生成右旋转除数
原文:https://www . geesforgeks . org/generating-numbers-这些数字是它们右旋转的除数/
给定一个数 m，找出所有有 m 位数并且是其右旋转除数的数。数字 N 的右旋转是将 N 的数字向右旋转一个位置，并将最低有效数字环绕，使其成为最高有效数字的结果。例如，4356 的右旋转是 6435。
示例:
Input: 2
Output:
11
22
33
44
55
66
77
88
99
Input: 6
Output:
102564
111111
128205
142857
153846
179487
205128
222222
230769
333333
444444
555555
666666
777777
888888
999999
128205 satisfies the condition as 128205 * 4 = 512820.
蛮力法 : 最简单的方法是遍历所有大于等于 10 m-1 小于 10 m 的数字，检查是否满足要求的条件。我们可以用常数时间来检查，所以整个过程的时间复杂度为 O(10 m ，只对 m 的小值可行 以下是上述 方法的实现:
C++
// C++ program to Generating numbers that
// are divisor of their right-rotations
#include <bits/stdc++.h>
using namespace std;
// Function to check if N is a
// divisor of its right-rotation
bool rightRotationDivisor(int N)
{
int lastDigit = N % 10;
int rightRotation = (lastDigit * pow(10 ,int(log10(N))))
+ floor(N / 10);
return (rightRotation % N == 0);
}
// Function to generate m-digit
// numbers which are divisor of
// their right-rotation
void generateNumbers(int m)
{
for (int i=pow(10,(m - 1));i<pow(10 , m);i++)
if (rightRotationDivisor(i))
cout<<i<<endl;
}
// Driver code
int main()
{
int m = 3;
generateNumbers(m);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to Generating numbers that
// are divisor of their right-rotations
public class GFG {
// Function to check if N is a
// divisor of its right-rotation
static boolean rightRotationDivisor(int N)
{
int lastDigit = N % 10;
int rightRotation = (int)(lastDigit * Math.pow(10 ,(int)(Math.log10(N)))
+ Math.floor(N / 10));
return (rightRotation % N == 0);
}
// Function to generate m-digit
// numbers which are divisor of
// their right-rotation
static void generateNumbers(int m)
{
for (int i= (int)Math.pow(10,(m - 1)); i < Math.pow(10 , m);i++)
if (rightRotationDivisor(i))
System.out.println(i);
}
// Driver code
public static void main(String args[])
{
int m = 3;
generateNumbers(m);
}
// This Code is contributed by ANKITRAI1
}
Python 3
# Python program to Generating numbers that are
# divisor of their right-rotations
from math import log10
# Function to check if N is a
# divisor of its right-rotation
def rightRotationDivisor(N):
lastDigit = N % 10
rightRotation = (lastDigit * 10 ** int(log10(N))
+ N // 10)
return rightRotation % N == 0
# Function to generate m-digit
# numbers which are divisor of
# their right-rotation
def generateNumbers(m):
for i in range(10 ** (m - 1), 10 ** m):
if rightRotationDivisor(i):
print(i)
# Driver code
m = 3
generateNumbers(m)
C
// C# program to Generating numbers that
// are divisor of their right-rotations
using System;
public class GFG{
// Function to check if N is a
// divisor of its right-rotation
static bool rightRotationDivisor(int N)
{
int lastDigit = N % 10;
int rightRotation = (int)(lastDigit * Math.Pow(10 ,(int)(Math.Log10(N)))
+ Math.Floor((double)N/10));
return (rightRotation % N == 0);
}
// Function to generate m-digit
// numbers which are divisor of
// their right-rotation
static void generateNumbers(int m)
{
for (int i= (int)Math.Pow(10,(m - 1)); i < Math.Pow(10 , m);i++)
if (rightRotationDivisor(i))
Console.WriteLine(i);
}
// Driver code
public static void Main()
{
int m = 3;
generateNumbers(m);
}
}
// This code is contributed by 29AjayKumar
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to Generating numbers that
// are divisor of their right-rotations
// Function to check if N is a
// divisor of its right-rotation
function rightRotationDivisor($N)
{
$lastDigit = $N % 10;
$rightRotation = ($lastDigit * pow(10 ,
(int)(log10($N)))) +
floor($N / 10);
return ($rightRotation % $N == 0);
}
// Function to generate m-digit
// numbers which are divisor of
// their right-rotation
function generateNumbers($m)
{
for ($i = pow(10, ($m - 1));
$i < pow(10 , $m); $i++)
if (rightRotationDivisor($i))
echo $i . "\n";
}
// Driver code
$m = 3;
generateNumbers($m);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// Javascript program to Generating numbers that
// are divisor of their right-rotations
// Function to check if N is a
// divisor of its right-rotation
function rightRotationDivisor(N)
{
let lastDigit = N % 10;
let rightRotation = (lastDigit * Math.pow(10 ,
Math.floor((Math.log10(N)))) + Math.floor(N / 10));
return (rightRotation % N == 0);
}
// Function to generate m-digit
// numbers which are divisor of
// their right-rotation
function generateNumbers(m)
{
for (let i= Math.floor(Math.pow(10,(m - 1))); i <
Math.floor(Math.pow(10 , m));i++)
if (rightRotationDivisor(i))
document.write(i+"<br>");
}
// Driver code
let m = 3;
generateNumbers(m);
// This code is contributed by avanitrachhadiya2155
</script>
Output:
111
222
333
444
555
666
777
888
999
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