题面

传送门

Sol

公式\[ a^b\equiv \begin{cases} a^{b\%\phi(p)}~~~~~~~~~~~gcd(a,p)=1\a^b~~~~~~~~~~~~~~~~~~gcd(a,p)\neq1,b<\phi(p)\a^{b\%\phi(p)+\phi(p)}~~~~gcd(a,p)\neq1,b\geq\phi(p) \end{cases}~~~~~~~(mod~p) \]
递归处理\(\varphi\)，一个数取\(\varphi\)约\(log\)次就成\(1\)了，暴力搞就行了

# 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;
}