BZOJ3884上帝与集合的正确用法
题目链接
解析
欧拉定理及扩展欧拉定理：
\[a^b \equiv
\begin{cases}
a^{b \ mod \ \phi(p)}, & \gcd (a, p) = 1 \\
a^b, & \gcd(a, p) \ne 1, &b < \phi(p) \\
a^{b \ mod \ \phi(p) + \phi(p)}, & \gcd(a, p) \ne 1, & b \ge \phi(p)
\end{cases}
\ (mod \ p)
\]
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
