FZU1759(SummerTrainingDay04-B 欧拉降幂公式)

Problem 1759 Super A^B mod C

Accept: 1056    Submit: 3444
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

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

Source

FZU 2009 Summer Training IV--Number Theory

 

 //2017-08-04
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define ll long long 

 using namespace std;

 const int N = ;
 char b[N];
 ll a, c;

 ll quick_pow(ll a, ll n, ll MOD){
     ll ans = ;
     while(n){
         if(n&)ans = ans*a%MOD;
         a = a*a%MOD;
         n>>=;
     }
     return ans;
 }

 ll phi(ll n){
     ll ans = n;
     for(ll i = ; i*i <= n; i++){
         if(n%i==){
             ans -= ans/i;
             while(n%i==)
                 n /= i;
         }
     }
     if(n > )ans = ans - ans/n;
     return ans;
 }

 int main()
 {
     while(scanf("%lld%s%lld", &a, b, &c)!=EOF){
         ll len = strlen(b);
         ll MOD = phi(c), num = ;
         for(ll i = ; i < len; i++)
             num = (num* + b[i]-'')%MOD;
         num += MOD;
         printf("%lld\n", quick_pow(a, num, c));
     }
     return ;
 }