GuGuFishtion

推式子

∑ a = 1 m ∑ b = 1 n ϕ ( a , b ) ϕ ( a ) ϕ ( b ) = ∑ a = 1 m ∑ b = 1 n g c d ( a , b ) ϕ ( g c d ( a , b ) ) = ∑ d = 1 m d ϕ ( d ) ∑ a = 1 m d ∑ b = 1 m d g c d ( a , b ) = = 1 = ∑ d = 1 m d ϕ ( d ) ∑ a = 1 m d ∑ b = 1 m d ∑ k ∣ g c d ( a , b ) μ ( k ) 我 们 约 定 n , m ， 满 足 n < m = ∑ d = 1 n d ϕ ( d ) ∑ k = 1 n d μ ( k ) ⌊ n k d ⌋ ⌊ m k d ⌋ \sum_{a = 1 } ^{m} \sum_{b = 1 } ^{n} \frac{\phi(a, b)}{\phi(a) \phi(b)}\\ = \sum_{a = 1} ^{m} \sum_{b = 1} ^{n} \frac{gcd(a, b)}{\phi(gcd(a, b))}\\ = \sum_{d = 1} ^{m} \frac{d}{\phi(d)} \sum_{a = 1} ^{\frac{m}{d}} \sum_{b = 1} ^{\frac{m}{d}} gcd(a, b) == 1\\ = \sum_{d = 1} ^{m} \frac{d}{\phi(d)} \sum_{a = 1} ^{\frac{m}{d}} \sum_{b = 1} ^{\frac{m}{d}} \sum_{k \mid gcd(a, b)} \mu(k)\\ 我们约定n, m，满足 n < m\\ = \sum_{d = 1} ^{n} \frac{d}{\phi(d)} \sum_{k = 1} ^{\frac{n}{d}} \mu(k) \lfloor\frac{n}{kd}\rfloor \lfloor \frac{m}{kd} \rfloor\\ a=1∑m​b=1∑n​ϕ(a)ϕ(b)ϕ(a,b)​=a=1∑m​b=1∑n​ϕ(gcd(a,b))gcd(a,b)​=d=1∑m​ϕ(d)d​a=1∑dm​​b=1∑dm​​gcd(a,b)==1=d=1∑m​ϕ(d)d​a=1∑dm​​b=1∑dm​​k∣gcd(a,b)∑​μ(k)我们约定n,m，满足n<m=d=1∑n​ϕ(d)d​k=1∑dn​​μ(k)⌊kdn​⌋⌊kdm​⌋

接下来我们只要筛选出前 1 e 6 1e6 1e6个数的 ϕ   a n d   μ \phi \ and\ \mu ϕ and μ，再通过两次数论分块即可整体复杂度是 O ( n n ) = O ( n ) 。 O(\sqrt n\sqrt n )= O(n)。 O(n ​n ​)=O(n)。

代码 

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

const int N = 1e6 + 10;

int prime[N], cnt;

int phi[N], mu[N], inv[N], sum[N], n, m, mod;

bool st[N];

void init() {
    phi[1] = mu[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime[cnt++] = i;
            mu[i] = -1;
            phi[i] = i - 1;
        }
        for(int j = 0; j < cnt && i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            mu[i * prime[j]] = -mu[i];
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
    for(int i = 1; i < N; i++) {
        mu[i] += mu[i - 1];
    }
}

ll calc(ll n, ll m) {
    ll ans = 0;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ans = (ans + (mu[r] - mu[l - 1]) * (n / l) % mod * (m / l) % mod) % mod;
    }
    return ans;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    init();
    int T = read();
    while(T--) {
        n = read(), m = read(), mod = read();
        if(n > m) swap(n, m);
        inv[1] = 1;
        for(int i = 2; i < N; i++) {
            inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
        }
        for(int i = 1; i < N; i++) {
            sum[i] = (sum[i - 1] + 1ll * i * inv[phi[i]] % mod) % mod;
        }
        ll ans = 0;
        for(int l = 1, r; l <= n; l = r + 1) {
            r = min(n / (n / l), m / (m / l));
            ll temp = calc(n / l, m / l);
            ans = (ans + (sum[r] - sum[l - 1] + mod) % mod * calc(n / l, m / l) % mod) % mod;
        }
        printf("%lld\n", ans);
    }
	return 0;
}