Portal -->loj6179

Solution

​　　 这题其实有一个式子一喵一样的版本在bzoj，但是那题是\(m\)特别大然后只有一组数据

　　 这题多组数据==

​　　

　　 首先根据\(\varphi(x)\)的通项\(\varphi(x)=x\prod\limits_{i=1}^{n}(1-\frac{1}{p1_i})=\prod\limits_{i=1}^{m}(p_i-1)p_i^{a_i-1}\)（其中\(n\)是\(x\)分解质因数之后没有去重的质因数列表\(p1\)的长度，\(m\)是去重之后质因数列表\(p\)的长度，\(x=\prod\limits_{i=1}^{m} p_i^{a_i}\)）我们有\(\varphi(i*j)=\varphi(i)*\varphi(j)*\frac{gcd(i,j)}{\varphi(gcd(i,j))}\)，具体就是因为\(\varphi(i)*\varphi(j)\)中\(gcd\)的质因子的部分被算了两次，但是除掉\(\varphi(gcd(i,j))\)之后又没有将\(gcd\)对\(a_i\)的贡献算上

​　　 然后我们就可以快乐推式子了，为了让接下来的式子更加简洁，我们默认\(n<=m\)：
\[ \begin{aligned} &\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\varphi(ij)\\ =&\sum\limits_{i=1}^n \sum\limits_{j=1}^m \varphi(i)\varphi(j)\frac{gcd(i,j)}{\varphi(gcd(i,j))}\\ =&\sum\limits_{d=1}^n\sum\limits_{i=1}^n\sum\limits_{j=1}^m\varphi(i)\varphi(j)\frac{d}{\varphi(d)}[gcd(i,j)=d]\\ =&\sum\limits_{d=1}^n\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}\varphi(id)\varphi(jd)\frac{d}{\varphi(d)}[gcd(i,j)=1]\\ =&\sum\limits_{d=1}^n\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}\varphi(id)\varphi(jd)\frac{d}{\varphi(d)}\sum\limits_{k|i,k|j}\mu(k)\\ =&\sum\limits_{d=1}^n\frac{d}{\varphi(d)}\sum\limits_{k=1}^{\lfloor\frac{n}{d}\rfloor}\mu(k)\sum\limits_{i=1}^{\lfloor\frac{n}{dk}\rfloor}\varphi(idk)\sum\limits_{j=1}^{\lfloor\frac{m}{dk}\rfloor}\varphi(jdk)\\ =&\sum\limits_{T=1}^n\sum\limits_{k|T}\mu(\frac{T}{k})\frac{k}{\varphi(k)}\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)\sum\limits_{j=1}^{\lfloor\frac{m}{T}\rfloor}\varphi(jT)\\ \end{aligned} \]
​　　 稍微说一下最后一步是相当于枚举\(dk\)，也就是令\(T=dk\)然后枚举\(T\)

​　　 然后我们可以令\(g(x)=\sum\limits_{k|x}\mu(\frac{x}{k})\frac{k}{\varphi(k)}\)，令\(s(i,j)=\sum\limits_{k=1}^j\varphi(ik)\)

　　 那么这个式子就可以写成：
\[ \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\varphi(ij)=\sum\limits_{i=1}^ng(i)\cdot s(i,\lfloor\frac{n}{i}\rfloor)\cdot s(i,\lfloor\frac{m}{i}\rfloor) \]
​
　　 
​　　 接下来看起来是怎么化也化不动了qwq但是我们发现\(g(i)\)和\(s(i,j)\)都可以在调和级数的复杂度内预处理出来，但是再接下来我们发现\(O(n)\)求解显然是不现实的

​　　 这时候当然是要大力分段啊，只不过光是普通的操作还是不行（前缀和这个东西很难搞），这里我们还需要一个黑科技，我们手动设定一个阈值\(TOP\)，然后对于\(i<=\frac{m}{TOP}\)的情况我们暴力算，对于\(i>\frac{m}{TOP}\)的情况，我们再预处理一个\(T[i][j][k]\)（也就是前缀和）：
\[ \begin{aligned} T[i][j][k]&=\sum\limits_{p=1}^kg(p)\sum\limits_{t=1}^i\varphi(tp)\sum\limits_{t=1}^i\varphi(tp)\\ &=\sum\limits_{p=1}^kg(p)\cdot s(p,i)\cdot s(p,j) \end{aligned} \]
​　　 然后当\(i>\frac{m}{TOP}\)的时候\(\lfloor\frac{m}{i}\rfloor<=TOP\)，所以我们\(i,j\)只要预处理到\(TOP\)然后直接用普通的分段操作来搞就好了

　　 代码大概长这个样子

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int N=1e5+10,MOD=998244353,TOP=35,inf=2147483647;
int p[N],g[N],miu[N],phi[N],inv[N];
vector<int>S[N];//s[i][j]=\sum_{k=1}^{j}phi[i*k]
vector<int> T[TOP+1][TOP+1];//T[i][j][k]=\sum_{p=1}^{k}\sum_{t=1}^{i}phi(t*p)\sum_{t=1}^{j}phi(t*p)
int vis[N];
int n,m,ans,T1;
void prework(int n){int cnt=0;miu[1]=1; phi[1]=1;for (int i=2;i<=n;++i){if (!vis[i])miu[i]=-1,p[++cnt]=i,phi[i]=i-1;for (int j=1;j<=cnt&&i*p[j]<=n;++j){vis[i*p[j]]=true;if (i%p[j]==0){miu[i*p[j]]=0; phi[i*p[j]]=phi[i]*p[j];break;}elsemiu[i*p[j]]=-miu[i],phi[i*p[j]]=phi[i]*phi[p[j]];}}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)for (int j=i;j<=n;j+=i){if (j==9)int debug=1;g[j]=(1LL*g[j]+1LL*miu[j/i]*(1LL*(i)*inv[phi[i]]%MOD)+MOD)%MOD;}for (int i=1;i<=n;++i){S[i].push_back(0);for (int j=1;j<=n/i;++j) S[i].push_back((S[i][j-1]+phi[i*j])%MOD);}for (int i=1;i<=TOP;++i)for (int j=1;j<=TOP;++j){T[i][j].push_back(0);for (int k=1;k<=n/i&&k<=n/j;++k)T[i][j].push_back((1LL*T[i][j][k-1]+1LL*g[k]*S[k][i]%MOD*S[k][j]%MOD)%MOD);}}
void solve(){if (n>m) swap(n,m);ans=0;for (int i=1;i<=m/TOP;++i)ans=(1LL*ans+1LL*g[i]*S[i][n/i]%MOD*S[i][m/i]%MOD)%MOD;for (int i=m/TOP+1,pos=0;i<=n;i=pos+1){pos=min(m/(m/i),(n/i)?n/(n/i):inf);ans=(1LL*ans+(1LL*T[n/i][m/i][pos]+MOD-T[n/i][m/i][i-1])%MOD)%MOD;}printf("%lld\n",ans);
}int main(){
#ifndef ONLINE_JUDGEfreopen("a.in","r",stdin);
#endifprework(N-10);//prework(10);scanf("%d",&T1);for (int o=1;o<=T1;++o){scanf("%d%d",&n,&m);solve();}
}