n)-)

        {

            int now=,cnt=;

            rep(j,,n-)

            {

                if(i&(<<j))

                {

                    cnt++;

                    now=now*fac[j]/gcd(now,fac[j]);

                }

            }

            if(cnt&)ret+=m/now;

            else ret-=m/now;

        }

        printf("%d\n",ret);

    }

    return ;

}