素数 n 模 n 的本原根
给定一个素数 n，任务是在模 n 下找到它的本原根。素数 n 的本原根是介于[1，n-1]之间的整数 r，使得 x 在[0，n-2]范围内的 r^x(mod n 的值不同。如果 n 是非质数，返回-1。
示例:
Input : 7
Output : Smallest primitive root = 3
Explanation: n = 7
3^0(mod 7) = 1
3^1(mod 7) = 3
3^2(mod 7) = 2
3^3(mod 7) = 6
3^4(mod 7) = 4
3^5(mod 7) = 5
Input : 761
Output : Smallest primitive root = 6
一个简单的解决方法就是尝试从 2 到 n-1 的所有数字。对于每个数字 r，计算 r^x(mod 的值，其中 x 在[0，n-2]的范围内。如果所有这些值都不同，则返回 r，否则继续下一个 r 值。如果所有 r 值都尝试过，则返回-1。
一个高效的解决方案是基于以下事实。 如果一个数 r 模 n 的乘法阶等于欧拉全能函数φ(n)(注意一个素数 n 的欧拉全能函数是 n-1)，那么它就是本原根。
1- Euler Totient Function phi = n-1 [Assuming n is prime]
1- Find all prime factors of phi.
2- Calculate all powers to be calculated further
using (phi/prime-factors) one by one.
3- Check for all numbered for all powers from i=2
to n-1 i.e. (i^ powers) modulo n.
4- If it is 1 then 'i' is not a primitive root of n.
5- If it is never 1 then return i;.
虽然一个素数可以有多个本原根，但我们只关心最小的一个。如果你想找到所有的根，那么继续这个过程直到 p-1，而不是通过找到第一个原始根来分解。
C++
// C++ program to find primitive root of a
// given number n
#include<bits/stdc++.h>
using namespace std;
// Returns true if n is prime
bool isPrime(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n%2 == 0 || n%3 == 0) return false;
for (int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return false;
return true;
}
/* Iterative Function to calculate (x^n)%p in
O(logy) */
int power(int x, unsigned int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Utility function to store prime factors of a number
void findPrimefactors(unordered_set<int> &s, int n)
{
// Print the number of 2s that divide n
while (n%2 == 0)
{
s.insert(2);
n = n/2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (int i = 3; i <= sqrt(n); i = i+2)
{
// While i divides n, print i and divide n
while (n%i == 0)
{
s.insert(i);
n = n/i;
}
}
// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
s.insert(n);
}
// Function to find smallest primitive root of n
int findPrimitive(int n)
{
unordered_set<int> s;
// Check if n is prime or not
if (isPrime(n)==false)
return -1;
// Find value of Euler Totient function of n
// Since n is a prime number, the value of Euler
// Totient function is n-1 as there are n-1
// relatively prime numbers.
int phi = n-1;
// Find prime factors of phi and store in a set
findPrimefactors(s, phi);
// Check for every number from 2 to phi
for (int r=2; r<=phi; r++)
{
// Iterate through all prime factors of phi.
// and check if we found a power with value 1
bool flag = false;
for (auto it = s.begin(); it != s.end(); it++)
{
// Check if r^((phi)/primefactors) mod n
// is 1 or not
if (power(r, phi/(*it), n) == 1)
{
flag = true;
break;
}
}
// If there was no power with value 1.
if (flag == false)
return r;
}
// If no primitive root found
return -1;
}
// Driver code
int main()
{
int n = 761;
cout << " Smallest primitive root of " << n
<< " is " << findPrimitive(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find primitive root of a
// given number n
import java.util.*;
class GFG
{
// Returns true if n is prime
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
{
return false;
}
if (n <= 3)
{
return true;
}
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
{
return false;
}
for (int i = 5; i * i <= n; i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false;
}
}
return true;
}
/* Iterative Function to calculate (x^n)%p in
O(logy) */
static int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y % 2 == 1)
{
res = (res * x) % p;
}
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Utility function to store prime factors of a number
static void findPrimefactors(HashSet<Integer> s, int n)
{
// Print the number of 2s that divide n
while (n % 2 == 0)
{
s.add(2);
n = n / 2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (int i = 3; i <= Math.sqrt(n); i = i + 2)
{
// While i divides n, print i and divide n
while (n % i == 0)
{
s.add(i);
n = n / i;
}
}
// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
{
s.add(n);
}
}
// Function to find smallest primitive root of n
static int findPrimitive(int n)
{
HashSet<Integer> s = new HashSet<Integer>();
// Check if n is prime or not
if (isPrime(n) == false)
{
return -1;
}
// Find value of Euler Totient function of n
// Since n is a prime number, the value of Euler
// Totient function is n-1 as there are n-1
// relatively prime numbers.
int phi = n - 1;
// Find prime factors of phi and store in a set
findPrimefactors(s, phi);
// Check for every number from 2 to phi
for (int r = 2; r <= phi; r++)
{
// Iterate through all prime factors of phi.
// and check if we found a power with value 1
boolean flag = false;
for (Integer a : s)
{
// Check if r^((phi)/primefactors) mod n
// is 1 or not
if (power(r, phi / (a), n) == 1)
{
flag = true;
break;
}
}
// If there was no power with value 1.
if (flag == false)
{
return r;
}
}
// If no primitive root found
return -1;
}
// Driver code
public static void main(String[] args)
{
int n = 761;
System.out.println(" Smallest primitive root of " + n
+ " is " + findPrimitive(n));
}
}
/* This code contributed by PrinciRaj1992 */
Python 3
# Python3 program to find primitive root
# of a given number n
from math import sqrt
# Returns True if n is prime
def isPrime( n):
# Corner cases
if (n <= 1):
return False
if (n <= 3):
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
return False
i = 5
while(i * i <= n):
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
""" Iterative Function to calculate (x^n)%p
in O(logy) */"""
def power( x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more
# than or equal to p
while (y > 0):
# If y is odd, multiply x with result
if (y & 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
# Utility function to store prime
# factors of a number
def findPrimefactors(s, n) :
# Print the number of 2s that divide n
while (n % 2 == 0) :
s.add(2)
n = n // 2
# n must be odd at this po. So we can
# skip one element (Note i = i +2)
for i in range(3, int(sqrt(n)), 2):
# While i divides n, print i and divide n
while (n % i == 0) :
s.add(i)
n = n // i
# This condition is to handle the case
# when n is a prime number greater than 2
if (n > 2) :
s.add(n)
# Function to find smallest primitive
# root of n
def findPrimitive( n) :
s = set()
# Check if n is prime or not
if (isPrime(n) == False):
return -1
# Find value of Euler Totient function
# of n. Since n is a prime number, the
# value of Euler Totient function is n-1
# as there are n-1 relatively prime numbers.
phi = n - 1
# Find prime factors of phi and store in a set
findPrimefactors(s, phi)
# Check for every number from 2 to phi
for r in range(2, phi + 1):
# Iterate through all prime factors of phi.
# and check if we found a power with value 1
flag = False
for it in s:
# Check if r^((phi)/primefactors)
# mod n is 1 or not
if (power(r, phi // it, n) == 1):
flag = True
break
# If there was no power with value 1.
if (flag == False):
return r
# If no primitive root found
return -1
# Driver Code
n = 761
print("Smallest primitive root of",
n, "is", findPrimitive(n))
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
C
// C# program to find primitive root of a
// given number n
using System;
using System.Collections.Generic;
class GFG
{
// Returns true if n is prime
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
{
return false;
}
if (n <= 3)
{
return true;
}
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
{
return false;
}
for (int i = 5; i * i <= n; i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false;
}
}
return true;
}
/* Iterative Function to calculate (x^n)%p in
O(logy) */
static int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y % 2 == 1)
{
res = (res * x) % p;
}
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Utility function to store prime factors of a number
static void findPrimefactors(HashSet<int> s, int n)
{
// Print the number of 2s that divide n
while (n % 2 == 0)
{
s.Add(2);
n = n / 2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (int i = 3; i <= Math.Sqrt(n); i = i + 2)
{
// While i divides n, print i and divide n
while (n % i == 0)
{
s.Add(i);
n = n / i;
}
}
// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
{
s.Add(n);
}
}
// Function to find smallest primitive root of n
static int findPrimitive(int n)
{
HashSet<int> s = new HashSet<int>();
// Check if n is prime or not
if (isPrime(n) == false)
{
return -1;
}
// Find value of Euler Totient function of n
// Since n is a prime number, the value of Euler
// Totient function is n-1 as there are n-1
// relatively prime numbers.
int phi = n - 1;
// Find prime factors of phi and store in a set
findPrimefactors(s, phi);
// Check for every number from 2 to phi
for (int r = 2; r <= phi; r++)
{
// Iterate through all prime factors of phi.
// and check if we found a power with value 1
bool flag = false;
foreach (int a in s)
{
// Check if r^((phi)/primefactors) mod n
// is 1 or not
if (power(r, phi / (a), n) == 1)
{
flag = true;
break;
}
}
// If there was no power with value 1.
if (flag == false)
{
return r;
}
}
// If no primitive root found
return -1;
}
// Driver code
public static void Main(String[] args)
{
int n = 761;
Console.WriteLine(" Smallest primitive root of " + n
+ " is " + findPrimitive(n));
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript program to find primitive root of a
// given number n
// Returns true if n is prime
function isPrime(n) {
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
/* Iterative Function to calculate (x^n)%p in
O(logy) */
function power(x, y, p) {
let res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Utility function to store prime factors of a number
function findPrimefactors(s, n) {
// Print the number of 2s that divide n
while (n % 2 == 0) {
s.add(2);
n = n / 2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (let i = 3; i <= Math.sqrt(n); i = i + 2) {
// While i divides n, print i and divide n
while (n % i == 0) {
s.add(i);
n = n / i;
}
}
// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
s.add(n);
}
// Function to find smallest primitive root of n
function findPrimitive(n) {
let s = new Set();
// Check if n is prime or not
if (isPrime(n) == false)
return -1;
// Find value of Euler Totient function of n
// Since n is a prime number, the value of Euler
// Totient function is n-1 as there are n-1
// relatively prime numbers.
let phi = n - 1;
// Find prime factors of phi and store in a set
findPrimefactors(s, phi);
// Check for every number from 2 to phi
for (let r = 2; r <= phi; r++) {
// Iterate through all prime factors of phi.
// and check if we found a power with value 1
let flag = false;
for (let it of s) {
// Check if r^((phi)/primefactors) mod n
// is 1 or not
if (power(r, phi / it, n) == 1) {
flag = true;
break;
}
}
// If there was no power with value 1.
if (flag == false)
return r;
}
// If no primitive root found
return -1;
}
// Driver code
let n = 761;
document.write(" Smallest primitive root of " + n + " is " + findPrimitive(n));
// This code is contributed by gfgking
</script>
输出:
Smallest primitive root of 761 is 6
本文由尼泰什·库马尔T3【萨希尔·查布拉(akku) 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
版权属于：月萌API www.moonapi.com，转载请注明出处
