莫比乌斯反演 + 欧拉函数

由 Gu(a,b) = \(\frac{\phi(ab) }{\phi (a)*\phi(b)}\)

可以推出 Gu(a, b) = \(\frac{gcd(a,b))}{\phi (gcd(a,b))}\)

然后这个东西用莫比乌斯反演，预处理前缀和值域分块求。。。

网上貌似很多题解不用分块也AC了，但是我不用就T orz

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline ll read(){ll X = 0, w = 0; char ch = 0;while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();return w ? -X : X;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){A ans = 1;for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;return ans;
}
const int N = 1000005;
int _, m, n, p, tot;
ll inv[N], prime[N], mo[N], phi[N], sum[N];
bool vis[N];void sieve(){mo[1] = 1, phi[1] = 1;for(int i = 2; i <= N; i ++){if(!vis[i]){prime[++tot] = i;mo[i] = -1, phi[i] = i - 1;}for(int j = 1; j <= tot && prime[j] * i <= N; j ++){vis[i * prime[j]] = true;if(i % prime[j] == 0){mo[i * prime[j]] = 0;phi[i * prime[j]] = phi[i] * prime[j];break;}else{mo[i * prime[j]] = -mo[i];phi[i * prime[j]] = phi[i] * phi[prime[j]];}}}for(int i = 1; i <= N; i ++){sum[i] = sum[i - 1] + mo[i];}
}ll calc(int x, int y){ll ret = 0;for(int l = 1, r; l <= min(x, y); l = r + 1){r = min(x / (x / l), y / (y / l));ret = (ret % p + (1LL * (x / l) * (y / l) % p * (sum[r] - sum[l - 1]) % p) % p) % p;}return ret;
}int main(){sieve();for(_ = read(); _; _ --){m = read(), n = read(), p = read();if(n > m) swap(n, m);inv[1] = 1;for(int i = 2; i <= n; i ++)inv[i] = (p - p / i) * inv[p % i] % p;ll ans = 0;for(int i = 1; i <= min(n, m); i ++){ans = (ans % p + (1LL * i * inv[phi[i]] % p * calc(m / i, n / i)) % p) % p;}printf("%lld\n", ans);}return 0;
}