Codeforces1295D. Same GCDs （欧拉函数）

https://codeforces.com/contest/1295/problem/D

设gcd（a，m）= n，那么找gcd（a +x ，m）= n个数，其实就等于找gcd（（a+x）/n，m/n) = 1的个数，等价于求m/n的欧拉函数

 #include<bits/stdc++.h>
 typedef long long ll;
 using namespace std;
 ll euler_phi(ll n) {
   ll m = ll(sqrt(n + 0.5));
   ll ans = n;
   for (ll i = ; i <= m; i++)
     if (n % i == ) {
       ans = ans / i * (i - );
       while (n % i == ) n /= i;
     }
   if (n > ) ans = ans / n * (n - );
   return ans;
 }
 ll gcd(ll a, ll b) {
   if (b == ) return a;
   return gcd(b, a % b);
 }
 int main()
 {
     int t;
     cin>>t;
     while(t--){
         ll a,m;
         cin>>a>>m;
         m=m/gcd(a,m);
         ll x = euler_phi(m);
         cout<<x<<endl;
     }
     return ;
 }