HDU 1211 RSA 逆元 快速模取幂

/**   * url: http://acm.hdu.edu.cn/showproblem.php?pid=1211 RSA  * Author: Johnsondu  * Time: 2012-7-28 18:00 around  * Stratege: 逆元(扩展欧几里德) , 快速模取幂  * Status: 63642132012-07-28 18:12:54Accepted12110MS256K951 BC++johnsondu*/#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <stack>using namespace std ;typedef __int64 lld ;lld p, q, e, l ;lld n, f, pd ;lld c, M ;void exGcd (lld a, lld b, lld &d, lld &x, lld &y){if (b == 0){x = 1 ;y = 0 ;d = a ;return ;}exGcd (b, a%b, d, x, y) ;lld tmp = x ;x = y ;y = tmp - a/b*y ;}lld ModPow (lld a, lld pn, lld Mod){int tmp = 1 ;while (pn){if (pn&1)tmp = tmp*a % Mod ;a = (a*a) % Mod ;pn >>= 1 ;}return tmp ;}int main () {lld x, y, d, a, b ;while (scanf ("%I64d%I64d%I64d%I64d", &p, &q, &e, &l) != EOF){n = p * q ;f = (p-1) * (q-1) ;exGcd (e, f, d, x, y) ;pd = (x % f + f) % f ;for (int i = 0; i < l; i ++){scanf ("%I64d", &c) ;M = ModPow (c, pd, n) ;printf ("%c", char(M)) ;}printf ("\n") ;}return 0 ;}