题干：

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

解题报告：

   裸题，仅供模板准备。

AC代码：

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int N=1000005;
typedef long long LL;
char str[N];
int phi(int n) {int rea = n;for(int i=2; i*i<=n; i++) {if(n % i == 0) {rea = rea - rea / i;while(n % i == 0) n /= i;}}if(n > 1)rea = rea - rea / n;return rea;
}
LL multi(LL a,LL b,LL m) {LL ans = 0;a %= m;while(b) {if(b & 1) {ans = (ans + a) % m;b--;}b >>= 1;a = (a + a) % m;}return ans;
}
LL quick_mod(LL a,LL b,LL m) {LL ans = 1;a %= m;while(b) {if(b & 1) {ans = multi(ans,a,m);b--;}b >>= 1;a = multi(a,a,m);}return ans;
}
void Solve(LL a,char str[],LL c) {LL len = strlen(str);LL ans = 0;LL p = phi(c);if(len <= 15) {for(int i=0; i<len; i++)ans = ans * 10 + str[i] - '0';} else {for(int i=0; i<len; i++) {ans = ans * 10 + str[i] - '0';ans %= p;}ans += p;}printf("%I64d\n",quick_mod(a,ans,c));
}
int main() {LL a,c;while(~scanf("%I64d%s%I64d",&a,str,&c))Solve(a,str,c);return 0;
}