南京網絡預選賽 The Preliminary Contest for ICPC Asia Nanjing 2019 B. super_log 歐拉降冪

模擬歐拉降冪

#include <bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <utility>
#include <cstdlib>
#include <cassert>
#include <ctime>
//#include <unordered_map>
#include <bitset> 
#include <stack>
#include <sstream>
#include <algorithm>
#include <functional>
#include <numeric>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <set>
#include <iomanip>
#include <cctype>
#define sf scanf
#define gr getchar()
#define pf printf
#define sz(type) (type).size()
#define pb push_back
#define mp make_pair
#define IT iterator
using namespace std;
typedef long long ll;
inline ll qpow(ll,ll,ll);
const int N = 1e6+5;
int prime[N],tot;
ll phi[N],maxn = 1e6;

void pre(int n){
	tot = 0;
	phi[1] = 1;
	memset(phi,0,sizeof(phi));
	for(int i=2;i<=n;i++){
		if(!phi[i]){
			prime[tot++]=i;
			phi[i]=1ll*(i-1);
		}
		for(int j=0;j<tot&&i*prime[j]<=n;j++){
			if(i%prime[j]!=0) phi[i*prime[j]]=phi[i]*phi[prime[j]];
			else{
				phi[i*prime[j]]=1ll*prime[j]*phi[i];
				break;
			}
		}
	}
}
inline ll euler(ll n){
	if(n <= 1ll*maxn) return phi[n];
	ll ans = n;
	for(int i=0;i<tot&&prime[i]*prime[i]<=n;i++){
		if(n%prime[i]==0){
			ans -= ans/prime[i];
			while(n%prime[i]==0) n/=prime[i];
		}
	}
	if(n > 1) ans -= ans/n;
	return ans;
}
inline ll check(ll x,ll n,ll mod){
	ll ans = 1;
	for(ll i=1;i<=n;i++){
		ans*=x;
		if(ans >= mod) return ans;
	}
	return ans;
}

inline ll solve(ll a,int n,ll p){
	if(n==0) return 1;
	if(p==1) return 1;
	ll exp = solve(a,n-1,euler(p)),ans;
	ans = check(a,exp,p);
	if(ans >= p) ans = qpow(a,exp,p) + p;
	return ans;
} 
int main(){
	int t,b;
	ll a,p;
	pre(maxn);
	sf("%d",&t);
	while(t--){
		sf("%lld%d%lld",&a,&b,&p);
		pf("%lld\n",solve(a,b,p)%p);
	}
    return 0;
}

inline ll qpow(ll x,ll n,ll mod) {
    ll ans = 1,a = x;
    while(n) {
        if(n&1) {
            ans = ans*a%mod;
        }
        a = a*a%mod;
        n>>=1;
    }
    return ans;
}