【HDU 4992】Primitive Roots

1.​​题目链接​​。原根的基本求法。根据下面的性质求：

                                         

流程：判断是不是存在原根？---->求出最小----->从最小得到全部。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#pragma warning(disable:4996)
const int maxn = 1e6 + 5;
int prime[maxn + 5];
bool check[maxn + 5];
int phi[maxn + 5];
int num_prime;
void init()
{
  memset(check, false, sizeof(check));
  phi[1] = 1;
  for (int i = 2; i <= maxn; i++)
  {
    if (!check[i])
    {
      prime[num_prime++] = i;
      phi[i] = i - 1;
    }
    for (int j = 0; j < num_prime; j++)
    {
      if (i*prime[j] > maxn) break;
      check[i*prime[j]] = true;
      if (i%prime[j] == 0)
      {
        phi[i*prime[j]] = phi[i] * prime[j];
        break;
      }
      else
      {
        phi[i*prime[j]] = phi[i] * (prime[j] - 1);
      }
    }
  }
}

LL gcd(LL a, LL b)
{
  return b ? gcd(b, a%b) : a;
}

void get(LL n, vector<LL>& fac)    //对n进行因式分解 
{
  fac.clear();
  for (LL i = 2; i*i <= n; i++)
    if (n%i == 0)
    {
      fac.push_back(i);
      while (n%i == 0) n /= i;
    }
  if (n > 1) fac.push_back(n);
}

LL qpow(LL x, LL n, LL mod)        //求x^n%mod
{
  LL ret = 1;
  for (; n; n >>= 1)
  {
    if (n & 1) ret = ret * x%mod;
    x = x * x%mod;
  }
  return ret;
}

vector<LL> fac;
vector<LL> ans;

bool ok(LL x)
{
  if (x % 2 == 0) x /= 2;
  if (x % 2 == 0) return false;
  for (int i = 0; prime[i] * prime[i] <= x; i++) if (x%prime[i] == 0)
  {
    while (x%prime[i] == 0) x /= prime[i];
    return x == 1;
  }
  return true;
}

LL get_g(LL p)        //得到一个正整数p的最小原根 
{
  for (int i = 2; i < p; i++)
  {
    bool flag = false;
    for (LL x : fac)
      if (qpow(i, phi[p] / x, p) == 1)
      {
        flag = true;
        break;
      }
    if (!flag&&qpow(i, phi[p], p) == 1)  return i;
  }
}

void GetAns(LL g, LL p, vector<LL>& ans)    //由最小原根p，得到某正整数p的全部原根 
{
  ans.clear();
  ans.push_back(g);
  for (int i = 2; i < phi[p]; i++)
    if (gcd(i, phi[p]) == 1) ans.push_back(qpow(g, i, p));
}

int main()
{
  init();
  LL p;
  while (~scanf("%lld", &p))
  {
    if (p == 2 || p == 4)
    {
      printf("%lld\n", p - 1);
      continue;
    }
    if (!ok(p))        //首先判断是否有原根 
    {
      puts("-1");
      continue;
    }
    get(phi[p], fac);
    LL g = get_g(p);        //定义g为最小原根
    GetAns(g, p, ans);
    sort(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); i++)
      printf("%lld%c", ans[i], i == ans.size() - 1 ? '\n' : ' ');
  }
}