cf Educational Codeforces Round 81 D. Same GCDs

原题：

D. Same GCDs
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given two integers a and m. Calculate the number of integers x such that 0≤x 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 Output

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

3
4 9
5 10
42 9999999967

Output

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.

中文：

给你两个数a和m，现在让你找出x的个数，使得gcd(a,m)=gcd(a+x,m). 其中gcd为最大公约数。

程序：

#include 
 
using namespace std;
const int maxn = 1e5 +5;
typedef long long ll;
 
 
 
 
ll phi(ll n)
{
     
    double result = n;
    for (ll p = 2; p * p <= n; ++p) {
     
        if (n % p == 0) {
     
            while (n % p == 0)
                n /= p;
            result *= (1.0 - (1.0 / (double)p));
        }
    }
    if (n > 1) result *= (1.0 - (1.0 / (double)n));
    return (ll)result;
}
 
 
ll t,a,m;
int main()
{
     
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--)
    {
     
        cin>>a>>m;
        cout<<phi(m/__gcd(a,m))<<endl;
    }
 
    return 0;
}
 
 

解答：

设
现在由最大公约数的定理可知 $gcd(a,b)=gcd(b,a\%b)$
所以$gcd(a+x,m)=gcd(m,(a+x)\%m)$
所以要找$gcd(m,(a+x)\%m)=d$的x个数，且要求$x\in [0,m)$
将因子d除掉，即$gcd(\frac{(a+x)\%m}{d},\frac{m}{d})=1$
因此，最后结果等价于寻找$\frac{m}{d}$的非因子的个数，直接欧拉函数就完事