给定数字N，任务是检查给定的数字是否为Proth Prime 。
Proth素数是Proth素数。
前几个Proth素数是–
3, 5, 13, 17, 41, 97, 113, 193, 241, 257, 353, 449, 577, 641, 673, 769, 929, 1153, 1217, …..
例子：
Input: 41
Output: 41 is Proth Prime
Input: 19
Output: 19 is not a Proth Prime
方法：
这个想法是使用Eratosthenes的Sieve查找最高达N的素数。然后检查给定的编号是否是Proth Number。如果number是Proth Number也是质数，则给定的数字是Proth Prime。
下面是上述算法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
int prime[1000000];
// Calculate all primes upto n.
void SieveOfEratosthenes(int n)
{
// Initialize all entries it as true.
// A value in prime[i] will finally
// false if i is Not a prime, else true.
for (int i = 1; i <= n + 1; i++)
prime[i] = true;
prime[1] = false;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true) {
// Update all multiples of p
// greater than or equal to
// the square of it numbers
// which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
// Utility function to check power of two
bool isPowerOfTwo(int n)
{
return (n && !(n & (n - 1)));
}
// Function to check if the Given
// number is Proth number or not
bool isProthNumber(int n)
{
int k = 1;
while (k < (n / k)) {
// check if k divides n or not
if (n % k == 0) {
// Check if n/k is power of 2 or not
if (isPowerOfTwo(n / k))
return true;
}
// update k to next odd number
k = k + 2;
}
// If we reach here means there
// exists no value of K such
// that k is odd number and n/k
// is a power of 2 greater than k
return false;
}
// Function to check whether the given
// number is Proth Prime or Not.
bool isProthPrime(int n)
{
// Check n for Proth Number
if (isProthNumber(n - 1)) {
// if number is prime, return true
if (prime[n])
return true;
else
return false;
}
else
return false;
}
// Driver Code
int main()
{
int n = 41;
// if number is proth number,
// calculate primes upto n
SieveOfEratosthenes(n);
for (int i = 1; i <= n; i++)
// Check n for Proth Prime
if (isProthPrime(i))
cout << i << endl;
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
static boolean[] prime = new boolean[1000000];
// Calculate all primes upto n.
static void SieveOfEratosthenes(int n)
{
// Initialize all entries it as true.
// A value in prime[i] will finally
// false if i is Not a prime, else true.
for (int i = 1; i <= n + 1; i++)
prime[i] = true;
prime[1] = false;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
// greater than or equal to
// the square of it numbers
// which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
// Utility function to check power of two
static boolean isPowerOfTwo(int n)
{
return (n > 0 && (n & (n - 1)) == 0);
}
// Function to check if the Given
// number is Proth number or not
static boolean isProthNumber(int n)
{
int k = 1;
while (k < (int)(n / k))
{
// check if k divides n or not
if (n % k == 0)
{
// Check if n/k is power of 2 or not
if (isPowerOfTwo((int)(n / k)))
return true;
}
// update k to next odd number
k = k + 2;
}
// If we reach here means there
// exists no value of K such
// that k is odd number and n/k
// is a power of 2 greater than k
return false;
}
// Function to check whether the given
// number is Proth Prime or Not.
static boolean isProthPrime(int n)
{
// Check n for Proth Number
if (isProthNumber(n - 1))
{
// if number is prime, return true
if (prime[n])
return true;
else
return false;
}
else
return false;
}
// Driver Code
public static void main(String args[])
{
int n = 41;
// if number is proth number,
// calculate primes upto n
SieveOfEratosthenes(n);
for (int i = 1; i <= n; i++)
// Check n for Proth Prime
if (isProthPrime(i))
System.out.println(i);
}
}
// This code is contributed by
// Surendra_GangwarPython3
# Python3 implementation of the
# above approach
import math as mt
prime = [0 for i in range(1000000)]
# Calculate all primes upto n.
def SieveOfEratosthenes(n):
# Initialize all entries it as true.
# A value in prime[i] will finally
# false if i is Not a prime, else true.
for i in range(1, n + 2):
prime[i] = True
prime[1] = False
for p in range(2, mt.ceil(n**(0.5))):
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p
# greater than or equal to
# the square of it numbers
# which are multiple of p and are
# less than p^2 are already been marked.
for i in range(p * p, n + 1, p):
prime[i] = False
# Utility function to check power of two
def isPowerOfTwo(n):
return (n and (n & (n - 1)) == False)
# Function to check if the Given
# number is Proth number or not
def isProthNumber(n):
k = 1
while (k < (n // k)):
# check if k divides n or not
if (n % k == 0):
# Check if n/k is power of 2 or not
if (isPowerOfTwo(n // k)):
return True
# update k to next odd number
k = k + 2
# If we reach here means there
# exists no value of K such
# that k is odd number and n/k
# is a power of 2 greater than k
return False
# Function to check whether the given
# number is Proth Prime or Not.
def isProthPrime(n):
# Check n for Proth Number
if (isProthNumber(n - 1)):
# if number is prime, return true
if (prime[n]):
return True
else:
return False
else:
return False
# Driver Code
n = 41
# if number is proth number,
# calculate primes upto n
SieveOfEratosthenes(n)
for i in range(1, n + 1):
# Check n for Proth Prime
if isProthPrime(i) == True:
print(i)
# This code is contributed by
# Mohit kumar 29C#
// C# implementation of the above approach
using System;
class GFG
{
static Boolean[] prime = new Boolean[1000000];
// Calculate all primes upto n.
static void SieveOfEratosthenes(int n)
{
// Initialize all entries it as true.
// A value in prime[i] will finally
// false if i is Not a prime, else true.
for (int i = 1; i <= n + 1; i++)
prime[i] = true;
prime[1] = false;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
// greater than or equal to
// the square of it numbers
// which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
// Utility function to check power of two
static Boolean isPowerOfTwo(int n)
{
return (n > 0 && (n & (n - 1)) == 0);
}
// Function to check if the Given
// number is Proth number or not
static Boolean isProthNumber(int n)
{
int k = 1;
while (k < (int)(n / k))
{
// check if k divides n or not
if (n % k == 0)
{
// Check if n/k is power of 2 or not
if (isPowerOfTwo((int)(n / k)))
return true;
}
// update k to next odd number
k = k + 2;
}
// If we reach here means there
// exists no value of K such
// that k is odd number and n/k
// is a power of 2 greater than k
return false;
}
// Function to check whether the given
// number is Proth Prime or Not.
static Boolean isProthPrime(int n)
{
// Check n for Proth Number
if (isProthNumber(n - 1))
{
// if number is prime, return true
if (prime[n])
return true;
else
return false;
}
else
return false;
}
// Driver Code
static public void Main(String []args)
{
int n = 41;
// if number is proth number,
// calculate primes upto n
SieveOfEratosthenes(n);
for (int i = 1; i <= n; i++)
// Check n for Proth Prime
if (isProthPrime(i))
Console.WriteLine(i);
}
}
// This code is contributed by Arnab KunduPHP
输出：
3
5
13
17
41
时间复杂度： O(n * log(log(n)))
参考：
- https://zh.wikipedia.org/wiki/Proth_number#Proth_primes
- https://www.geeksforgeeks.org/program-to-check-whether-a-number-is-proth-number-or-not/