ZOJ 2836 Number Puzzle （ 容斥原理 ）

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
#define CLR( a, b ) memset( a, b, sizeof(a) )
int m, n, A[11];

LL gcd( LL a, LL b )
{
    return b == 0 ? a : gcd( b, a % b );
}
LL lcm( LL a, LL b )
{
    return a / gcd( a, b ) * b;
}


void solve()
{
    LL ans = 0;
    for( int i = 1; i < ( 1 << n ); ++i )
    {
        LL mult = 1;
        LL bits = 0;
        for( int j = 0; j < n; ++j )
        {
            if( ( 1 << j ) & i )
            {
                mult = lcm( mult, A[j] );
                bits++;
            }
        }
        if( bits & 1 )
            ans += m / mult;
        else
            ans -= m / mult;
    }
    printf( "%lld\n", ans );
}

void Orz()
{
    while( ~scanf( "%d %d", &n, &m ) )
    {
        for( int i = 0; i < n; ++i )
            scanf( "%d", &A[i] );
        solve();
    }
}
int main()
{
    Orz();
    return 0;
}