当(gcd(a,m)=1)时就是欧拉定理

后面两种情况对(gcd)无要求，注意(c)与(phi(m))的关系

求(2^{2^{2^{2^{cdots}}}}(mod　p))
简化为求(xequiv 2^x　(mod 　p)))

(Solution)

因为(2^{2^{2^{2^{cdots}}}} > phi (p))
且(phi(p))在不断缩小，所以用第三个公式即可
用(ans(p))表示对(p)取模的答案，显然(ans(p)=2^{ans(phi(p))+phi(p)})
(phi(p))在不断缩小，直到为(1)时，则任何数(mod　1=0)，所以到达递归边界，返回(0)即可

(Code)

#include <cstdio>
#define int long long

typedef long long ll;
const int N = 1e7+71;

int p_(ll x, int y, int z) {
    int ret = 1;
    for (; y; (x *= x) %= z, y >>= 1) if (y & 1) (ret *= x) %= z;
    return ret;
}
signed phi[N];
ll solve(int p) {
    if (p == 1) return 0;
    return p_(2, solve(phi[p]) + phi[p], p);
}
int t, p;

signed main() {
    phi[1] = 1;
    for (signed i = 2; i < N; i++) if (!phi[i]) {
        for (signed j = i; j < N; j += i) {
            if (!phi[j]) phi[j] = j;
            phi[j] = phi[j] / i * (i-1);
        }
    }
    for (scanf("%lld", &t); t--; ) {
        scanf("%lld", &p);
        printf("%lld
", solve(p));
    }
}