Super A^B mod C

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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 42 10 1000

Sample Output

124

處理大數的方法挺經典的，把別人代碼貼出來以示膜拜

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;}