B. Product（2019ICPC西安邀请赛）（杜教筛）

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=1nkd2ϕ(i)−1)T=kd∑T=1n∑k∣Tk×Tk(∑i=1nT2ϕ(i)−1)∑T=1nTσ0(T)(∑i=1nT2ϕ(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=1∑nj=1∑nk=1∑ngcd(i,j)[k∣gcd(i,j)]k=1∑ni=1∑knj=1∑kngcd(ik,jk)k=1∑nki=1∑knj=1∑kngcd(i,j)k=1∑nkd=1∑kndi=1∑kdnj=1∑kdn[gcd(i,j)=1]k=1∑nkd=1∑knd⎝⎛i=1∑kdn2ϕ(i)−1⎠⎞T=kdT=1∑nk∣T∑k×kT⎝⎛i=1∑Tn2ϕ(i)−1⎠⎞T=1∑nTσ0(T)⎝⎛i=1∑Tn2ϕ(i)−1⎠⎞

设f(n)=∑i=1niσ0(i)f(n) = \sum\limits_{i = 1} ^{n} i \sigma_0(i)f(n)=i=1∑niσ0(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}{d}} i \times d\\ \sum_{d = 1} ^{n} d \frac{ (\frac{n}{d} + 1) \frac{n}{d}}{2} i=1∑nid∣i∑d=1∑ni=1∑dni×dd=1∑nd2(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;
}

B. Product（2019ICPC西安邀请赛）（杜教筛）相关推荐

51nod 1220 约数之和【莫比乌斯反演+杜教筛】
首先由这样一个式子:$ d(ij)=\sum_{p|i}\sum_{q|j}[gcd(p,q)==1]\frac{pj}{q} $大概感性证明一下吧我不会证然后开始推: \[ \sum_{i=1 ...
2018 ICPC 徐州网络赛 D. Easy Math（思维，反演，杜教筛）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划题目链接 https://nanti.jisuanke.com/t/A2003 Problem 计算 ...
Luogu P6055 [RC-02] GCD（莫比乌斯反演，杜教筛）（这题乐死我了，真就图一乐呗）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划 Weblink https://www.luogu.com.cn/problem/P6055 Prob ...
2018ACM四川省赛G.Grisaia（超棒的杜教筛好题）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划 G. Grisaia(灰色的果实好耶<灰色的果实(The Fruit of Grisaia)&g ...
【bzoj4176】Lucas的数论莫比乌斯反演+杜教筛
题目描述去年的Lucas非常喜欢数论题,但是一年以后的Lucas却不那么喜欢了. 在整理以前的试题时,发现了这样一道题目"求Sigma(f(i)),其中1<=i<=N" ...
【bzoj4916】神犇和蒟蒻杜教筛
题目描述很久很久以前,有一只神犇叫yzy; 很久很久之后,有一只蒟蒻叫lty; 输入请你读入一个整数N;1<=N<=1E9,A.B模1E9+7; 输出请你输出一个整数A=\sum_{ ...
【51nod2026】Gcd and Lcm（杜教筛）
题目传送门:51nod 我们可以先观察一下这个$f(x)=\sum_{d|x}\mu(d) \cdot d$. 首先它是个积性函数,并且$f(p^k)=1-p \ (k>0)$,这说明函数$f( ...
bzoj 3512: DZY Loves Math IV【欧拉函数+莫比乌斯函数+杜教筛】
参考:http://blog.csdn.net/wzf_2000/article/details/54630931 有这样一个显然的结论:当$ |\mu(n)|==1 $时,\( \phi(nk) ...
【51nod】1239 欧拉函数之和杜教筛
[题意]给定n,求Σφ(i),n<=10^10. [算法]杜教筛 [题解] 定义$s(n)=\sum_{i=1}^{n}\varphi(i)$ 杜教筛$\sum_{i=1}^{n}(\varph ...
[NOI2016]循环之美(杜教筛)
首先要求每个数互不相等,故有$x\perp y$. 可以发现$\frac{x}{y}$在$k$进制下为纯循环小数的充要条件为$x\cdot k^{len}\equiv x(mod\ y)$,即$y\p ...

B. Product（2019ICPC西安邀请赛）（杜教筛）

Product

B. Product（2019ICPC西安邀请赛）（杜教筛）相关推荐

最新文章

热门文章