Super A^B mod C （快速幂+欧拉函数+欧拉定理）

题目链接：http://acm.fzu.edu.cn/problem.php?pid=1759

题目：Problem Description

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).

Input

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 4 2 10 1000

Sample Output

1 24

思路：由于B的数据太大，因而我们需要借助欧拉定理来进行降幂，然后再用快速幂解决。

代码实现如下：

#include <cstdio>
#include <cstring>

const int maxn = 1e6 + 7;
long long a, b, c;
char s[maxn];

int euler(int x) {
    int ans = x;
    for(int i = 2; i * i <= x; i++) {
        if(x % i == 0) {
            ans = ans / i * (i - 1);
            while(x % i == 0) x /= i;
        }
    }
    if(x > 1) ans = ans / x * (x - 1);
    return ans;
}

long long ModPow(int n, int p, int mod) {
    long long rec = 1;
    while(p) {
        if(p & 1) rec = rec * n % mod;
        n = (long long) n * n % mod;
        p >>= 1;
    }
    return rec;
}

int main() {
    while(~scanf("%lld%s%lld", &a, s, &c)) {
        b = 0;
        int rec = euler(c);
        int len = strlen(s);
        for(int i = 0; i < len; i++) {
            b = b % rec;
            b = ((s[i] - '0') + 10 * b) % rec;
        }
        b += rec;
        printf("%lld
", ModPow(a, b, c));
    }
    return 0;
}