原根二连 HDU 4992 && poj 1284 Primitive Roots

原根存在的充要条件

n = 1,2,4,p^r (p为奇素数，r为任意正整数)

原根的性质

若n存在原根，则原根个数为φ(φ(n))

若g是n的一个原根，则g^d是n的原根的充要条件为gcd(d,φ(n)) = 1

一个数的全体原根乘积模n余1

一个数的全体原根和模n余μ(n-1)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
bool isprime[1000001];
LL prime[100001];
LL primesize = 0,n,ans[10001],p;
LL res[1000005];
LL gcd(LL a,LL b){
  if (b == 0) return a;
  return gcd(b,a % b);
}

LL Fast_Mod(LL a,LL b,LL p){
   LL res = 1,base = a;
   while (b){
       if (b&1) res = (res*base) % p;
       base = (base*base) % p;
       b = b >> 1;
   }
   return res;
}

void initPrime(){
  memset(isprime,1,sizeof(isprime));
  isprime[1] = false;
  LL ListSize = 1000000;
  for (LL i=2;i<= ListSize ; i++){
    if (isprime[i]) {
      primesize ++; prime[primesize] = i;
    }
    for (int j=1;j<=primesize && i*prime[j] <= ListSize; j++){
      isprime[i*prime[j]] = false;
      if (i % prime[j] == 0) break;
    }
 }
}

LL phi(LL n){
    LL i,rea=n;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            rea=rea-rea/i;
            while(n%i==0)  n/=i;
        }
    }
    if(n>1)
        rea=rea-rea/n;
    return rea;
}

bool exist(LL n){
  if (n % 2 == 0) n /= 2;
  if (isprime[n]) return 1;
  for (int i=3;i*i<=n;i+=2){
    if (n % i == 0){
      while (n % i == 0) n /= i;
      return (n == 1);
    }
  }
  return 0;
}

int main()
{
    // freopen("test.in","r",stdin);
    initPrime();
    while(scanf("%lld",&p)!=-1)
    {
        int cnt=0;
        if (p == 2) {
          cout << 1 << endl; continue;
        }
        if (p == 4){
          cout << 3 << endl; continue;
        }
        if (!exist(p)){
          cout << -1 << endl; continue;
        }
        n = phi(p);
        for(int i=1;i<=primesize&&prime[i]*prime[i]<=n;i++)
        {
            if(n%prime[i]==0)
            {
                n/=prime[i];
                cnt ++;
                ans[cnt]=prime[i];
                while(n%prime[i]==0)
                {
                    n/=prime[i];
                }
            }
        }
        if(n!=1){
          cnt ++;
          ans[cnt]=n;
        }
        n = phi(p);
        LL g;
        for (int i=2;i<p;i++){
          if (gcd(i,p) == 1){
            int ok = 1;
            for (int j=1;j<=cnt;j++){
              if (Fast_Mod(i,n/ans[j],p) == 1){
                ok = 0; break;
              }
            }
            if (ok){
                g = i;
                break;
            }
          }
        }
        LL total = 0;
        for (int i=1;i<=n;i++){
          if (gcd(i,n) == 1){
            total ++;
            res[total] = Fast_Mod(g,i,p);
          }
        }
        sort(res+1,res+total+1);
        for (int i=1;i<=total-1;i++){
          cout << res[i] << " ";
        }
        cout << res[total] << endl;
    }

    return 0;
}

View Code

#include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef long long LL; bool isprime[1000001]; LL prime[100001]; LL primesize = 0,n,ans[10001],p; LL gcd(LL a,LL b){ if (b == 0) return a; return gcd(b,a % b); } LL Fast_Mod(LL a,LL b,LL p){ LL res = 1,base = a; while (b){ if (b&1) res = (res*base) % p; base = (base*base) % p; b = b >> 1; } return res; } void initPrime(){ memset(isprime,1,sizeof(isprime)); isprime[1] = false; LL ListSize = 1000000; for (LL i=2;i<= ListSize ; i++){ if (isprime[i]) { primesize ++; prime[primesize] = i; } for (int j=1;j<=primesize && i*prime[j] <= ListSize; j++){ isprime[i*prime[j]] = false; if (i % prime[j] == 0) break; } } } LL phi(LL n){ LL i,rea=n; for(i=2;i*i<=n;i++) { if(n%i==0) { rea=rea-rea/i; while(n%i==0) n/=i; } } if(n>1) rea=rea-rea/n; return rea; } int main() { // freopen("test.in","r",stdin); initPrime(); while(scanf("%lld",&p)!=-1) { int cnt=0; n = phi(p); for(int i=1;i<=primesize&&prime[i]*prime[i]<=n;i++) { if(n%prime[i]==0) { n/=prime[i]; cnt ++; ans[cnt]=prime[i]; while(n%prime[i]==0) { n/=prime[i]; } } } if(n!=1){ cnt ++; ans[cnt]=n; } bool find = 0; n = phi(p); for (int i=2;i<p;i++){ if (gcd(i,p) == 1){ int ok = 1; for (int j=1;j<=cnt;j++){ if (Fast_Mod(i,n/ans[j],p) == 1){ ok = 0; break; } } if (ok){ find = 1; break; } } } if (find){ cout << phi(phi(p)) << endl; } else { cout << 0 << endl; } } return 0; }

原根二连 HDU 4992 && poj 1284 Primitive Roots

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