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农安县住房城乡建设局网站,宣传推广的十种方式,上海网站建设聚众网络,北京出大大事了题目 T(T<10)组样例，每次给出一个n(2<n<1e18)， 询问多少对，满足答案对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…

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;
}