P4240 毒瘤之神的考验

题目

P4240 毒瘤之神的考验

神仙题\(emmm\)

前置

首先有一个很神奇的性质：

\(\varphi(ij)=\dfrac{\varphi(i)\varphi(j)gcd(i,j)}{\varphi(gcd(i,j))}\)

证明：

\[\begin{aligned}
\varphi(i)\varphi(j) &= i\prod\limits_{p|i}\frac{p-1}{p}j\prod\limits_{p|j}\frac{p-1}{p}\\
&= ij\prod\limits_{p|ij}\frac{p-1}{p}\prod\limits_{q|gcd(i,j)}\frac{p-1}{p}
\end{aligned}\]

\[\therefore \varphi(i)\varphi(j)gcd(i,j)=\varphi(ij)\varphi(gcd(i,j))
\]

具体做法

求\(\sum_{i=1}^n\sum_{j=1}^m\varphi(i,j)\)

我们最终是能得到：$$\begin{aligned}

\sum\limits_{i=1}^{{n}\sum\limits_{j=1}}{m}\varphi(ij)=\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}$$

网上好像很多人的证明都有些错误

证明：

\[\begin{aligned}\\
Ans=&\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\varphi(ij)\\
=&\sum\limits_{i=1}^n \sum\limits_{j=1}^m \varphi(i)\varphi(j)\frac{gcd(i,j)}{\varphi(gcd(i,j))}\\
=&\sum\limits_{d=1}^n\sum\limits_{i=1}^n\sum\limits_{j=1}^m\varphi(i)\varphi(j)\frac{d}{\varphi(d)}[gcd(i,j)=d]\\
=&\sum\limits_{d=1}^n\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}\varphi(id)\varphi(jd)\frac{d}{\varphi(d)}[gcd(i,j)=1]\\
=&\sum\limits_{d=1}^n\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}\varphi(id)\varphi(jd)\frac{d}{\varphi(d)}\sum\limits_{k|i,k|j}\mu(k)\\
=&\sum\limits_{d=1}^n\frac{d}{\varphi(d)}\sum\limits_{k=1}^{\lfloor\frac{n}{d}\rfloor}\mu(k)\sum\limits_{i=1}^{\lfloor\frac{n}{dk}\rfloor}\varphi(idk)\sum\limits_{j=1}^{\lfloor\frac{m}{dk}\rfloor}\varphi(jdk)\\
=&\sum\limits_{T=1}^n\sum\limits_{k|T}\mu(\frac{T}{k})\frac{k}{\varphi(k)}\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)\sum\limits_{j=1}^{\lfloor\frac{m}{T}\rfloor}\varphi(jT)\\
\end{aligned}\]

看我们最终得到的式子，多次查询完全在线做，参数这么多时间肯定承受不住

\(\sum\limits_{k|T}\mu(\frac{T}{k})\frac{k}{\varphi(k)}\)设有函数\(F(x)=\sum\limits_{k|x}\mu(\frac{x}{k})\frac{k}{\varphi(k)}\)再普通不过就不多讲了

\(\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)\)参数有两个，设有函数\(G(y,x)=\sum\limits_{i=1}^{x}\varphi(iy)\)

显然\(G(y,x)=G(y,x-1)+\varphi(xy)\)

整个式子设为函数\(S(y,z,x)=\sum\limits_{T=1}^x\sum\limits_{k|T}\mu(\frac{T}{k})\frac{k}{\varphi(k)}\sum\limits_{i=1}^{y}\varphi(iT)\sum\limits_{j=1}^{z}\varphi(jT)\)

显然\(S(y,z,x)=S(y,z,x-1)+F[x]*G(x,y)*G(x,z)\)

\(F(x)(x<=n),G(y,x)(x<=n,y<=B),S(y,z,x)(x<=n,y,z<=B)\)

B是自己取的任意参数，或许你会想\(\frac{n}{T}>B\)怎么办，\(\Rightarrow T<\frac{n}{B}\)，这部分暴力算，后面分块

#include<cstring>
#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
const int B=35;
const int maxn=1e5+9;
const LL p=998244353;
inline int Read(){
    int x(0),f(1); char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-') f=-1; c=getchar();
    }
    while(c>='0'&&c<='9')
        x=(x<<3)+(x<<1)+c-'0',c=getchar();
    return x*f;
}
bool visit[maxn];
int mu[maxn],phi[maxn],prime[maxn];
int *G[maxn],*S[B+1][B+1],F[maxn];
int inv[maxn];

inline void First(LL N){
    mu[1]=phi[1]=inv[1]=1;
    int tot(0);
    for(int i=2;i<=N;++i){
        if(!visit[i]){
            prime[++tot]=i,
            mu[i]=-1,
            phi[i]=i-1;
        }
        for(int j=1;j<=tot&&i*prime[j]<=N;++j){
            visit[i*prime[j]]=true;
            if(i%prime[j]==0){
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }else{
                phi[i*prime[j]]=phi[i]*phi[prime[j]],
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
    for(int i=2;i<=N;++i)
        inv[i]=p-1ll*p/i*inv[p%i]%p;
    for(int i=1;i<=N;++i)
        for(int j=1;j*i<=N;++j)
            F[i*j]=1ll*(1ll*F[i*j]+1ll*i*inv[phi[i]]%p*mu[j]%p)%p;
    for(int i=1;i<=N;++i){
        G[i]=new int [N/i+1];
        G[i][0]=0;
        for(int j=1;j<=N/i;++j)
            G[i][j]=1ll*(1ll*G[i][j-1]+1ll*phi[j*i])%p;
    }
    for(int j=1;j<=B;++j)
        for(int k=1;k<=B;++k){
        	int len(N/(max(j,k)));
        	S[j][k]=new int [len+1];
        	S[j][k][0]=0;
        	for(int i=1;i<=len;++i)
        		S[j][k][i]=1ll*(1ll*S[j][k][i-1]+1ll*F[i]*G[i][j]%p*G[i][k]%p)%p;
        }
}
inline LL Solve(int n,int m){
    if(n>m)
        swap(n,m);
    LL ans(0);
    for(int i=1;i<=m/B;++i)
        ans=(ans+1ll*F[i]*G[i][n/i]%p*G[i][m/i]%p)%p;
    for(int l=m/B+1,r;l<=n;l=r+1){
        r=min(n/(n/l),m/(m/l));
        ans=(ans+1ll*(S[n/l][m/l][r]-S[n/l][m/l][l-1]+p)%p)%p;
    }
    return ans;
}
int main(){
    int T=Read();
    First(100000);
    while(T--){
        int n(Read()),m(Read());
        printf("%lld\n",Solve(n,m));
    }
    return 0;
}/*
3
1 1
2 2
3 3

1 5 19
*/