kaprekar 编号
Kaprekar 数是这样一个数，它的平方被分成两部分，这样两部分的和等于原始数，并且没有一部分的值为 0。(来源: Wiki ) 给定一个号码，任务是检查是否是 Kaprekar 号码。 示例:
Input : n = 45
Output : Yes
Explanation : 452 = 2025 and 20 + 25 is 45
Input : n = 13
Output : No
Explanation : 132 = 169\. Neither 16 + 9 nor 1 + 69 is equal to 13
Input : n = 297
Output : Yes
Explanation: 2972 = 88209 and 88 + 209 is 297
Input : n = 10
Output : No
Explanation: 102 = 100\. It is not a Kaprekar number even if
sum of 100 + 0 is 100\. This is because of the condition that
none of the parts should have value 0.
- 求 n 的平方，并计算平方中的位数。
- 在不同的位置拆分正方形，看看任何拆分中两个部分的和是否等于 n。
下面是这个想法的实现。
C++
//C++ program to check if a number is Kaprekar number or not
#include<bits/stdc++.h>
using namespace std;
// Returns true if n is a Kaprekar number, else false
bool iskaprekar(int n)
{
if (n == 1)
return true;
// Count number of digits in square
int sq_n = n * n;
int count_digits = 0;
while (sq_n)
{
count_digits++;
sq_n /= 10;
}
sq_n = n*n; // Recompute square as it was changed
// Split the square at different poitns and see if sum
// of any pair of splitted numbers is equal to n.
for (int r_digits=1; r_digits<count_digits; r_digits++)
{
int eq_parts = pow(10, r_digits);
// To avoid numbers like 10, 100, 1000 (These are not
// Karprekar numbers
if (eq_parts == n)
continue;
// Find sum of current parts and compare with n
int sum = sq_n/eq_parts + sq_n % eq_parts;
if (sum == n)
return true;
}
// compare with original number
return false;
}
// Driver code
int main()
{
cout << "Printing first few Kaprekar Numbers"
" using iskaprekar()\n";
for (int i=1; i<10000; i++)
if (iskaprekar(i))
cout << i << " ";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check if a number is
// Kaprekar number or not
class GFG
{
// Returns true if n is a Kaprekar number, else false
static boolean iskaprekar(int n)
{
if (n == 1)
return true;
// Count number of digits in square
int sq_n = n * n;
int count_digits = 0;
while (sq_n != 0)
{
count_digits++;
sq_n /= 10;
}
sq_n = n*n; // Recompute square as it was changed
// Split the square at different poitns and see if sum
// of any pair of splitted numbers is equal to n.
for (int r_digits=1; r_digits<count_digits; r_digits++)
{
int eq_parts = (int) Math.pow(10, r_digits);
// To avoid numbers like 10, 100, 1000 (These are not
// Karprekar numbers
if (eq_parts == n)
continue;
// Find sum of current parts and compare with n
int sum = sq_n/eq_parts + sq_n % eq_parts;
if (sum == n)
return true;
}
// compare with original number
return false;
}
// Driver method
public static void main (String[] args)
{
System.out.println("Printing first few Kaprekar Numbers" +
" using iskaprekar()");
for (int i=1; i<10000; i++)
if (iskaprekar(i))
System.out.print(i + " ");
}
}
计算机编程语言
# Python program to check if a number is Kaprekar number or not
import math
# Returns true if n is a Kaprekar number, else false
def iskaprekar( n):
if n == 1 :
return True
#Count number of digits in square
sq_n = n * n
count_digits = 1
while not sq_n == 0 :
count_digits = count_digits + 1
sq_n = sq_n / 10
sq_n = n*n # Recompute square as it was changed
# Split the square at different poitns and see if sum
# of any pair of splitted numbers is equal to n.
r_digits = 0
while r_digits< count_digits :
r_digits = r_digits + 1
eq_parts = (int) (math.pow(10, r_digits))
# To avoid numbers like 10, 100, 1000 (These are not
# Karprekar numbers
if eq_parts == n :
continue
# Find sum of current parts and compare with n
sum = sq_n/eq_parts + sq_n % eq_parts
if sum == n :
return True
# compare with original number
return False
# Driver method
i=1
while i<10000 :
if (iskaprekar(i)) :
print i," ",
i = i + 1
# code contributed by Nikita Tiwari
C
// C# program to check if a number is
// Kaprekar number or not
using System;
class GFG {
// Returns true if n is a Kaprekar
// number, else false
static bool iskaprekar(int n)
{
if (n == 1)
return true;
// Count number of digits
// in square
int sq_n = n * n;
int count_digits = 0;
while (sq_n != 0) {
count_digits++;
sq_n /= 10;
}
// Recompute square as it was changed
sq_n = n * n;
// Split the square at different poitns
// and see if sum of any pair of splitted
// numbers is equal to n.
for (int r_digits = 1; r_digits < count_digits;
r_digits++)
{
int eq_parts = (int)Math.Pow(10, r_digits);
// To avoid numbers like 10, 100, 1000
// These are not Karprekar numbers
if (eq_parts == n)
continue;
// Find sum of current parts and compare
// with n
int sum = sq_n / eq_parts + sq_n % eq_parts;
if (sum == n)
return true;
}
// compare with original number
return false;
}
// Driver method
public static void Main()
{
Console.WriteLine("Printing first few "
+ "Kaprekar Numbers using iskaprekar()");
for (int i = 1; i < 10000; i++)
if (iskaprekar(i))
Console.Write(i + " ");
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to check if a number
// is Kaprekar number or not
// Returns true if n is a Kaprekar
// number, else false
function iskaprekar($n)
{
if ($n == 1)
return true;
// Count number of digits
// in square
$sq_n = $n * $n;
$count_digits = 0;
while ($sq_n)
{
$count_digits++;
$sq_n = (int)($sq_n / 10);
}
$sq_n1 = $n * $n; // Recompute square
// as it was changed
// Split the square at different
// points and see if sum of any
// pair of splitted numbers is equal to n.
for ($r_digits = 1;
$r_digits < $count_digits;
$r_digits++)
{
$eq_parts = pow(10, $r_digits);
// To avoid numbers like
// 10, 100, 1000 (These are not
// Karprekar numbers
if ($eq_parts == $n)
continue;
// Find sum of current parts
// and compare with n
$sum = (int)($sq_n1 / $eq_parts) +
$sq_n1 % $eq_parts;
if ($sum == $n)
return true;
}
// compare with original number
return false;
}
// Driver code
echo "Printing first few Kaprekar " .
"Numbers using iskaprekar()\n";
for ($i = 1; $i < 10000; $i++)
if (iskaprekar($i))
echo $i . " ";
// This code is contributed by mits
?>
java 描述语言
<script>
// Javascript program to check if a number
// is Kaprekar number or not
// Returns true if n is a Kaprekar
// number, else false
function iskaprekar(n)
{
if (n == 1)
return true;
// Count number of digits
// in square
let sq_n = n * n;
let count_digits = 0;
while (sq_n)
{
count_digits++;
sq_n = parseInt(sq_n / 10);
}
let sq_n1 = n * n; // Recompute square
// as it was changed
// Split the square at different
// points and see if sum of any
// pair of splitted numbers is equal to n.
for (let r_digits = 1;
r_digits < count_digits;
r_digits++)
{
let eq_parts = Math.pow(10, r_digits);
// To avoid numbers like
// 10, 100, 1000 (These are not
// Karprekar numbers
if (eq_parts == n)
continue;
// Find sum of current parts
// and compare with n
let sum = parseInt((sq_n1 / eq_parts) +
sq_n1 % eq_parts);
if (sum == n)
return true;
}
// compare with original number
return false;
}
// Driver code
document.write("Printing first few Kaprekar " +
"Numbers using iskaprekar()<br>");
for (let i = 1; i < 10000; i++)
if (iskaprekar(i))
document.write(i + " ");
// This code is contributed by _saurabh_jaiswal
</script>
输出:
Printing first few Kaprekar Numbers using iskaprekar()
1 9 45 55 99 297 703 999 2223 2728 4879 4950 5050 5292 7272 7777 9999
参考: 【https://en.wikipedia.org/wiki/Kaprekar_number】 相关文章: Kaprekar 常量 本文由Sahil Chhabra(KILLER)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。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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