bzoj 1420 Discrete Root - 原根 - exgcd - BSGS

题目传送门

　　戳我来传送

题目大意

　　给定$k, p, a$，求$x^{k}\equiv a \pmod{p}$在模$p$意义下的所有根。

　　考虑模$p$下的某个原根$g$。

　　那么$x  = g^{ind_{g}x}, a = g^{ind_{g}a}$。

　　所以原方程转化为$g^{k\cdot ind_{g}x}\equiv g^{ind_{g}a} \pmod{p}$。

　　所以方程等价于$k\cdot ind_{g}x \equiv ind_{g}a \pmod{\varphi(p)}$。

　　用exgcd解出$ind_{g}x$的所有可能的解，再用快速幂算出$x$即可。

Code

 /**
  * bzoj
  * Problem#1420
  * Accepted
  * Time: 44ms
  * Memory: 2032k
  */
 #include <bits/stdc++.h>
 #include <ext/pb_ds/hash_policy.hpp>
 #include <ext/pb_ds/assoc_container.hpp>
 using namespace std;
 using namespace __gnu_pbds;
 typedef bool boolean;

 typedef class HashMap {
     private:
         static const int M = ;
     public:
         int ce;
         int h[M], key[M], val[M], next[M];

         HashMap() {    }

         void insert(int k, int v) {
             int ha = k % M;
             for (int i = h[ha]; ~i; i = next[i])
                 if (key[i] == k) {
                     val[i] = v;
                     return;
                 }
             ++ce, key[ce] = k, val[ce] = v, next[ce] = h[ha], h[ha] = ce;
         }

         int operator [] (int k) {
             int ha = k % M;
             for (int i = h[ha]; ~i; i = next[i])
                 if (key[i] == k)
                     return val[i];
             return -;
         }

         void clear() {
             ce = -;
             memset(h, -, sizeof(h));
         }
 }HashMap; 

 void exgcd(int a, int b, int& d, int& x, int& y) {
     if (!b)
         d = a, x = , y = ;
     else {
         exgcd(b, a % b, d, y, x);
         y -= (a / b) * x;
     }
 }

 int qpow(int a, int pos, int p) {
     int pa = a, rt = ;
     for ( ; pos; pos >>= , pa = pa * 1ll * pa % p)
         if (pos & )
             rt = rt * 1ll * pa % p;
     return rt;
 }

 int inv(int a, int n) {
     int d, x, y;
     exgcd(a, n, d, x, y);
     return (x < ) ? (x + n) : (x);
 }

 int p, k, a;
 int g;

 inline void init() {
     scanf("%d%d%d", &p, &k, &a);
 }

 HashMap mp;
 int ind(int a, int b, int p) {
     mp.clear();
     int cs = sqrt(p - 0.5), ac = qpow(a, cs, p), iac = b * 1ll * inv(ac, p) % p, pw = ;
     for (int i = cs - ; ~i; i--)
         iac = iac * 1ll * a % p, mp.insert(iac, i);
     for (int i = ; i < p; i += cs, pw = pw * 1ll * ac % p)
         if (~mp[pw])
             return mp[pw] + i;
     return -;
 }

 int top = ;
 int fac[];
 void getfactors(int x) {
     for (int i = ; i * i <= x; i++) {
         if (!(x % i)) {
             fac[++top] = i;
             while (!(x % i))    x /= i;
         }
     }
     if (x != )    fac[++top] = x;
 }

 boolean isroot(int x) {
     int pos = p - ;
     for (int i = ; i <= top; i++)
         if (qpow(x, pos / fac[i], p) == )
             return false;
     return true;
 }

 inline void solve() {
     if (p ==  || !a) {
         printf("1\n0");
         return;
     }
     if (p == )
         g = ;
     else {
         getfactors(p - );
         for (g = ; !isroot(g); g++);
     }

     int inda = ind(g, a, p), d, x, y, cir;
     exgcd(k, p - , d, x, y);
     if (inda % d) {
         puts("");
         return;
     }
     cir = (p - ) / d;
     x = x * 1ll * (inda / d) % cir;
     (x <= ) ? (x += cir) : ();
     vector<int> res;
     for (; x < p; x += cir)
         res.push_back(qpow(g, x, p));
     sort(res.begin(), res.end());
     printf("%d\n", (signed) res.size());
     for (int i = ; i < (signed) res.size(); i++)
         printf("%d\n", res[i]);
 }

 int main() {
     init();
     solve();
     return ;
 }