P4240 毒瘤之神的考验

Description

\(\mathcal{P}\text{ortal.}\)

Solution

首先想到要把 \(\varphi(ij)\) 拆开，这里有个公式

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

考虑证明，有

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

也就是 \(i,j\) 的并加上 \(i,j\) 的交。 接下来进入推柿子环节

\[\begin{aligned} \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m} \varphi(ij) &= \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m} \frac{\varphi(i)\varphi(j)\gcd(i,j)}{\varphi(\gcd(i,j))} \\&= \sum\limits_{d=1}^{n}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m} \frac{\varphi(i)\varphi(j)d[\gcd(i,j)=d]}{\varphi(d)} \\&= \sum\limits_{d=1}^{n}\frac{d}{\varphi(d)}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m }\varphi(i)\varphi(j)[\gcd(i,j)=d] \\&= \sum\limits_{d=1}^{n}\frac{d}{\varphi(d)}\sum_{t=1}^{n/d}\mu(t)\cdot \sum\limits_{i=1}^{n/(dt)}\sum\limits_{j=1}^{m/(dt)}\varphi(idt)\varphi(jdt) \\&= \sum\limits_{k=1}^{n}\sum\limits_{d|k} \frac{d\cdot\mu(k/d)}{\varphi(d)} \sum\limits_{i=1}^{n/k }\varphi(ik)\sum\limits_{j=1}^{m/k }\varphi(jk) \end{aligned} \]

记 \(\displaystyle f(k)=\sum\limits_{d|k} \frac{d\cdot\mu(k/d)}{\varphi(d)},g(k,n)=\sum\limits_{i=1}^{n}\varphi(ik)\)，容易发现这两个函数都容易在调和级数的复杂度内预处理出来。然而 \(\mathcal O(nT)\) 仍然是不现实的。如果是 CF 我就直接上了。

考虑怎么优化，还是将优化重点放在整除分块上

\[\sum_{i=1}^n\sum_{j=1}^m\varphi(ij)=\sum_{k=1}^nf(k)\cdot g(k,n/k)\cdot g(k,m/k) \]

记 \(\displaystyle h(a,b,n)=\sum_{k=1}^nf(k)\cdot g(k,a)\cdot g(k,b)\)，于是可以整除分块

\[\sum_{i=1}^n\sum_{j=1}^m\varphi(ij)=\sum_{n/l=n/r,\ m/l=m/r}h(n/r,m/r,r)-h(n/r,m/r,l-1) \]

问题是，预处理 \(h\) 的复杂度是很高的，这里可以考虑 根号分治 —— 设定阈值 \(B\)，将 \(a,b\leqslant B\) 的 \(h\) 预处理出来，不难发现这是 \(\mathcal O(nB^2)\) 的。当查询的时候，若 \(n/r\leqslant B\) 就直接 \(\mathcal O(1)\) 查询，否则有 \(r\leqslant n/B\)，干脆不差分直接暴力算就是 \(\mathcal O(n/B)\) 的。

复杂度 \(\mathcal O\left(n\ln n+nB^2+\left(n^{1/2}+n/B\right)T\right)\).

块长可以用均值不等式来算，总复杂度算出来大概是 \(\mathcal O(6\cdot 10^7)\).

Code

# include <cstdio>
# include <cctype>
# define print(x,y) write(x), putchar(y)

template <class T>
inline T read(const T sample) {
    T x=0; char s; bool f=0;
    while(!isdigit(s=getchar())) f|=(s=='-');
    for(; isdigit(s); s=getchar()) x=(x<<1)+(x<<3)+(s^48);
    return f? -x: x;
}
template <class T>
inline void write(T x) {
    static int writ[50], w_tp=0;
    if(x<0) putchar('-'), x=-x;
    do writ[++w_tp]=x-x/10*10, x/=10; while(x);
    while(putchar(writ[w_tp--]^48), w_tp);
}

# include <vector>
# include <iostream>
using namespace std;

const int B = 20;
const int maxn = 1e5+5;
const int mod = 998244353;

int inv(int x,int y=mod-2,int r=1) {
    for(; y; y>>=1, x=1ll*x*x%mod)
        if(y&1) r=1ll*r*x%mod; return r;
}

bool is[maxn];
vector <int> g[maxn]; int h[B+1][B+1][maxn];
int phi[maxn], mu[maxn], pc, p[maxn], f[maxn];

int beelzebul(int n,int m) {
    int ans=0; if(n>m) swap(n,m);
    for(int i=1;i<=m/B+1;++i)
        ans = (ans+1ll*f[i]*g[i][n/i]%mod*g[i][m/i]%mod)%mod;
    for(int l=m/B+2, r; l<=n; l=r+1) {
        r = min(n, min(n/(n/l),m/(m/l)));
        ans = (0ll+ans+h[n/r][m/r][r]-h[n/r][m/r][l-1])%mod;
    }
    return (ans+mod)%mod;
}

int func(int i,int a,int b) {
    if(a>=g[i].size() || b>=g[i].size()) return 0;
    return 1ll*f[i]*g[i][a]%mod*g[i][b]%mod;
}

void sieve() {
    phi[1]=mu[1]=1;
    for(int i=2;i<=maxn-5;++i) {
        if(!is[i]) p[++pc]=i,
        mu[i]=-1, phi[i]=i-1;
        for(int j=1; j<=pc && i*p[j]<=maxn-5; ++j) {
            is[i*p[j]] = true, mu[i*p[j]]=-mu[i];
            if(i%p[j]==0) {
                phi[i*p[j]] = phi[i]*p[j];
                mu[i*p[j]]=0; break;
            } phi[i*p[j]] = phi[i]*(p[j]-1);
        }
    }
    for(int i=1;i<=maxn-5;++i) {
        const int coe = 1ll*i*inv(phi[i])%mod;
        g[i].emplace_back(0);
        for(int j=i;j<=maxn-5;j+=i)
            f[j] = (1ll*coe*mu[j/i]+f[j])%mod,
            g[i].emplace_back((g[i][j/i-1]+phi[j])%mod);
    }
    for(int a=1;a<=B;++a) for(int b=1;b<=B;++b) 
        for(int i=1;i<=maxn-5;++i) 
            h[a][b][i] = (h[a][b][i-1]+func(i,a,b))%mod;
}

int main() {
    sieve();
    for(int T=read(9); T; --T) {
        int n=read(9), m=read(9);
        print(beelzebul(n,m),'\n');
    }
    return 0;
}