检查给定的数字是否是堂兄素数
给定两个正整数 n1 和 n2，任务是检查两者是否都是表哥素数。如果两个数字都是表亲素数，则打印“是”，否则打印“否”。 表亲素数: 在数学中，表亲素数是相差 4 的素数。假设“p”是一个质数，如果(p + 4)也是一个质数，那么这两个质数将被称为表亲质数。 100 以下的表亲素数为:
(3, 7), (7, 11), (13, 17), (19, 23), (37, 41), (43, 47), (67, 71), (79, 83), (97, 101)
示例:
Input: n1 = 7, n2 = 11
Output: YES
Both are prime numbers and they differ by 4.
Input: n1 = 5, n2 = 11
Output: NO
Both are prime numbers but they differ by 6.
方法:一个简单的解决方法是检查两个数是否都是素数并且相差 4。如果它们是素数，相差 4，那么它们将是表亲素数，否则不是。 以下是上述办法的实施情况:
卡片打印处理机（Card Print Processor 的缩写）
// CPP program to check Cousin prime
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number is prime or not
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;
}
// Returns true if n1 and n2 are Cousin primes
bool isCousinPrime(int n1, int n2)
{
// Check if they differ by 4 or not
if (abs(n1 - n2) != 4)
return false;
// Check if both are prime number or not
else
return (isPrime(n1) && isPrime(n2));
}
// Driver code
int main()
{
// Get the 2 numbers
int n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
Java 语言（一种计算机语言，尤用于创建网站）
// JAVA program to check Cousin prime
import java.util.*;
class GFG {
// Function to check if a number is prime or not
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;
}
// Returns true if n1 and n2 are Cousin primes
static boolean isCousinPrime(int n1, int n2)
{
// Check if they differ by 4 or not
if (Math.abs(n1 - n2) != 4)
return false;
// Check if both are prime number or not
else
return (isPrime(n1) && isPrime(n2));
}
// Driver code
public static void main(String[] args)
{
// Get the 2 numbers
int n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
System.out.println("YES");
else
System.out.println("NO");
}
}
计算机编程语言
# Python program to check Cousin prime
import math
# Function to check whether a
# number is prime or not
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
for i in range(5, int(math.sqrt(n)+1), 6):
if n % i == 0 or n %(i + 2) == 0:
return False
return True
# Returns true if n1 and n2 are Cousin primes
def isCousinPrime( n1, n2) :
# Check if they differ by 4 or not
if(not (abs(n1-n2)== 4)):
return False
# Check if both are prime number or not
else:
return (isPrime(n1) and isPrime(n2))
# Driver code
# Get the 2 numbers
n1 = 7
n2 = 11
# Check the numbers for cousin prime
if (isCousinPrime(n1, n2)):
print("YES")
else:
print("NO")
C
// C# Code for Cousin Prime Numbers
using System;
class GFG {
// Function to check if the given
// number is prime or not
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;
}
// Returns true if n1 and n2 are Cousin primes
static bool isCousinPrime(int n1, int n2)
{
// Check if the numbers differ by 4 or not
if (Math.Abs(n1 - n2) != 4) {
return false;
}
else {
return (isPrime(n1) && isPrime(n2));
}
}
// Driver program
public static void Main()
{
// Get the 2 numbers
int n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PhP program to check Cousin prime
// Function to check if the given
// Number is prime or not
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 || $n % 3 == 0)
return false;
for ($i = 5; $i * $i <= $n; $i = $i + 6)
if ($n % $i == 0 || $n % ($i + 2) == 0)
return false;
return true;
}
// Returns true if n1 and n2 are Cousin primes
function isCousinPrime($n1, $n2)
{
// Check if the given number differ by 4 or not
if(abs($n1-$n2)!=4)
return false;
// Check if both numbers are prime or not
else
return (isPrime($n1) && isPrime($n2));
}
// Driver code
// Get the 2 numbers
$n1 = 7;
$n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime($n1, $n2))
echo "YES";
else
echo "NO";
?>
java 描述语言
<script>
// Javascript program to check Cousin prime
// Function to check if a number is prime or not
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 (var i = 5; i * i <= n; i = i + 6) {
if (n % i == 0 || n % (i + 2) == 0) {
return false;
}
}
return true;
}
// Returns true if n1 and n2 are Cousin primes
function isCousinPrime(n1, n2)
{
// Check if they differ by 4 or not
if(Math.abs(n1 - n2) != 4)
return false;
// Check if both are prime number or not
else
return (isPrime(n1) && isPrime(n2));
}
// Driver code
// Get the 2 numbers
var n1 = 7, n2 = 11;
// Check the numbers for cousin prime
if (isCousinPrime(n1, n2))
document.write( "YES");
else
document.write( "NO" );
</script>
Output:
YES
时间复杂度: O(n1 1/2 )
辅助空间: O(1)
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