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 = 1000010;
char b[N];
ll a, c;

ll quick_pow(ll a, ll n, ll MOD){
    ll ans = 1;
    while(n){
        if(n&1)ans = ans*a%MOD;
        a = a*a%MOD;
        n>>=1;
    }
    return ans;
}

ll phi(ll n){
    ll ans = n;
    for(ll i = 2; i*i <= n; i++){
        if(n%i==0){
            ans -= ans/i;
            while(n%i==0)
                n /= i;
        }
    }
    if(n > 1)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 = 0;
        for(ll i = 0; i < len; i++)
            num = (num*10 + b[i]-'0')%MOD;
        num += MOD;
        printf("%lld
", quick_pow(a, num, c));
    }
    return 0;
}