求幂塔函数

算出幂塔函数\(a^a\)共b个a乘幂的结果模m的值。

求a^(a(a^(a(a^...)))),a的a次方的a次方的a次方..%m

我只能想到欧拉降幂递归求解,但是要优化,不然会超时(下面那个优化我也没看懂mod的函数为什么要用在快速幂和递归return里面

在这个大佬的博客看到的https://www.cnblogs.com/uid001/p/11443528.html

#include <cstdio>
typedef long long LL;
LL phi[1000010];
LL maxn = 1000005;
void phi_table() {
    phi[0] = 0, phi[1] = 1;
    for(LL i = 2; i < maxn; ++i) phi[i] = i;
    for(LL i = 2; i < maxn; ++i) {
        if(phi[i] == i) {
            for(LL j = i; j < maxn; j += i)
                phi[j] = phi[j] / i * (i - 1);
        }
    }
}
LL mod(LL a, LL b) {
    return a < b ? a : a % b + b;
}
LL pow_mod(LL a, LL b, LL p) {
    LL ret = 1;
    while(b) {
        if(b & 1) ret = mod(ret * a, p);
        a = mod(a * a, p);
        b >>= 1;
    }
    return ret;
}
LL euler(LL a, LL b, LL c) {
    if(b == 1 || c == 1) return mod(a, c);
    return pow_mod(a, euler(a, b - 1, phi[c]), c);
}
int main() {
    phi_table();
    int t;
    LL a, b, m;
    scanf("%d", &t);
    while(t--) {
        scanf("%lld %lld %lld", &a, &b, &m);
        if(b == 0 || a == 1) printf("%lld\n", 1ll % m);
        else {
            LL ans = euler(a, b, m);
            printf("%d\n", ans % m);
        }
    }
    return 0;
}