原文:
给定一个数字 n，任务是检查给定的数字是否为prost prime。 一个原质数是一个是原质数。 前几个原素是–
3, 5, 13, 17, 41, 97, 113, 193, 241, 257, 353, 449, 577, 641, 673, 769, 929, 1153, 1217, … ..
例:
input: 41
output: 41 is proth prime
input: 19
output: 19 is not a proth prime
方法: 想法是使用厄拉多塞的筛子来寻找阿普顿的素数。然后检查给定的号码是否为。如果数是一个并且也是一个素数，那么给定的数就是原素数。 以下是上述算法的实现:
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 语言(一种计算机语言，尤用于创建网站)
// 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_gangwar
python 3
# 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 29
c
// 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 kundu
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
output:
3
5
13
17
41
时间复杂度: o(nlog(log(n))) 参考文献:*
- https://en . wikipedia . org/wiki/proth _ number # proth _ primes
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