#题面
传送门
#Sol
公式$$
ab\equiv
\begin
a{b%\phi(p)}gcd(a,p)=1\
a^bgcd(a,p)\neq1,b<\phi(p)\
a^{b%\phi(p)+\phi(p)}(modp)gcd(a,p)\neq1,b\geq\phi(p)
\end
\[
递归处理$\varphi$，一个数取$\varphi$约$log$次就成$1$了，暴力搞就行了
```cpp
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 1);
IL ll Read(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()){
if(c == '#') exit(0);
z = c == '-' ? -1 : 1;
}
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int m, prime[_], num, phi[_];
bool isprime[_];
IL void Sieve(){
isprime[1] = 1; phi[1] = 1;
for(RG int i = 2; i < _; ++i){
if(!isprime[i]) prime[++num] = i, phi[i] = i - 1;
for(RG int j = 1; j <= num && prime[j] * i < _; ++j){
isprime[i * prime[j]] = 1;
if(i % prime[j]) phi[i * prime[j]] = phi[i] * (prime[j] - 1);
else{ phi[i * prime[j]] = phi[i] * prime[j]; break; }
}
}
}
IL int Pow(RG ll x, RG ll y, RG ll p){
RG int flg2 = 0, flg1 = 0; RG ll cnt = 1;
for(; y; y >>= 1){
if(y & 1) flg1 |= (cnt * x >= p || flg2), cnt = cnt * x % p;
flg2 |= (x * x >= p); x = x * x % p;
}
return cnt + flg1 * p;
}
IL int Calc(RG int p){
if(p == 1) return Pow(2, 1, p);
return Pow(2, Calc(phi[p]), p);
}
int main(RG int argc, RG char* argv[]){
Sieve(); for(RG int T = Read(); T; --T) m = Read(), printf("%d\n", Calc(m) % m);
return 0;
}
```
\]