簡化題意,就是求出 A^(A^(A^(A^(A^(A)^))))(B+ 1層) % C 的值
ackerman函數的三階,A,B,C三個數都在1e6以內,這裏就要用到歐拉降冪
A^(B) % C 1.若 B >= C A^(B % φ(C) + φ(C))
2.若B < C (A^B) %C
這裏的φ(n)是歐拉函數,也就是小於n且與n互質的數的個數
因爲是b + 1層的次方 所以遞歸求解
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <ctype.h>
#define LL long long
#define ULL unsigned long long
#define mod 1000000007
#define INF 0x7ffffff
#define mem(a,b) memset(a,b,sizeof(a))
#define MODD(a,b) (((a%b)+b)%b)
using namespace std;
const double eps = 1e-2;
const int maxn = 1e6 + 5;
int phi[maxn],prime[maxn],vis[maxn];
int a;
void orla(int x)
{
int index = 0;
for(int i = 2; i < x; i++){
if(!vis[i]) {prime[index++] = i;phi[i] = i - 1;}
for(int j = 0; j < x && i * prime[j] < x; j++){
vis[i * prime[j]] = 1;
if(i % prime[j] == 0){
phi[i * prime[j]] = (phi[i] * prime[j]);
break;
}
else phi[i * prime[j]] = (phi[i] * phi[prime[j]]);
}
}
}
LL quick_pow(LL a,LL b,LL p)
{
LL ans = 1;
int flag = 0;
while(b){
if(b & 1){
ans *= a;
if(ans >= p){flag = 1;ans %= p;}
}
b >>= 1;
if(b == 0) break;
a = a * a;
if(a >= p){flag = 1; a %= p;}
}
return flag ? ans + p : ans;
}
LL solve(LL cnt,LL p)
{
if(p == 1 || cnt == 1) return a < p ? a : a % p + p;
LL k = phi[p];
return quick_pow(a,solve(cnt - 1,k),p);
}
int main()
{
orla(maxn);
int n,T;
scanf("%d",&T);
while(T--){
int b,c;
scanf("%d%d%d",&a,&b,&c);
if(b == 0) printf("%lld\n",(LL) 1 % c);
else printf("%lld\n",solve(b,c) % c);
}
return 0;
}