Number Puzzle

Number Puzzle

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given a list of integers (A₁, A₂, ..., A_n), 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 A₁, A₂, ..., A_n(1 <= A_i <= 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+10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}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=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,fac[10];
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        rep(i,0,n-1)scanf("%d",&fac[i]);
        int ret=0;
        rep(i,1,(1<<n)-1)
        {
            int now=1,cnt=0;
            rep(j,0,n-1)
            {
                if(i&(1<<j))
                {
                    cnt++;
                    now=now*fac[j]/gcd(now,fac[j]);
                }
            }
            if(cnt&1)ret+=m/now;
            else ret-=m/now;
        }
        printf("%d
",ret);
    }
    return 0;
}