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

处理大数的方法挺经典的，把别人代码贴出来以示膜拜

//A^B %C=A^( B%phi(C)+phi(C) ) %C
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include<string>
#include<cmath>
using namespace std;
typedef __int64 ll;
int phi(int x)
{
  int i,j;
  int num = x;
  for(i = 2; i*i <= x; i++)
  {
    if(x % i == 0)
    {
      num = (num/i)*(i-1);
      while(x % i == 0)
      {
        x = x / i;
      }
    }
  }
  if(x != 1) num = (num/x)*(x-1);
  return num;
}
ll quickpow(ll m,ll n,ll k)
{
  ll ans=1;
  while(n)
  {
    if(n&1) ans=(ans*m)%k;
    n=(n>>1);
    m=(m*m)%k;
  }
  return ans;
}
char tb[1000015];
int main()
{
  ll a,nb;
  int c;
  while(scanf("%I64d%s%d",&a,tb,&c)!=EOF)
  {
    int PHI=phi(c);
    ll res=0;
    for(int i=0;tb[i];i++)
    {
      res=(res*10+tb[i]-'0');
      if(res>c)break;
    }
    if(res<=PHI)
    {
      printf("%I64d
",quickpow(a,res,c));
    }
    else
    {
      res=0;
      for(int i=0;tb[i];i++)
      {
        res=(res*10+tb[i]-'0')%PHI;
      }
      printf("%I64d
",quickpow(a,res+PHI,c));
    }
  }
  return 0;
}