给定一个值n，找出最大回文数，该数是两个n位数字的乘积。
例子 ：
Input : n = 2
Output : 9009
9009 is the largest number which is product of two
2-digit numbers. 9009 = 91*99.
Input : n = 3
Output : 906609以下是查找所需号码的步骤。
- 找出n位数字的下限。例如，对于n = 2，lower_limit为10。
- 找出n位数字的上限。例如，对于n = 2，upper_limit为99。
- 考虑所有数字对，只要数字在[lower_limit，upper_limit]范围内
下面是上述步骤的实现。
C++
// C++ problem to find out the
// largest palindrome number which
// is product of two n digit numbers
#include
using namespace std;
// Function to calculate largest
// palindrome which is product of
// two n-digits numbers
int larrgestPalindrome(int n)
{
int upper_limit = pow(10,n) - 1;
// largest number of n-1 digit.
// One plus this number is lower
// limit which is product of two numbers.
int lower_limit = 1 + upper_limit / 10;
// Initialize result
int max_product = 0;
for (int i = upper_limit;
i >= lower_limit;
i--)
{
for (int j = i; j >= lower_limit; j--)
{
// calculating product of
// two n-digit numbers
int product = i * j;
if (product < max_product)
break;
int number = product;
int reverse = 0;
// calculating reverse of
// product to check whether
// it is palindrome or not
while (number != 0)
{
reverse = reverse * 10 +
number % 10;
number /= 10;
}
// update new product if exist
// and if greater than previous one
if (product == reverse &&
product > max_product)
max_product = product;
}
}
return max_product;
}
// Driver code
int main()
{
int n = 2;
cout << larrgestPalindrome(n);
return 0;
} Java
// Java problem to find out the
// largest palindrome number
// which is product of two
// n digit numbers.
import java.lang.Math;
class GFG
{
// Function to calculate largest
// palindrome which isproduct of
// two n-digits numbers
static int larrgestPalindrome(int n)
{
int upper_limit = (int)Math.pow(10, n) - 1;
// largest number of n-1 digit.
// One plus this number
// is lower limit which is
// product of two numbers.
int lower_limit = 1 + upper_limit / 10;
// Initialize result
int max_product = 0;
for (int i = upper_limit; i >= lower_limit; i--)
{
for (int j = i; j >= lower_limit; j--)
{
// calculating product of two
// n-digit numbers
int product = i * j;
if (product < max_product)
break;
int number = product;
int reverse = 0;
// calculating reverse of product
// to check whether it is
// palindrome or not
while (number != 0)
{
reverse = reverse * 10 + number % 10;
number /= 10;
}
// update new product if exist and if
// greater than previous one
if (product == reverse && product > max_product)
max_product = product;
}
}
return max_product;
}
// Driver code
public static void main (String[] args)
{
int n = 2;
System.out.print(larrgestPalindrome(n));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python problem to find
# out the largest palindrome
# number which is product of
# two n digit numbers.
# Function to calculate largest
# palindrome which is
# product of two n-digits numbers
def larrgestPalindrome(n):
upper_limit = (10**(n))-1
# largest number of n-1 digit.
# One plus this number
# is lower limit which is
# product of two numbers.
lower_limit = 1 + upper_limit//10
max_product = 0 # Initialize result
for i in range(upper_limit,lower_limit-1, -1):
for j in range(i,lower_limit-1,-1):
# calculating product of
# two n-digit numbers
product = i * j
if (product < max_product):
break
number = product
reverse = 0
# calculating reverse of
# product to check
# whether it is palindrome or not
while (number != 0):
reverse = reverse * 10 + number % 10
number =number // 10
# update new product if exist and if
# greater than previous one
if (product == reverse and product > max_product):
max_product = product
return max_product
# Driver code
n = 2
print(larrgestPalindrome(n))
# This code is contributed
# by Anant Agarwal.C#
// C# problem to find out the
// largest palindrome number
// which is product of two
// n digit numbers.
using System;
class GFG
{
// Function to calculate largest
// palindrome which isproduct of
// two n-digits numbers
static int larrgestPalindrome(int n)
{
int upper_limit = (int)Math.Pow(10, n) - 1;
// largest number of n-1 digit.
// One plus this number
// is lower limit which is
// product of two numbers.
int lower_limit = 1 + upper_limit / 10;
// Initialize result
int max_product = 0;
for (int i = upper_limit; i >= lower_limit; i--)
{
for (int j = i; j >= lower_limit; j--)
{
// calculating product of two
// n-digit numbers
int product = i * j;
if (product < max_product)
break;
int number = product;
int reverse = 0;
// calculating reverse of product
// to check whether it is
// palindrome or not
while (number != 0)
{
reverse = reverse * 10 + number % 10;
number /= 10;
}
// update new product if exist and if
// greater than previous one
if (product == reverse && product > max_product)
max_product = product;
}
}
return max_product;
}
// Driver code
public static void Main ()
{
int n = 2;
Console.Write(larrgestPalindrome(n));
}
}
// This code is contributed by nitin mittal.PHP
= $lower_limit;
$i--)
{
for ($j = $i;
$j >= $lower_limit;
$j--)
{
// calculating product of
// two n-digit numbers
$product = $i * $j;
if ($product < $max_product)
break;
$number = $product;
$reverse = 0;
// calculating reverse of
// product to check whether
// it is palindrome or not
while ($number != 0)
{
$reverse = $reverse * 10 +
$number % 10;
$number = (int)($number / 10);
}
// update new product if exist
// and if greater than previous one
if ($product == $reverse &&
$product > $max_product)
$max_product = $product;
}
}
return $max_product;
}
// Driver code
$n = 2;
echo(larrgestPalindrome($n));
// This code is contributed by Ajit.
?>输出 ：
9009