D. Same GCDs----------------------------思维+结论(重要)

You are given two integers a and m. Calculate the number of integers x such that 0≤x<m and gcd(a,m)=gcd(a+x,m).

Note: gcd(a,b) is the greatest common divisor of a and b.

Input
The first line contains the single integer T (1≤T≤50) — the number of test cases.

Next T lines contain test cases — one per line. Each line contains two integers a and m (1≤a<m≤1010).

Output
Print T integers — one per test case. For each test case print the number of appropriate x-s.

Example
inputCopy
3
4 9
5 10
42 9999999967
outputCopy
6
1
9999999966
Note
In the first test case appropriate x-s are [0,1,3,4,6,7].

In the second test case the only appropriate x is 0.

解析:
已知gcd(a,m)=gcd(a+x,m).那么x的个数就是φ(m/d) ，d=gcd(a,m)
那么x=φ（m/gcd(a,m)）

记住这个结论
推理过程:https://www.cnblogs.com/wangjunrui/p/12242400.html

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int t;
ll a,b;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld %lld",&a,&b);
		ll n=b/__gcd(a,b);
		ll k=n;
		for(ll i=2;i*i<=n;i++)
		{
			if(n%i==0)
			{
				k-=(k/i);
				while(n%i==0) n/=i;
			}
		}
		if(n!=1) k-=(k/n);
		cout<<k<<endl;
	}
}