【FZU - 1759】Super A^B mod C （数论，快速幂，快速乘，欧拉降幂，指数循环节，模板）

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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;
}