两个 n 位数乘积的最大回文
原文:https://www . geesforgeks . org/maximum-回文-product-两位数-数字/
给定值 n，找出最大的回文数，它是两个 n 位数的乘积。
示例:
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
以下是找到所需号码的步骤。
- 找到 n 位数的下限。例如，对于 n = 2，lower_limit 为 10。
- 找出 n 位数的上限。例如，对于 n = 2，upper_limit 为 99。
- 考虑所有的数对，其中任何一个数都在范围内[下限，上限]
下面是以上步骤的实现。
C++
// C++ problem to find out the
// largest palindrome number which
// is product of two n digit numbers
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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-digit numbers
function larrgestPalindrome($n)
{
$upper_limit = 0;
// Loop to calculate upper bound
// (largest number of n-digit)
for ($i = 1; $i <= $n; $i++)
{
$upper_limit *= 10;
$upper_limit += 9;
}
// largest number of n-1 digit
// One plus this number
// is lower limit which is
// product of two numbers.
$lower_limit = 1 + (int)($upper_limit / 10);
// Initialize result
$max_product = 0;
for ($i = $upper_limit;
$i >= $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.
?>
java 描述语言
<script>
// Javascript problem to find out the
// largest palindrome number
// which is product of two
// n digit numbers.
// Function to calculate largest
// palindrome which isproduct of
// two n-digits numbers
function larrgestPalindrome(n)
{
let upper_limit = Math.pow(10, n) - 1;
// largest number of n-1 digit.
// One plus this number
// is lower limit which is
// product of two numbers.
let lower_limit = 1 +
parseInt(upper_limit / 10, 10);
// Initialize result
let max_product = 0;
for (let i = upper_limit; i >= lower_limit; i--)
{
for (let j = i; j >= lower_limit; j--)
{
// calculating product of two
// n-digit numbers
let product = i * j;
if (product < max_product)
break;
let number = product;
let reverse = 0;
// calculating reverse of product
// to check whether it is
// palindrome or not
while (number != 0)
{
reverse = reverse * 10 + number % 10;
number = parseInt(number / 10, 10);
}
// update new product if exist and if
// greater than previous one
if (product == reverse &&
product > max_product)
max_product = product;
}
}
return max_product;
}
let n = 2;
document.write(larrgestPalindrome(n));
</script>
输出:
9009
这篇文章中使用的方法简单明了。如果你找到更好的方法，请评论。 本文由希瓦姆·普拉丹(anuj_charm) 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用contribute.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 发现有不正确的地方请写评论，或者想分享更多以上讨论话题的信息
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