HDU 1796 How many integers can you find （状态压缩 + 容斥原理）

//HDU 1796
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define LL __int64
using namespace std ;

LL sh[12] ;

LL gcd(LL a,LL b)
{
    return b == 0 ? a : gcd(b,a%b) ;
}
int main()
{
    LL a,b ;
    while(~scanf("%I64d %I64d",&a,&b))
    {
        int c ,j = 0;
        for(int i = 0 ; i < b ; i++)
        {
            scanf("%d",&c) ;
            if(c > 0 && c < a)
            sh[j++] = c ;
        }
        b = j;
    //    sort(sh,sh+b);
        LL ans = 0 ;
        for(int i = 1 ; i < (1 << b) ; i++)//(1 << b)-1种情况
        {
            LL num = 0 ,lcm = 1;
            for(int j = 0 ; j < b ; j++)
            {
                if(i & (1 << j))
                {
                    num ++ ;
                    lcm = lcm*sh[j]/gcd(lcm,sh[j]) ;
                }
            }
            if(num & 1)
                ans += (a-1)/lcm ;
            else
                ans -= (a-1)/lcm ;
        }
        printf("%I64d\n",ans) ;
    }
    return 0 ;
}