hdu 2588 GCD

Problem Description

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

分析：求gcd(x,n)>=m的个数 。如果gcd(x,n)=m,那么x=q*m,n=p*m,p与q互素。求gcd(x,n)>=m,枚举n/m,即p，求p所有欧拉函数和。

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;

int lur(int nn)
{
    int res=nn;
    for(int i=2;i*i<=nn;i++)
    {
        if(nn%i==0)
        {
            res=res/i*(i-1);
            nn/=i;
            while(nn%i==0)
            nn/=i; 
        }
    }
    if(nn>1)
    res=res/nn*(nn-1);
    return res;
}

int main()
{
    int t,m,n;
    cin>>t;
    while(t--)
    {
        int sum=0;
        cin>>n>>m;
        if(m==1)
        {
            cout<<n<<endl;
            continue;
        }
        for(int i=1;i*i<=n;i++)// 枚举m 我试着i [m,n]超时了,对于两个数相乘形式，枚举其中一个，这么做好像快些。
        { 
            if(n%i==0)          
            {
                if(i>=m)
                sum+=lur(n/i);// n/i 即p
                if(n/i!=i&&(n/i)>=m)
                sum+=lur(i);
            }
        }
        cout<<sum<<endl;
     } 
     return 0;
 }