D. Same GCDs, Educational Codeforces Round 81 (Rated for Div. 2),欧拉函数

D. Same GCDs, Educational Codeforces Round 81 (Rated for Div. 2)

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.

思路：
首先提取a,m的最大公约数g，令 $a = xg,m = yg$

需要使得 $gcd(xg,yg) = gcd(xg+z,yg)$

易知z为g的倍数，令 $z = kg,k \in [0,y)$

那么满足所有 $gcd(x+k,y) = 1$的k都可以，即求[x,y+x)中与y互素的数的个数

因为a < m，那么可知x < y，则要分别算[x,y]，(y,x+y)中与y互素的数的个数

因为 $gcd(x,y) = gcd(y,x\%y)$，故当 $x+k \in (y,x+y)$ 时， $gcd(x+k,y) = gcd(y,x+k-y)$，而 $x+k-y \in (0,x)$

所以(y,x+y)中与y互素的数的个数与(0,x)中与y互素的数的个数相等，故总的答案为[1,y]中与y互素的数的个数

#include<bits/stdc++.h>
#define ll long long
using namespace std;
long long gcd(long long a,long long b)
{
	return b?gcd(b,a%b):a;
}
long long oula(long long n)
{
	long long rea = n;
	for(long long i = 2;i * i <= n;++i)
	{
		if(n % i == 0)
		{
			rea -= rea/i;
			while(n % i == 0)
				n /= i;
		}
	}
	if(n > 1)
		rea -= rea/n;
	return rea;
}
int t;
ll a,b;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&a,&b);
		ll g = gcd(a,b);
		a/=g,b/=g;
		ll ans = oula(b);
		printf("%lld\n",ans);
	}
	return 0;
}