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B. Product(2019ICPC西安邀请赛)(杜教筛)

發布時間:2023/12/4 编程问答 37 豆豆
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Product

∑i=1n∑j=1n∑k=1ngcd?(i,j)[k∣gcd?(i,j)]∑k=1n∑i=1nk∑j=1nkgcd?(ik,jk)∑k=1nk∑i=1nk∑j=1nkgcd?(i,j)∑k=1nk∑d=1nkd∑i=1nkd∑j=1nkd[gcd?(i,j)=1]∑k=1nk∑d=1nkd(∑i=1nkd2?(i)?1)T=kd∑T=1n∑k∣Tk×Tk(∑i=1nT2?(i)?1)∑T=1nTσ0(T)(∑i=1nT2?(i)?1)\sum_{i = 1} ^{n} \sum_{j = 1} ^{n} \sum_{k = 1} ^{n} \gcd(i, j)[k \mid \gcd(i, j)]\\ \sum_{k = 1} ^{n} \sum_{i = 1} ^{\frac{n}{k}} \sum_{j = 1} ^{\frac{n}{k}} \gcd(ik, jk)\\ \sum_{k = 1} ^{n} k \sum_{i = 1} ^{\frac{n}{k}} \sum_{j = 1} ^{\frac{n}{k}} \gcd(i, j)\\ \sum_{k = 1} ^{n} k \sum_{d = 1} ^{\frac{n}{k}} d \sum_{i = 1} ^{\frac{n}{kd}} \sum_{j = 1} ^{\frac{n}{kd}} [\gcd(i, j) = 1]\\ \sum_{k = 1} ^{n} k \sum_{d = 1} ^{\frac{n}{k}} d \left(\sum_{i = 1} ^{\frac{n}{kd}} 2 \phi(i) - 1\right)\\ T = kd\\ \sum_{T = 1} ^{n} \sum_{k \mid T} k \times \frac{T}{k} \left(\sum_{i = 1} ^{\frac{n}{T}} 2 \phi(i) - 1\right)\\ \sum_{T = 1} ^{n} T \sigma_0(T) \left(\sum_{i = 1} ^{\frac{n}{T}} 2 \phi(i) - 1\right) i=1n?j=1n?k=1n?gcd(i,j)[kgcd(i,j)]k=1n?i=1kn??j=1kn??gcd(ik,jk)k=1n?ki=1kn??j=1kn??gcd(i,j)k=1n?kd=1kn??di=1kdn??j=1kdn??[gcd(i,j)=1]k=1n?kd=1kn??d???i=1kdn??2?(i)?1???T=kdT=1n?kT?k×kT????i=1Tn??2?(i)?1???T=1n?Tσ0?(T)???i=1Tn??2?(i)?1???

f(n)=∑i=1niσ0(i)f(n) = \sum\limits_{i = 1} ^{n} i \sigma_0(i)f(n)=i=1n?iσ0?(i)。
∑i=1ni∑d∣i∑d=1n∑i=1ndi×d∑d=1nd(nd+1)nd2\sum_{i = 1} ^{n} i \sum_{d \mid i}\\ \sum_{d = 1} ^{n} \sum_{i = 1} ^{\frac{n}ozvdkddzhkzd} i \times d\\ \sum_{d = 1} ^{n} d \frac{ (\frac{n}ozvdkddzhkzd + 1) \frac{n}ozvdkddzhkzd}{2} i=1n?idi?d=1n?i=1dn??i×dd=1n?d2(dn?+1)dn??

#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10;int f[N], num[N], phi[N], prime[N], primes[N], cnt;bool st[N];int n, m, p, mod;inline int add(int x, int y) {return x + y < mod ? x + y : x + y - mod; }inline int sub(int x, int y) {return x >= y ? x - y : x - y + mod; }void init() {f[1] = phi[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;phi[i] = i - 1;f[i] = 2;num[i] = 1;primes[i] = i;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];f[i * prime[j]] = f[i / primes[i]] * (num[i] + 2);num[i * prime[j]] = num[i] + 1;primes[i * prime[j]] = primes[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);f[i * prime[j]] = f[i] * 2;num[i * prime[j]] = 1;primes[i * prime[j]] = prime[j];}}for (int i = 1; i < N; i++) {phi[i] = add(phi[i], phi[i - 1]);f[i] = add(1ll * i * f[i] % mod, f[i - 1]);} }unordered_map<int, int> Phi;int Djs_phi(int n) {if (n < N) {return phi[n];}if (Phi.count(n)) {return Phi[n];}int ans = 1ll * n * (n + 1) / 2 % mod;for (int l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = sub(ans, Djs_phi(n / l) * (r - l + 1) % mod);}return Phi[n] = ans; }unordered_map<int, int> f1;int calc1(int l, int r) {return 1ll * (l + r) * (r - l + 1) / 2 % mod; }int F(int n) {if (n < N) {return f[n];}if (f1.count(n)) {return f1[n];}int ans = 0;for (int l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = add(ans, 1ll * (n / l) * (n / l + 1) / 2 % mod * calc1(l, r) % mod);}return f1[n] = ans; }int calc(int n) {return sub(2ll * Djs_phi(n) % mod, 1); }int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % p;}n >>= 1;a = 1ll * a * a % p;}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("ou.txt", "w", stdout);scanf("%d %d %d", &n, &m, &p);mod = p - 1;init();int ans = 0;for (int l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = add(ans, 1ll * sub(F(r), F(l - 1)) * calc(n / l) % mod);}printf("%d\n", quick_pow(m, ans));return 0; }

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