uva10692-指数循环节

题目链接：https://vjudge.net/problem/UVA-10692

求 a1 ^ (a2 ^ (a3 ^ (... ^ an) )) % m 的值

利用指数循环节的降幂公式 a ^ b % mod = a ^ (b ^ phi(mod) + phi(mod)) % mod 递归的进行计算即可，phi()为欧拉函数，可以预处理出来

代码：

# include 
# include 
# include 
# include 
# include 
using namespace std;
typedef long long LL;
const int maxn = 10 + 5;
const int maxm = 1e4 + 5;
char s[maxn];
int a[maxn];
int phi[maxm];
int m, n;

int fast_pow(int x, int n, int mod) {
    int r = 1;
    while (n) {
        if (n & 1) r = (LL)r * x % mod;
        x = (LL)x * x % mod;
        n >>= 1;
    }
    return r;
}

void phi_table() {
    memset(phi, 0, sizeof phi);
    phi[1] = 1;
    for (int i = 2; i < maxm; ++i) if (!phi[i])
        for (int j = i; j < maxm; j += i) {
            if (!phi[j]) phi[j] = j;
            phi[j] = phi[j] / i * (i - 1);
        }
}

int solve(int id, int mod) {
    if (id == n - 1) return a[id] % mod;
    else return fast_pow(a[id], solve(id + 1, phi[mod]) + phi[mod], mod) % mod;
}

int main(void)
{
    phi_table();
    int Case = 0;
    while (scanf("%s", s) == 1 && strcmp(s, "#")) {
        sscanf(s, "%d", &m);
        scanf("%d", &n);
        for (int i = 0; i < n; ++i) scanf("%d", a + i);
        printf("Case #%d: %d\n", ++Case, solve(0, m));
    }

    return 0;
}

/*
10 4 2 3 4 5
100 2 5 2
53 3 2 3 2
#
*/