GCD HDU - 2588（欧拉函数）

The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.
Input
The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.
Output
For each test case,output the answer on a single line.
Sample Input
3
1 1
10 2
10000 72
Sample Output
1
6
260
这题也很暴力的说，直接找出n的所有素因子以及相应的个数，然后暴力组合所有n的因子，然后和m判断大小，如果满足条件，就用欧拉函数找出个数即可，不需要任何的优化。

#include
#include
#include
#include
#include
#define N 200005
#define mod 1000000007
using namespace std;
int n,m;
int phi(int x)//欧拉函数
{
    int ans=x;
    for(int i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            ans-=ans/i;
            while(x%i==0)
                x/=i;
        }
        if(x==1)
            break;
    }
    if(x>1)
        ans-=ans/x;
    return ans;
}
int fac[20];
int e[20];
int cnt;
void getFac(int x)//晒出x的因子和相应个数
{
    for(int i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            fac[cnt]=i;
            e[cnt]=0;
            while(x%i==0)
            {
                x/=i;
                e[cnt]++;
            }
            cnt++;
        }
    }
    if(x>1)
    {
        fac[cnt]=x;
        e[cnt]=1;
        cnt++;
    }
}
int P(int a,int b)//手写个快速幂。。。。
{
    int ans=1;
    while(b)
    {
        if(b&1)
            ans*=a;
        a*=a;
        b>>=1;
    }
    return ans;
}
int ans;
void dfs(int temp,int cur)//暴力枚举因子即可
{
    if(cur==cnt)
    {
        if(temp>=m)
            ans+=phi(n/temp);
        return ;
    }
    int now;
    for(int i=0;i<=e[cur];i++)
    {
        now=temp*P(fac[cur],i);
        dfs(now,cur+1);
    }
}
int main()
{
    int t;
    //cout<
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        if(m==1)
            printf("%d\n",n);
        else
        {
            ans=0;
            cnt=0;
            getFac(n);
            dfs(1,0);
            printf("%d\n",ans);
        }
    }
    return 0;
}