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 + ;
 long long a, b, c;
 char s[maxn];

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

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

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