K - Super A^B mod C

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

这题的b非常大，只能用字符串处理；

需要一个降幂公式

然后带入欧拉公式，快速幂，快速乘法就可以了

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
typedef __int64 ll;
typedef long double ld;
//const ll mod=1e9+7;
using namespace std;
const double pi=acos(-1.0);
vector<int>v;
char x[1000005];
ll multi(ll a,ll b,ll c)//快速乘
{
	ll ans=0;
	while(b)
	{
		if(b&1)
		ans=(ans+a)%c;
		a=(a+a)%c;
		b>>=1;
	}
	return ans;
}
ll pow(ll a,ll b,ll c)//快速幂
{
	ll ans=1,bas=a;
	while(b)
	{
		if(b&1)
		ans=multi(ans,bas,c);
		bas=multi(bas,bas,c);
		b>>=1;
	}
	return ans;
}
ll p[1000100];
ll ola(ll n){ //欧拉函数
    ll i, j, r, aa;
    r = n;
    aa = n;
    mm(p,0);
    for(i=2; i<=sqrt(n); i++)
	{

            if(aa%i==0)
			{
                r = r/i*(i-1);
                while(aa%i==0)
                    aa /= i;
            }
    }
    if(aa>1)
        r = r/aa*(aa-1);
    return r;
}
int main()
{
	ll a,b,mod;
	while(~sf("%I64d %s %I64d",&a,&x,&mod))
	{
		ll cas=ola(mod);
		ll ans=0;
		int num=strlen(x);
		for(int i=0;i<num;i++)//求那个b%phi(c)
		{
			ans=ans*10+x[i]-48;
			ans%=cas;
		}
		if(ans<0) ans+=mod;
		pf("%I64d\n",pow(a,ans+cas,mod));
	}
}