计算可以被10整除的旋转

给定数字N ，任务是计算给定数字的所有可被10整除的旋转。
例子：

Input: N = 10203 
Output: 2 
Explanation: 
There are 5 rotations possible for the given number. They are: 02031, 20310, 03102, 31020, 10203 
Out of these rotations, only 20310 and 31020 are divisible by 10. So 2 is the output. 
Input: N = 135 
Output: 0 
 

幼稚的方法：针对此问题的幼稚的方法是形成所有可能的旋转。已知对于数量为K的数量，该数量N的可能转数为K。因此，找到所有旋转，并为每个旋转检查数字是否可被10整除。这种方法的时间复杂度是二次的。
高效方法：高效方法背后的概念是，为了检查一个数字是否可以被10整除，我们只需检查最后一个数字是否为0。因此，其思想是简单地遍历给定的数字并找到计数为0。如果0的计数为F ，那么很明显，在给定数N的末尾， K旋转中的F个旋转将为0。
下面是上述方法的实现：

C++

// C++ implementation to find the
// count of rotations which are
// divisible by 10
 
#include 
using namespace std;
 
// Function to return the count of
// all the rotations which are
// divisible by 10.
int countRotation(int n)
{
    int count = 0;
 
    // Loop to iterate through the
    // number
    do {
        int digit = n % 10;
 
        // If the last digit is 0,
        // then increment the count
        if (digit == 0)
            count++;
        n = n / 10;
    } while (n != 0);
 
    return count;
}
 
// Driver code
int main()
{
    int n = 10203;
    cout << countRotation(n);
}

C#

// CSharp implementation to find the
// count of rotations which are
// divisible by 10
 
using System;
class Solution {
 
    // Function to return the count
    // of all rotations which are
    // divisible by 10.
    static int countRotation(int n)
    {
        int count = 0;
 
        // Loop to iterate through the
        // number
        do {
            int digit = n % 10;
 
            // If the last digit is 0,
            // then increment the count
            if (digit % 2 == 0)
                count++;
            n = n / 10;
        } while (n != 0);
 
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 10203;
        Console.Write(countRotation(n));
    }
}

Java

// Java implementation to find the
// count of rotations which are
// divisible by 10
 
class GFG {
 
    // Function to return the count
    // of all rotations which are
    // divisible by 10.
    static int countRotation(int n)
    {
        int count = 0;
 
        // Loop to iterate through the
        // number
        do {
            int digit = n % 10;
 
            // If the last digit is 0,
            // then increment the count
            if (digit == 0)
                count++;
            n = n / 10;
        } while (n != 0);
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 10203;
 
        System.out.println(countRotation(n));
    }
}

Python

# Python3 implementation to find the
# count of rotations which are
# divisible by 10
 
# Function to return the count of
# all rotations which are divisible
# by 10.
def countRotation(n):
    count = 0;
 
    # Loop to iterate through the
    # number
    while n > 0:
        digit = n % 10
 
        # If the last digit is 0,
        # then increment the count
        if(digit % 2 == 0):
            count = count + 1
        n = int(n / 10)
     
    return count;   
   
# Driver code 
if __name__ == "__main__" :
   
    n = 10203; 
    print(countRotation(n));

Javascript

 
// Javascript implementation to find the
// count of rotations which are
// divisible by 10
 
// Function to return the count of
// all the rotations which are
// divisible by 10.
function countRotation(n)
{
    let count = 0;
 
    // Loop to iterate through the
    // number
    do {
        let digit = n % 10;
 
        // If the last digit is 0,
        // then increment the count
        if (digit == 0)
            count++;
        n = parseInt(n / 10);
    } while (n != 0);
 
    return count;
}
 
// Driver code
let n = 10203;
document.write(countRotation(n));
 

输出：

2

时间复杂度： O(N) ，其中N是数字的长度。