SGU 261. Discrete Roots

给定\(p, k, A\)，满足\(k, p\)是质数，求

\[x^k \equiv A \mod p\]

不会。。。

upd:3:29

两边取指标，是求

\[k\text{ind}_x\equiv \text{ind}_A\mod p-1\]

的解数，先求最小的解，然后暴力求之后的就行了。

 #include <iostream>
 #include <map>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <cmath>
 using namespace std;
 typedef long long LL;

 vector<LL> f, as;
 LL fast_pow(LL base, LL index, LL mod) {
     LL ret = ;
     for(; index; index >>= , base = base * base % mod)
         if(index & ) ret = ret * base % mod;
     return ret;
 }
 bool test_Primitive_Root(LL g, LL p) {
     for(LL i = ; i < f.size(); ++i)
         if(fast_pow(g, (p - ) / f[i], p) == )
             return ;
     return ;
 }
 LL get_Primitive_Root(LL p) {
     f.clear();
     LL tmp = p - ;
     for(LL i = ; i <= tmp / i; ++i)
         if(tmp % i == )
             for(f.push_back(i); tmp % i == ; tmp /= i);
     if(tmp != ) f.push_back(tmp);
     for(LL g = ; ; ++g) {
         if(test_Primitive_Root(g, p))
             return g;
     }
 }
 LL get_Discrete_Logarithm(LL x, LL n, LL m) {
     map<LL, int> rec;
     LL s = (LL)(sqrt((double)m) + 0.5), cur = ;
     for(LL i = ; i < s; rec[cur] = i, cur = cur * x % m, ++i);
     LL mul = cur;
     cur = ;
     for(LL i = ; i < s; ++i) {
         LL more = n * fast_pow(cur, m - , m) % m;
         if(rec.count(more))
             return i * s + rec[more];
         cur = cur * mul % m;
     }
     return -;
 }
 LL ext_Euclid(LL a, LL b, LL &x, LL &y) {
     if(b == ) {
         x = , y = ;
         return a;
     } else {
         LL ret = ext_Euclid(b, a % b, y, x);
         y -= x * (a / b);
         return ret;
     }
 }
 void solve_Linear_Mod_Equation(LL a, LL b, LL n) {
     LL x, y, d;
     as.clear();
     d = ext_Euclid(a, n, x, y);
     if(b % d == ) {
         x %= n, x += n, x %= n;
         as.push_back(x * (b / d) % (n / d));
         for(LL i = ; i < d; ++i)
             as.push_back((as[] + i * n / d) % n);
     }
 }

 int main() {
 #ifndef ONLINE_JUDGE
     freopen("data.in", "r", stdin); freopen("data.out", "w", stdout);
 #endif

     LL p, k, a;
     cin >> p >> k >> a;
     if(a == ) {
         puts("1\n0");
         return ;
     }
     LL g = get_Primitive_Root(p);
     LL q = get_Discrete_Logarithm(g, a, p);
     solve_Linear_Mod_Equation(k, q, p - );
     for(int i = ; i < as.size(); ++i)
         as[i] = fast_pow(g, as[i], p);
     sort(as.begin(), as.end());
     printf("%d\n", as.size());
     for(int i = ; i < as.size(); ++i) {
         printf("%lld%c", as[i], i == as.size() -  ? '\n' : ' ');
     }
     return ;
 }