题目链接
解析
欧拉定理及扩展欧拉定理：
[a^b equiv
egin{cases}
a^{b mod phi(p)}, & gcd (a, p) = 1 \
a^b, & gcd(a, p)
e 1, &b < phi(p) \
a^{b mod phi(p) + phi(p)}, & gcd(a, p)
e 1, & b ge phi(p)
end{cases}
(mod p)
]
然后其实可以把第一个和第三个合起来得到(a^b equiv a^{b mod phi(p) + phi(p)} (mod p),b ge phi(p))
令题目要求的(2^{2^{2^{2^{..}}}}mod p)为(f(p))，那么(f(p) = 2^{f(phi(p)) + phi(p)} (mod p))
于是线性筛出(phi)，递归地求(f)即可，边界条件是(f(1) = 0)
代码
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#define MAXP 10000005
typedef long long LL;
int phi[MAXP], T, P;
void make_phi();
int calc(int);
int main() {
make_phi();
std::ios::sync_with_stdio(false);
std::cin >> T;
while (T--) {
std::cin >> P;
std::cout << calc(P) << std::endl;
}
return 0;
}
void make_phi() {
static char isn_prime[MAXP];
static std::vector<int> prime;
phi[1] = 1;
for (int i = 2; i <= MAXP; ++i) {
if (!isn_prime[i]) {
prime.push_back(i);
phi[i] = i - 1;
}
for (int j = 0; j < prime.size() && (LL)i * prime[j] <= MAXP; ++j) {
isn_prime[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; }
}
}
}
int qpower(int x, int y, int p) {
int res = 1;
while (y) {
if (y & 1) res = (LL)res * x % p;
x = (LL)x * x % p;
y >>= 1;
}
return res;
}
int calc(int p) {
if (p == 1) return 0;
return qpower(2, calc(phi[p]) + phi[p], p);
}
//Rhein_E