How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Now
you get a number N, and a M-integers set, you should find out how many
integers which are small than N, that they can divided exactly by any
integers in the set. For example, N=12, and M-integer set is {2,3}, so
there is another set {2,3,4,6,8,9,10}, all the integers of the set can
be divided exactly by 2 or 3. As a result, you just output the number 7.

Input

There
are a lot of cases. For each case, the first line contains two integers
N and M. The follow line contains the M integers, and all of them are
different from each other. 0<N<2^31,0<M<=10, and the M
integer are non-negative and won’t exceed 20.

Output

For each case, output the number.

Sample Input

12 2
2 3

Sample Output

7

分析：容斥原理，注意long long；

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[20],all;
int main()
{
    int i,j;
    while(~scanf("%d%d",&m,&n))
    {
        --m;
        all=0;
        rep(i,0,n-1)
        {
            scanf("%d",&j);
            if(j)fac[all++]=j;
        }
        ll ret=0;
        rep(i,1,(1<<all)-1)
        {
            ll now=1,cnt=0;
            rep(j,0,all-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("%lld\n",ret);
    }
    return 0;
}