查找下一个回文素数
原文:https://www.geeksforgeeks.org/find-next-palindrome-prime/
找出最小的回文数，它也是质数并且大于给定的数 N. 举例:
Input : N = 7
Output :11
11 is the smallest palindrome prime which
is greater than N.
Input : N = 112
Output : 131
一种简单的方法是从 N+1 开始循环。对于每个数字，检查是否是回文和质数。 一个有效的解决方案是基于以下观察。所有偶数的回文都是 11 的倍数。 我们可以证明如下: 11% 11 = 0 1111% 11 = 0 111111% 11 = 0 1111111% 11 = 0 所以: 1001% 11 =(1111–11 * 10)% 11 = 0 100001% 11 =(11111–11111) % 11 = 0 对于任何偶数位数的回文: abcdcba % 11 =(a * 10000001+b * 100001 * 10+c * 1001 * 100+d * 11 * 1000)% 11 = 0 所有偶数位数的回文都是 11 的倍数。 所以其中，11 是唯一一个素数 如果(8 < = N < = 11)返回 11 对于其他，我们只考虑奇数位的回文。
C++
// CPP program to find next palindromic
// prime for a given number.
#include <iostream>
#include <string>
using namespace std;
bool isPrime(int num)
{
if (num < 2 || num % 2 == 0)
return num == 2;
for (int i = 3; i * i <= num; i += 2)
if (num % i == 0)
return false;
return true;
}
int primePalindrome(int N)
{
// if(8<=N<=11) return 11
if (8 <= N && N <= 11)
return 11;
// generate odd length palindrome number
// which will cover given constraint.
for (int x = 1; x < 100000; ++x) {
string s = to_string(x), r(s.rbegin(), s.rend());
int y = stoi(s + r.substr(1));
// if y>=N and it is a prime number
// then return it.
if (y >= N && isPrime(y))
return y;
}
return -1;
}
// Driver code
int main()
{
cout << primePalindrome(112);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find next palindromic
// prime for a given number.
import java.lang.*;
class Geeks {
static boolean isPrime(int num)
{
if (num < 2 || num % 2 == 0)
return num == 2;
for (int i = 3; i * i <= num; i += 2)
if (num % i == 0)
return false;
return true;
}
static int primePalindrome(int N)
{
// if(8<=N<=11) return 11
if (8 <= N && N <= 11)
return 11;
// generate odd length palindrome number
// which will cover given constraint.
for (int x = 1; x < 100000; ++x) {
String s = Integer.toString(x);
StringBuffer buffer = new StringBuffer(s);
buffer.reverse();
int y = Integer.parseInt(s +
buffer.substring(1).toString());
// if y>=N and it is a prime number
// then return it.
if (y >= N && isPrime(y) == true)
return y;
}
return -1;
}
// Driver code
public static void main(String args[])
{
System.out.print(primePalindrome(112));
}
}
Python 3
# Python3 program to find next palindromic
# prime for a given number.
import math as mt
def isPrime(num):
if (num < 2 or num % 2 == 0):
return num == 2
for i in range(3, mt.ceil(mt.sqrt(num + 1))):
if (num % i == 0):
return False
return True
def primePalindrome(N):
# if(8<=N<=11) return 11
if (8 <= N and N <= 11):
return 11
# generate odd length palindrome number
# which will cover given constraint.
for x in range(1, 100000):
s = str(x)
d = s[::-1]
y = int(s + d[1:])
# if y>=N and it is a prime number
# then return it.
if (y >= N and isPrime(y)):
return y
# Driver code
print(primePalindrome(112))
# This code is contributed by
# Mohit kumar 29
C
// C# program to find next palindromic
// prime for a given number.
using System;
using System.Text;
using System.Collections;
class Geeks {
static bool isPrime(int num)
{
if (num < 2 || num % 2 == 0)
return num == 2;
for (int i = 3; i * i <= num; i += 2)
if (num % i == 0)
return false;
return true;
}
static int primePalindrome(int N)
{
// if(8<=N<=11) return 11
if (8 <= N && N <= 11)
return 11;
// generate odd length palindrome number
// which will cover given constraint.
for (int x = 1; x < 100000; ++x) {
string s = x.ToString();
char[] buffer = s.ToCharArray();
Array.Reverse(buffer);
int y = Int32.Parse(s + new string(buffer).Substring(1));
// if y>=N and it is a prime number
// then return it.
if (y >= N && isPrime(y) == true)
return y;
}
return -1;
}
// Driver code
public static void Main()
{
Console.WriteLine(primePalindrome(112));
}
}
// This code is contributed by Mithun Kumar.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find next palindromic
// prime for a given number.
function isPrime($num)
{
if ($num < 2 || $num % 2 == 0)
return $num == 2;
for ($i = 3; $i * $i <= $num; $i += 2)
if ($num % $i == 0)
return false;
return true;
}
function primePalindrome($N)
{
// if(8<=N<=11) return 11
if (8 <= $N && $N <= 11)
return 11;
// generate odd length palindrome number
// which will cover given constraint.
for ($x = 1; $x < 100000; ++$x)
{
$s = strval($x);
$r = strrev($s);
$y = intval($s.substr($r, 1));
// if y>=N and it is a prime number
// then return it.
if ($y >= $N && isPrime($y) == true)
return $y;
}
return -1;
}
// Driver code
print(primePalindrome(112));
// This code is contributed by mits
?>
java 描述语言
<script>
// Javascript program to find next palindromic
// prime for a given number.
function isPrime(num)
{
if (num < 2 || num % 2 == 0)
return num == 2;
for (i = 3; i * i <= num; i += 2)
if (num % i == 0)
return false;
return true;
}
function primePalindrome(N)
{
// if(8<=N<=11) return 11
if (8 <= N && N <= 11)
return 11;
// generate odd length palindrome number
// which will cover given constraint.
for (let x = 1; x < 100000; ++x)
{
let s = String(x);
let r = s.split("").reverse().join("");
let y = parseInt(s + r.substr(1));
// if y>=N and it is a prime number
// then return it.
if (y >= N && isPrime(y) == true)
return y;
}
return -1;
}
// Driver code
document.write(primePalindrome(112));
// This code is contributed by gfgking
</script>
Output:
131
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