ZOJ 2836 Number Puzzle （容斥原理）

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理解的好辛苦啊，，ym final爷，几分钟就能A出来！！二进制真是好东西啊

#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 0x3f3f3f3f3f3f3f3fLLusing 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;    }

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