BZOJ3884 上帝与集合的正确用法

Description

Solution

令\(f(p)=2^{2^{2^{2^{2^{\cdots}}}}}\mod p\)。由于\(b>\phi(p)\)时\(a^b\equiv a^{(b\,mod\,\phi(p))+\phi(p)}(mod\;p)\)。所以\(f(p)=2^{f(\phi(p))+\phi(p)}\mod p\)。

由于易证\(\phi(\phi(p))\leq\frac p2\)，所以时间为\(O(\log p)\)。

Code

#include <cstdio>
typedef long long LL;
const int N = 10000050;
int phi
, pr[N / 10], cnt;
int pow_mod(int a, int b, int p) {
int ans = 1;
for (; b; b >>= 1, a = (LL)a * a % p)
if (b & 1) ans = (LL)ans * a % p;
return ans;
}
int calc(int p) {
// 2 ^ (2 ^ (...)) % p
if (p == 1) return 0;
return pow_mod(2, calc(phi[p]) + phi[p], p);
}
int main() {
for (int i = 1; i < N; ++i) phi[i] = i;
for (int i = 2; i < N; ++i) {
if (phi[i] == i)
--phi[pr[cnt++] = i];
for (int j = 0; j < cnt && (LL)pr[j] * i < N; ++j) {
if (i % pr[j])
phi[i * pr[j]] = phi[i] * phi[pr[j]];
else {
phi[i * pr[j]] = phi[i] * pr[j];
break;
}
}
}
int T, p;
scanf("%d", &T);
while (T--) {
scanf("%d", &p);
printf("%d\n", calc(p));
}
}