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

题意：1<=x<=n,1<=m<=n,gcd(x,n)>=m,求满足条件的x的个数

解法：

          先找出N的约数x，并且gcd（x，N）>= M，结果为所有N/x的欧拉函数之和。

          因为x是Ｎ的约数，所以ｇｃｄ（ｘ，Ｎ）＝ｘ >= M；

　　　设ｙ＝Ｎ／ｘ，ｙ的欧拉函数为小于ｙ且与ｙ互质的数的个数。

　　　设与ｙ互质的的数为ｐ１，ｐ２，ｐ３，…，ｐ４

　　　那么ｇｃｄ（ｘ* pi，N）= x >= M。

          也就是说只要找出所有符合要求的y的欧拉函数之和就是答案了。

超时代码：

#include
#include
#include
using namespace std;
//求欧拉函数 
int Euler(int n)
{
	int ret=n;
	for(int i=2;i<=sqrt(n);i++)
	{
		if(n%i==0)
		{
			ret=ret/i*(i-1);
			while(n%i==0)
			  n/=i;
		}
	}
	if(n>1)
	   ret=ret/n*(n-1);
	return ret;
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n,m;
		cin>>n>>m;
		int ans=0;
		for(int i=2;i<=n;i++)
		{
			if(n%i==0&&i>=m)
			{
				ans+=Euler(n/i);
			}
		}
		cout<

 

ac代码：

#include
#include
#include
using namespace std;
//求欧拉函数 
int Euler(int n)
{
	int ret=n;
	for(int i=2;i<=sqrt(n);i++)
	{
		if(n%i==0)
		{
			ret=ret/i*(i-1);
			while(n%i==0)
			  n/=i;
		}
	}
	if(n>1)
	   ret=ret/n*(n-1);
	return ret;
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n,m;
		cin>>n>>m;
		int ans=0;
		int sq_n=sqrt(n);
		for(int i=2;i<=sq_n;i++)
		{
			if(n%i==0)
			{
				if(i>=m)   ans+=Euler(n/i);
				if(n/i>=m) ans+=Euler(i);
			}
		}
		if(n!=1&&sq_n*sq_n==n&&sq_n>=m) ans-=Euler(n/sq_n);
		cout<