HDU-2588-GCD （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

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;
long long int oula(int n)
{
	long long int res=1;
	for(int t=2;t*t<=n;t++)
	{
		if(n%t==0)
		{
			n/=t;
	   		res*=t-1;
		 while(n%t==0)
		 {
		 	n/=t;
		 	res*=t;
		  }
		 }

	}
	if(n>1)
	{
		res*=n-1;
	}
	return res;
}

int main()
{
	int n;
	cin>>n;
	int a,b;
	long long int ans;
	for(int t=0;t<n;t++)
	{
		ans=0;
		scanf("%d%d",&a,&b);
		int j;
		for( j=1;j*j<a;j++)
		{
			if(a%j==0)
			{

			if(j>=b)
			{
				ans+=oula(a/j);
			}
			if(a/j>=b)
			{
				ans+=oula(j);
			}
			}
		}
		if(j*j==a&&j>=b)
		{
			ans+=oula(j);
		}
		cout<<ans<<endl;
	}	

	return 0;
}