uva 10692

题目链接：uva 10692 - Huge Mods

题目大意：给出一个数的次方形式，就它模掉M的值。

解题思路：依据剩余系的性质，最后一定是行成周期的，所以就有ab=abmod(phi[M])+phi[M](phi[M]为M的欧拉函数)，这样就能够依据递归去求解。

#include <cstdio>
#include <cstring>
#include <cmath>

const int maxn = 15;

int A[maxn], k;

int pow_mod (int a, int n, int M) {
    int ans = 1;

    while (n) {
        if (n&1)
            ans = ans * a % M;
        a = a * a % M;
        n /= 2;
    }
    return ans;
}

int euler_phi(int n) {
    int m = (int)sqrt(n+0.5);
    int ans = n;
    for (int i = 2; i <= m; i++) {
        if (n % i == 0) {
            ans = ans / i * (i-1);
            while (n%i==0)
                n /= i;
        }
    }

    if (n > 1)
        ans = ans / n * (n - 1);
    return ans;
}

int solve (int d, int M) {
    if (d == k - 1)
        return A[d]%M;

    int phi = euler_phi(M);
        int c = solve (d+1, phi) + phi;
    return pow_mod(A[d], c, M);
}

int main () {
    int cas = 1;
    char str[maxn];

    while (scanf("%s", str) == 1 && strcmp(str, "#")) {
        int M;
        sscanf(str, "%d", &M);
        scanf("%d", &k);
        for (int i = 0; i < k; i++)
            scanf("%d", &A[i]);

        printf("Case #%d: %d
", cas++, solve(0, M));
    }
    return 0;
}