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
1 #include<bits/stdc++.h> 2 typedef long long ll; 3 using namespace std; 4 ll euler_phi(ll n) { 5 ll m = ll(sqrt(n + 0.5)); 6 ll ans = n; 7 for (ll i = 2; i <= m; i++) 8 if (n % i == 0) { 9 ans = ans / i * (i - 1); 10 while (n % i == 0) n /= i; 11 } 12 if (n > 1) ans = ans / n * (n - 1); 13 return ans; 14 } 15 ll gcd(ll a, ll b) { 16 if (b == 0) return a; 17 return gcd(b, a % b); 18 } 19 int main() 20 { 21 int t; 22 cin>>t; 23 while(t--){ 24 ll a,m; 25 cin>>a>>m; 26 m=m/gcd(a,m); 27 ll x = euler_phi(m); 28 cout<<x<<endl; 29 } 30 return 0; 31 }