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));
    }
};