题目
对于100%的数据，T<=1000,p<=10^7
题解
来捉这道神题
欧拉定理的一般形式：
\\[a^{m} \\equiv a^{m \\mod \\varphi(p) + [m \\ge \\varphi(p)]\\varphi(p)} \\pmod p
\\]
我们令
\\[ans(p) = 2^{2^{2^{...}}} \\mod p
\\]
那么有
\\[ans(p) = 2^{ans(\\varphi(p)) + \\varphi(p)} \\mod p
\\]
\\(O(\\log p)\\)递归即可
#include<iostream>
#include<cstdio>
#include<cmath>
#include<bitset>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<\' \'; puts("");
using namespace std;
const int maxn = 10000005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == \'-\') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
bitset<maxn> isn;
int p[maxn],phi[maxn],pi;
void init(){
phi[1] = 1;
for (int i = 2; i <= 10000000; i++){
if (!isn[i]) p[++pi] = i,phi[i] = i - 1;
for (int j = 1; j <= pi && i * p[j] <= 10000000; j++){
isn[i * p[j]] = true;
if (i % p[j] == 0){
phi[i * p[j]] = phi[i] * p[j];
break;
}
phi[i * p[j]] = phi[i] * (p[j] - 1);
}
}
}
int qpow(int a,int b,int p){
int ans = 1;
for (; b; b >>= 1,a = 1ll * a * a % p)
if (b & 1) ans = 1ll * ans * a % p;
return ans;
}
int Ans(int p){
if (p == 1) return 0;
return qpow(2,Ans(phi[p]) + phi[p],p);
}
int main(){
init();
int T = read(),p;
while (T--){
p = read();
printf("%d\\n",Ans(p));
}
return 0;
}