Goffi and GCD(欧拉函数 及其 性质)

Goffi and GCD

题意：

因为$gcd(n-i,n)，gcd(n-j,n)$都是n的因子（且不为n）所以乘起来不会超过$n^2$所以等价于求

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n[gcd(n-i,n)gcd(n-j,n)=n] \\ \sum _{i=1}^n\sum _{j=1}^n\sum _{x|n}\sum _{y|n}[xy=n][gcd(n-i,n)=x][gcd(n-j,n)=y] \\ \sum _{x|n}\sum _{y|n}[xy=n]\sum _{i=1}^n[gcd(i,n)=x]\sum _{j=1}^n[gcd(j,n)=x] \\ \sum _{x|n}\sum _{y|n}[xy=n]\varphi (\frac{n}{x})\varphi (\frac{n}{y}) \\ \sum _{x|n}\varphi (\frac{n}{x})\varphi (x) \end{gathered}$$

Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
ll eular(ll x) {
    
    //利用欧拉函数的性质求phi(x)
    ll ans = x;
    for(ll i=2; i*i<=x; i++) {
    
    
        if(x % i == 0) {
    
    
            ans = ans/i*(i-1);
            while(x % i == 0) x /= i;
        }
    }
    if(x > 1) ans = ans/x*(x-1);
    return ans; 
}
int main() {
    
    
    ll n, m;
    while(scanf("%lld %lld", &n, &m) != EOF) {
    
    
        ll ans = 0;
        if(m == 2 || n == 1) ans = 1;
        else  if(m == 1){
    
    
            for(ll i=1; i*i<=n; i++) {
    
    
                if(n % i == 0) {
    
    
                    if(n/i != i) ans = (ans+eular(i)*eular(n/i)*2%mod)%mod;
                    else ans = (ans+eular(i)*eular(i)%mod)%mod;
                }
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}

End