2018 Multi-University Training Contest 7

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
", ans);
    }
    return 0;
}