Atcoder 212G Power Pair

Problem Statement

Given is a prime number \(P\).

How many pairs of integers \((x,y)\)satisfy the following conditions?

• \(0≤x≤P−1\)
• \(0≤y≤P−1\)
• There exists a positive integer \(n\) such that \(x^n≡y(modP)\)

Since the answer may be enormous, print it modulo 998244353.

Constraints

• \(2≤P≤10^12\)
• \(P is a prime number.\)

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Input

Input is given from Standard Input in the following format:

P

Output

Print the answer modulo 998244353.

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Sample Input 1 Copy

Copy

3

Sample Output 1 Copy

Copy

4

Four pairs \((x,y)=(0,0),(1,1),(2,1),(2,2)\)satisfy the conditions.

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Sample Input 2 Copy

Copy

11

Sample Output 2 Copy

Copy

64

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Sample Input 3 Copy

Copy

998244353

Sample Output 3 Copy

Copy

329133417

题目翻译

给定质数\(P\),求满足\(0≤x≤P−1\)和\(0≤y≤P−1\)，且\(x^n≡y(modP)\)的方案数，答案对998244353取模

题目解析

首先是本题显然可以考虑原根

假设原根为\(G\),那么\(x,y\)可以写成\(G^a,G^b\)，原方程变成\(an=b(mod \ P-1)\)

假设枚举\(a\),求\(b\)的方案数。\((a,b)\)有解当\(gcd(P-1,a)|b\)

所以\(b\)的方案数为\(\frac{P-1}{gcd(P-1,a)}\),总方案数为\(\sum_{a=1}^{P-1}\frac{P-1}{gcd(P-1,a)}\)

但是\(O(P)\)的复杂度不能通过\(P<=10^{12}\)

考虑到存在很多的\(a\)，\(gcd(P-1,a)\)的值相同。因为\(P-1\)的因数个数为\(O(\sqrt{P-1})\),所以\(gcd(P-1,a)\)的值只有\(O(\sqrt{n})\)个

考虑枚举\(P-1\)的因数，令\(g=gcd(P-1,a)\)，答案为\(\sum_{g}\frac{P-1}{g}*φ(\frac{P-1}{g})\)

这里\(φ(x)\)不适用\(O(P)\)的线性筛，可以通过推导式：

\(φ(n)=n(1-\frac{1}{d_1-1})...(1-\frac{1}{d_k-1})\)

预处理\(O(\sqrt{P-1})\)。令\(P-1\)因数数为\(d\)，求总方案复杂度为\(O(d^2)\)

P.S:根据官方题解，小于\(10^{12}\)因数最多的数是963761198400，有6720个因数。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include<map>
#include<cmath>
using namespace std;
typedef long long ll;
ll P;
ll fac[2000005],ans;
int cnt,Mod=998244353;
void getFactor(){
    for (ll i=1;i<=sqrt(P);i++){
        if (P%i==0){
            fac[++cnt]=i;
            if (P/i!=i)
                fac[++cnt]=P/i;
        }
    }
}
ll getPhi(ll n){
    ll phi=n;
    for (ll i=2;i<=sqrt(n);i++){
        if (n%i==0){
            phi=phi/i*(i-1);
            while (n%i==0) n/=i;
        }
    }
    if (n>1) phi=phi/n*(n-1);
    return phi%Mod;
}
int main(){
    cin>>P;
    P--;
    getFactor();
    for (int i=1;i<=cnt;i++){
        ans=(ans+(P/fac[i])%Mod*getPhi(P/fac[i])%Mod)%Mod;
    }
    cout<<ans+1;//考虑x=0,y=0
}