Same GCDs

发表于 分类于 ，

    求

    $$\sum_{x=0}^{m-1} [\ \gcd(a, m) = \gcd(a + x, m)\ ]$$

    多组测试数据, $T \leq 50,\ 1 \leq a < m \leq 10^{10}$

Same GCDs

Created by LXC on Wed Apr 3 02:49:51 2024

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

ranting: 1800

tag: math, number theory

problem

求
$$\sum_{x=0}^{m-1} [\ \gcd(a, m) = \gcd(a + x, m)\ ]$$

多组测试数据, $T \leq 50,\ 1 \leq a < m \leq 10^{10}$

solution

$g = gcd(a, m)$

$gcd(a+x,m) = g$
所以$g|a+x \Rightarrow g|x$，令$u = \frac{a}{g}, v = \frac{m}{g}$，只需要求$gcd(u+t,v)=1$的个数，其中$ 0 \le t < v$

由于$gcd(u+t,v) = gcd((u+t) \bmod v,v)$，而$0 \le (u+t) \bmod v < v$

只需要求$gcd(t,v)=1, 0\le t < v$的个数

等价与求1到v中与v互质的个数，就是欧拉函数$\phi(v)$

答案就是$\phi(\frac{m}{gcd(a, m)})$

求$\phi(n)$可以在$O(\sqrt n)$完成

code

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#include <bits/stdc++.h>
// #define SINGLE_INPUT
#define ll long long
#define ull unsigned long long
#define N 500005
#define MOD 998244353
using namespace std;

template<class t,class u> ostream& operator<<(ostream& os,const pair<t,u>& p) {
    return os<<'['<<p.first<<", "<<p.second<<']';
}
template<class t> ostream& operator<<(ostream& os,const vector<t>& v) {
    os<<'['; int s = 1;
    for(auto e:v) { if (s) s = 0; else os << ", "; os << e; }
    return os<<']';
}
template<class t,class u> ostream& operator<<(ostream& os,const map<t,u>& mp){
    os<<'{'; int s = 1;
    for(auto [x,y]:mp) { if (s) s = 0; else os << ", "; os<<x<<": "<<y; }
    return os<<'}';
}

ll phi(ll n) {
    ll rt = n;
    for (ll i=2; i*i<=n; i++) {
        if (n%i) continue;
        while (n%i == 0) n/=i;
        rt /= i;
        rt *= i-1;
    }
    if (n != 1) {
        rt /= n;
        rt *= n-1;
    }
    return rt;
}

void sol() {
    ll a, m;
    cin >> a >> m;
    cout << phi(m/gcd(a, m)) << endl;
}

int main() {
    cout << setprecision(15) << fixed;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifndef SINGLE_INPUT
    int t;
    cin >> t;
    while (t--) {
        sol();
    }
#else
    sol();
#endif
    return 0;
}