FZU 1759 Super A^B mod C （欧拉降幂）

Problem:
给一个很大的B，求A^B mod C的值。
Solution:
通过欧拉降幂把B缩小，其中要用到高精度取余和快速幂取余。
欧拉降幂：

#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<vector>
#include<fstream>
#include<list>
#include<iomanip>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
#define ms(s) memset(s,0,sizeof(s))

const double PI = 3.141592653589;
const int INF = 0x3fffffff;

long long euler_phi(long long n) {
    long long m = sqrt(n);
    long long ans = n;
    for(long long i = 2; i <= m; i++) {
        if(n%i == 0) { //用变量保存i即最大的素因数
            ans = ans / i * (i-1);
            while(n%i == 0)
                n /= i;
        }
        if(n == 1)  break;
    }
    if(n > 1)
        ans = ans / n * (n-1);
    return ans;
}

long long q_mod(long long a, long long b, long long m){
    long long ans = 1;
    for( ; b; b >>= 1){
        if(b & 1)
            ans = ans*a % m;
        a = a*a % m;
    }
    return ans;
}

long long long_mod(string a, long long b){
    int flag = 1, idx = 0;
    if(a[0] == '-'){
        flag = -1;  idx++;
    }
    long long r = a[idx++] - 48;
    while(idx < a.length()){
        r = (r*10 + a[idx]-48) % b;
        idx++;
    }
    return r*flag;
}

int main() {
//    freopen("/Users/really/Documents/code/input","r",stdin);
//    freopen("/Users/really/Documents/code/output","w",stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    ll a, b, c, pc;
    string b_in;
    while(cin >> a >> b_in >> c) {
        pc = euler_phi(c);
        b = long_mod(b_in, pc) + pc;
        cout << q_mod(a, b, c) << endl;
    }


    return 0;
}