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;
}