ZOJ 2836 Number Puzzle （容斥原理）

#include "string"
#include "iostream"
#include "cstdio"
#include "cmath"
#include "set"
#include "queue"
#include "vector"
#include "cctype"
#include "sstream"
#include "cstdlib"
#include "cstring"
#include "stack"
#include "ctime"
#include "algorithm"
#define pa pair<int,int>
#define Pi M_PI
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef long long LL;
const int M=15;

int m, n, A[M];

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 )//用二进制来1,0来表示第几个素因子是否被用到,
//如n=3，三个因子是2,3,5，则i=3时二进制是011，表示第2、3个因子被用到
{
LL mult = 1;
LL bits = 0;
for( int j = 0; j < n; ++j )
{
if( ( 1 << j ) & i ) //判断第几个因子目前被用到
{
mult = lcm( mult, A[j] );
bits++;
}
}
if( bits & 1 )//容斥原理，奇加偶减
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;

}