TopCoder TCHS 16 Divisibility

容斥原理  第三题     要注意求最小公倍数时会超 long long 的问题

代码：

#include<iostream>

#include<cstring>

#include<cstdio>

#include<string>

#include<set>

#include<map>

#include<cmath>

#include<algorithm>

#include<vector>

#include<cmath>

#include<queue>

#include<stack>



//#define ull unsigned long long

#define ll long long



using namespace std;



const int INF=0x3f3f3f3f;

const int MOD=1000000007;

const ll LMOD=1000000007;

const double eps=1e-6;

const int N=1050;

ll gcd(ll x,ll y)

{

    if(x%y==0)

    return y;

    return gcd(y,x%y);

}

ll lcm(ll x,ll y)

{

    return x*y/gcd(x,y);

}

ll solve(ll n,vector<int> &a)

{

    ll sum=0;

    for(int k=1;k<(1<<(int)a.size());++k)

    {

        int num=0;

        ll LCM=1;

        for(int i=0;i<(int)a.size();++i)

        {

            if((k&(1<<i))>0)

            {

                ++num;

                LCM=lcm(LCM,(ll)a[i]);

                if(LCM>n)

                break;

            }

        }

        if(LCM>n)

        continue;



        if((num&1)>0)

        {

            sum+=(n/LCM);

        }else

        {

            sum-=(n/LCM);

        }

    }

    //cout<<sum<<endl;



    return sum;

}

class Divisibility

{

    public:

    int numDivisible(int L, int R, vector <int> a)

    {

        return (int)(solve(R,a)-solve(L-1,a));

    }

};