计算所有回文数，这是一个回文数的平方

给定两个正整数L和R(表示为字符串)，其中$1\leq L \leq R\leq10^{18}$ 。任务是找到包含范围[L，R]的超级回文总数。
如果回文既是回文又是回文的平方，则称为超级回文。
例子：

Input: L = "4", R = "1000"
Output: 4
Explanation: 4, 9, 121, and 484 are super-palindromes.

Input : L = "100000", R = "10000000000"
Output : 11

方法：
可以说$P = R^{2}$是一种超级回文。
现在，因为R是回文，的R的数字的第一半可以被用于确定R将一直到两种可能性。让我成为R中数字的前半部分。例如。如果i = 123 ，则R = 12321或R = 123321 。因此，我们可以遍历所有这些数字。同样，每个可能性在R中可以具有奇数或偶数个数字。
因此，我们遍历每个i直到10 ⁵并创建关联的回文数R ，并检查R ²是否为回文数。
同样，我们将分别处理奇数和偶数回文，并且只要回文超过R就会中断。
现在，因为$P \leq 10^{18}$ ， 和$R \leq (10^{18})^{\frac{1}{2}} \equiv 10^{9}$和$R = i||i^{'}$ (在串联上)，其中i ^‘是i的倒数(双向)，因此我们的LIMIT不会大于$i\leq10^{5}$ 。
下面是上述方法的实现：

C++

// C++ implementation of the 
// above approach
#include  
using namespace std;
  
// check if a number is a palindrome
bool ispalindrome(int x)
{
    int ans = 0;
    int temp = x;
    while (temp > 0)
    {
        ans = 10 * ans + temp % 10;
        temp = temp / 10;
    }
    return ans == x;
}
  
// Function to return required count 
// of palindromes
int SuperPalindromes(int L, int R)
{
    // Range [L, R]
  
    // Upper limit
    int LIMIT = 100000;
  
    int ans = 0;
  
    // count odd length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
        string s = to_string(i); // if s = '1234'
  
        string rs = s.substr(0, s.size() - 1);
        reverse(rs.begin(), rs.end());
  
        // then, t = '1234321'
        string p = s + rs; 
        int p_sq     = pow(stoi(p), 2);
        if (p_sq > R)
            break;
        if (p_sq >= L and ispalindrome(p_sq))
            ans = ans + 1;
    }
  
    // count even length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
        string s = to_string(i); // if s = '1234'
  
        string rs = s;
        reverse(rs.begin(), rs.end());
        string p = s + rs; // then, t = '12344321'
        int p_sq = pow(stoi(p), 2);
        if (p_sq > R)
            break;
        if (p_sq >= L and ispalindrome(p_sq))
            ans = ans + 1;
    }
  
    // Return count of super-palindromes
    return ans;
}
  
// Driver Code
int main()
{
    string L = "4";
    string R = "1000";
      
    // function call to get required answer
    printf("%d\n", SuperPalindromes(stoi(L), 
                                  stoi(R)));
    return 0;
} 
  
// This code is contributed 
// by Harshit Saini

Java

// Java implementation of the
// above approach
import java.lang.*;
  
class GFG
{
  
// check if a number is a palindrome
public static boolean ispalindrome(int x)
{
    int ans = 0;
    int temp = x;
    while (temp > 0)
    {
        ans = 10 * ans + temp % 10;
        temp = temp / 10;
    }
    return ans == x;
}
  
// Function to return required 
// count of palindromes
public static int SuperPalindromes(int L, 
                                   int R)
{
    // Range [L, R]
  
    // Upper limit
    int LIMIT = 100000;
  
    int ans = 0;
  
    // count odd length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        String s = Integer.toString(i);
  
        StringBuilder rs = new StringBuilder();
        rs.append(s.substring(0,
                     Math.max(1, s.length() - 1)));
        String srs = rs.reverse().toString();
  
        // then, t = '1234321'
        String p = s + srs;
        int p_sq = (int)(Math.pow(
                         Integer.parseInt(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // count even length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        String s = Integer.toString(i); 
  
        StringBuilder rs = new StringBuilder();
        rs.append(s);
        rs = rs.reverse();
  
        String p = s + rs; // then, t = '12344321'
        int p_sq = (int)(Math.pow(
                         Integer.parseInt(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // Return count of super-palindromes
    return ans;
}
  
// Driver program
public static void main(String [] args)
{
    String L = "4";
    String R = "1000";
  
    // function call to get required answer
    System.out.println(SuperPalindromes(
       Integer.parseInt(L), Integer.parseInt(R)));
}
}
  
// This code is contributed 
// by Harshit Saini

Python3

# Python implementation of the above approach
  
# check if a number is a palindrome
def ispalindrome(x):
    ans, temp = 0, x
    while temp > 0:
        ans = 10 * ans + temp % 10
        temp = temp // 10
    return ans == x
  
# Function to return required count of palindromes
def SuperPalindromes(L, R):
    # Range [L, R]
    L, R = int(L), int(R)
  
    # Upper limit
    LIMIT = 100000
  
    ans = 0
  
    # count odd length palindromes
    for i in range(LIMIT):
        s = str(i)  # if s = '1234'
        p = s + s[-2::-1]  # then, t = '1234321'
        p_sq = int(p) ** 2
        if p_sq > R:
            break
        if p_sq >= L and ispalindrome(p_sq):
            ans = ans + 1
  
    # count even length palindromes
    for i in range(LIMIT):
        s = str(i)  # if s = '1234'
        p = s + s[::-1]  # then, t = '12344321'
        p_sq = int(p) ** 2
        if p_sq > R:
            break
        if p_sq >= L and ispalindrome(p_sq):
            ans = ans + 1
  
    # Return count of super-palindromes
    return ans
  
# Driver program
L = "4"
R = "1000"
  
# function call to get required answer
print(SuperPalindromes(L, R))
  
# This code is written by
# Sanjit_Prasad

C#

// C# implementation of the 
// above approach
using System;
  
class GFG
{
  
// check if a number is a palindrome
static bool ispalindrome(int x)
{
    int ans = 0;
    int temp = x;
    while (temp > 0)
    {
        ans = 10 * ans + temp % 10;
        temp = temp / 10;
    }
    return ans == x;
}
  
// utility function used for 
// reversing a string
static string Reverse( string s )
{
    char[] charArray = s.ToCharArray();
    Array.Reverse( charArray );
    return new string( charArray );
}
  
// Function to return required 
// count of palindromes
static int SuperPalindromes(int L, int R)
{
    // Range [L, R]
  
    // Upper limit
    int LIMIT = 100000;
  
    int ans = 0;
  
    // count odd length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        string s = i.ToString();
  
        string rs = s.Substring(0,
                       Math.Max(1, s.Length - 1));
        rs = Reverse(rs);
  
        // then, t = '1234321'
        string p = s + rs;
        int p_sq = (int)(Math.Pow(
                            Int32.Parse(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // count even length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        string s = i.ToString();
  
        string rs = Reverse(s);
  
        string p = s + rs; // then, t = '12344321'
        int p_sq = (int)(Math.Pow(
                            Int32.Parse(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // Return count of super-palindromes
    return ans;
}
  
// Driver Code
public static void Main()
{
    string L = "4";
    String R = "1000";
  
    // function call to get required answer
    Console.WriteLine(SuperPalindromes(
            Int32.Parse(L), Int32.Parse(R)));
}
}
  
// This code is contributed 
// by Harshit Saini

PHP

 0)
    {     
        $ans = (10 * $ans) +
               ($temp % 10);
        $temp = (int)($temp / 10);
    }
  
    return $ans == $x;
}
  
// Function to return required
// count of palindromes
function SuperPalindromes($L, $R)
{
    // Range [L, R]
    $L = (int)$L;
    $R = (int)$R;
  
    // Upper limit
    $LIMIT = 100000;
  
    $ans = 0;
  
    // count odd length palindromes
    for($i = 0 ;$i < $LIMIT; $i++)
    {
  
        $s = (string)$i; // if s = '1234'
        $rs = substr($s, 0, strlen($s) - 1);
        $p = $s.strrev($rs); // then, t = '1234321'
        $p_sq = (int)$p ** 2;
        if ($p_sq > $R)
        {
            break;
        }
        if ($p_sq >= $L and ispalindrome($p_sq))
        {
            $ans = $ans + 1;
        }
    } 
  
    // count even length palindromes
    for($i = 0 ;$i < $LIMIT; $i++)
    {
        $s = (string)$i; // if s = '1234'
        $p = $s.strrev($s); // then, t = '12344321'
          
        $p_sq = (int)$p ** 2;
  
        if ($p_sq > $R)
        {
            break;
        }
        if ($p_sq >= $L and ispalindrome($p_sq))
        {
            $ans = $ans + 1;
        }
    }
  
    // Return count of super-palindromes
    return $ans;
}
  
  
// Driver Code
$L = "4";
$R = "1000";
  
// function call to get required answer
echo SuperPalindromes($L, $R);
  
// This code is contributed 
// by Harshit Saini 
?>

输出：

4

时间复杂度： O(N * log(N))，其中N为上限，log(N)项来自检查候选人是否为回文症。