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;}