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

view code//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\n",quickpow(a,res,c));

    }

    else

    {

      res=0;

      for(int i=0;tb[i];i++)

      {

        res=(res*10+tb[i]-'0')%PHI;

      }

      printf("%I64d\n",quickpow(a,res+PHI,c));

    }

  }

  return 0;

}