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The 2023 ICPC Asia Regionals Online Contest (1) E. Magical Pair（数论欧拉函数）

news 来源：原创 2025/4/22 6:00:02

题目

T(T<=10)组样例，每次给出一个n(2<=n<=1e18)，

询问多少对 $(x,y)(0<=x,y<=n^2-n)$ ，满足 $x^y\equiv y^x(mod \ n)$

答案对998244353取模，保证n-1不是998244353倍数

思路来源

OEIS、SSerxhs、官方题解

2023 ICPC 网络赛第一场简要题解 - 知乎

题解

~~官方题解还没有补，OEIS打了个表然后就通过了~~

这里给一下SSerxhs教的做法吧（图源：我、tanao学弟）

SSerxhs代码

我的理解

首先，证一下这个和 $\sum_{i=0}^{p-2}b_{i}^2$ 是等价的，其中bi为满足 $x^y\equiv i$ 的(x,y)的对数

关于这部分，题解里给的中国剩余定理的构造，也很巧妙

剩下的部分就很神奇了，首先需要注意到各个素因子的贡献是独立的，可以连积

对于求某个素因子的幂次的贡献时，用到了解的分布是均匀的性质

代码1（OEIS）

#include<bits/stdc++.h>
using namespace std;
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
map<ll,ll>ans;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}
 
ll n;
 
In ll mul(ll a, ll b, ll mod) 
{
    ll d = (long double)a / mod * b + 1e-8; 
    ll r = a * b - d * mod;
    return r < 0 ? r + mod : r;
}
In ll quickpow(ll a, ll b, ll mod)
{
    ll ret = 1;
    for(; b; b >>= 1, a = mul(a, a, mod))
        if(b & 1) ret = mul(ret, a, mod);
    return ret;
}
 
In bool test(ll a, ll d, ll n)
{
    ll t = quickpow(a, d, n);
    if(t == 1) return 1;
    while(d != n - 1 && t != n - 1 && t != 1) t = mul(t, t, n), d <<= 1;
    return t == n - 1;                
}
int a[] = {2, 3, 5, 7, 11};
In bool miller_rabin(ll n)
{
    if(n == 1) return 0;
    for(int i = 0; i < 5; ++i)     
    {
        if(n == a[i]) return 1;
        if(!(n % a[i])) return 0;
    }
    ll d = n - 1;
    while(!(d & 1)) d >>= 1;
    for(int i = 1; i <= 5; ++i)   
    {
        ll a = rand() % (n - 2) + 2;
        if(!test(a, d, n)) return 0;
    }
    return 1;
}
 
In ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
In ll f(ll x, ll a, ll mod) {return (mul(x, x, mod) + a) % mod;}
const int M = (1 << 7) - 1;     
ll pollard_rho(ll n)                   
{
    for(int i = 0; i < 5; ++i) if(n % a[i] == 0) return a[i];
    ll x = rand(), y = x, t = 1, a = rand() % (n - 2) + 2;
    for(int k = 2;; k <<= 1, y = x) 
    {
        ll q = 1;
        for(int i = 1; i <= k; ++i) 
        {
            x = f(x, a, n);
            q = mul(q, abs(x - y), n);
            if(!(i & M))   
            {
                t = gcd(q, n);
                if(t > 1) break;    
            }
        }
        if(t > 1 || (t = gcd(q, n)) > 1) break;  
    }
    return t;
}
In void find(ll x)
{
    if(x == 1 ) return;
    if(miller_rabin(x)) {ans[x]++;return;}
    ll p = x;
    while(p == x) p = pollard_rho(x);
    while(x % p == 0) x/=p;
    find(p); find(x);
}
const ll mod=998244353;
ll modpow(ll x,ll n){
	x%=mod;
	if(!x)return 0;
	ll res=1;
	for(;n;n>>=1,x=1ll*x*x%mod){
		if(n&1)res=1ll*res*x%mod;
	}
	return res;
}
ll cal(ll p,ll e){
	//printf("p:%lld e:%lld\n",p,e);
	return (modpow(p,e+1)+modpow(p,e)-1+mod)%mod*modpow(p,2*e-1)%mod;
}
int main()
{
    srand(time(0));
    int T = read();
    while(T--)
    {
    	ans.clear();
        n = read();
        ll m=n-1;
        find(m);
        ll phi=m%mod,res=1;
        for(auto &v:ans){
        	ll p=v.first,e=0;
        	while(m%p==0)m/=p,e++;
        	res=res*cal(p,e)%mod;
        }
        res=(res+phi*phi%mod)%mod;
        printf("%lld\n",res);
    }
    return 0;
}

代码2（SSerxhs代码）

#include<bits/stdc++.h>
using namespace std;
#define In inline
typedef long long ll;
typedef double db;
typedef pair<ll,int> P;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
map<ll,int>ans;
vector<P>ans2;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}
 
ll n;
 
In ll mul(ll a, ll b, ll mod) 
{
    ll d = (long double)a / mod * b + 1e-8; 
    ll r = a * b - d * mod;
    return r < 0 ? r + mod : r;
}
In ll quickpow(ll a, ll b, ll mod)
{
    ll ret = 1;
    for(; b; b >>= 1, a = mul(a, a, mod))
        if(b & 1) ret = mul(ret, a, mod);
    return ret;
}
 
In bool test(ll a, ll d, ll n)
{
    ll t = quickpow(a, d, n);
    if(t == 1) return 1;
    while(d != n - 1 && t != n - 1 && t != 1) t = mul(t, t, n), d <<= 1;
    return t == n - 1;                
}
int a[] = {2, 3, 5, 7, 11};
In bool miller_rabin(ll n)
{
    if(n == 1) return 0;
    for(int i = 0; i < 5; ++i)     
    {
        if(n == a[i]) return 1;
        if(!(n % a[i])) return 0;
    }
    ll d = n - 1;
    while(!(d & 1)) d >>= 1;
    for(int i = 1; i <= 5; ++i)   
    {
        ll a = rand() % (n - 2) + 2;
        if(!test(a, d, n)) return 0;
    }
    return 1;
}
 
In ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
In ll f(ll x, ll a, ll mod) {return (mul(x, x, mod) + a) % mod;}
const int M = (1 << 7) - 1;     
ll pollard_rho(ll n)                   
{
    for(int i = 0; i < 5; ++i) if(n % a[i] == 0) return a[i];
    ll x = rand(), y = x, t = 1, a = rand() % (n - 2) + 2;
    for(int k = 2;; k <<= 1, y = x) 
    {
        ll q = 1;
        for(int i = 1; i <= k; ++i) 
        {
            x = f(x, a, n);
            q = mul(q, abs(x - y), n);
            if(!(i & M))   
            {
                t = gcd(q, n);
                if(t > 1) break;    
            }
        }
        if(t > 1 || (t = gcd(q, n)) > 1) break;  
    }
    return t;
}
In void find(ll x)
{
    if(x == 1 ) return;
    if(miller_rabin(x)) {ans[x]++;return;}
    ll p = x;
    while(p == x) p = pollard_rho(x);
    while(x % p == 0) x/=p;
    find(p); find(x);
}
const ll mod=998244353;
ll modpow(ll x,ll n){
	x%=mod;
	if(!x)return 0;
	ll res=1;
	for(;n;n>>=1,x=1ll*x*x%mod){
		if(n&1)res=1ll*res*x%mod;
	}
	return res;
}
ll cal(ll p,ll e){
	//printf("p:%lld e:%lld\n",p,e);
	return (modpow(p,e+1)+modpow(p,e)-1+mod)%mod*modpow(p,2*e-1)%mod;
}
ll sol(){
	ll ta=1;//tt=1;
	for(auto &x:ans2){
		ll p=x.first,ans=0;
		int k=x.second;
		p%=mod;
		vector<ll> f(k+1),pw(k+1),num(k+1);
		pw[0]=1;
		rep(i,1,k)pw[i]=pw[i-1]*p%mod;
		rep(i,0,k-1)num[i]=(pw[k-i]+mod-pw[k-i-1])%mod;
		num[k]=1;
		rep(i,0,k){
			ll ni=num[i];
			rep(j,0,k){
				ll nj=num[j];
				f[min(k,i+j)]=(f[min(k,i+j)]+ni*nj%mod)%mod;
			}
		}
		rep(i,0,k){
			ll tmp=f[i]*modpow(num[i],mod-2)%mod;
			ans=(ans+tmp*tmp%mod*num[i]%mod)%mod;
		}
		ta=ta*ans%mod;
	}
	return ta;
}
int main()
{
    srand(time(0));
    int T = read();
    while(T--)
    {
    	ans.clear();
        ans2.clear();
        n = read();
        ll m=n-1;
        find(m);
        //ll phi=m%mod,res=1;
        for(auto &v:ans){
        	ll p=v.first,e=0;
        	while(m%p==0)m/=p,e++;
        	ans2.push_back(P(p,e));
        	//res=res*cal(p,e)%mod;
        }
        m=(n-1)%mod;
        ll res=sol();
        res=(res+m*m%mod)%mod;
        printf("%lld\n",res);
        //res=(res+phi*phi%mod)%mod;
        //printf("%lld\n",res);
    }
    return 0;
}