GuGuFishtion

推式子

$$\begin{gathered} \sum _{a=1}^m\sum _{b=1}^n\frac{?(a,b)}{?(a)?(b)} \\ =\sum _{a=1}^m\sum _{b=1}^n\frac{gcd(a,b)}{?(gcd(a,b))} \\ =\sum _{d=1}^m\frac{d}{?(d)}\sum _{a=1}^{\frac{m}{d}}\sum _{b=1}^{\frac{m}{d}}gcd(a,b)==1 \\ =\sum _{d=1}^m\frac{d}{?(d)}\sum _{a=1}^{\frac{m}{d}}\sum _{b=1}^{\frac{m}{d}}\sum _{k\mid gcd(a,b)}\mu (k) \\ 我們約定n,m，滿足n<m \\ =\sum _{d=1}^n\frac{d}{?(d)}\sum _{k=1}^{\frac{n}{d}}\mu (k)?\frac{n}{kd}??\frac{m}{kd}? \end{gathered}$$

接下來我們只要篩選出前$1e6$個數(shù)的$?and\mu$，再通過兩次數(shù)論分塊即可整體復雜度是$O(\sqrt{n}\sqrt{n})=O(n)。$

代碼

/*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 | 1using 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; }

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