计算可被 4 整除的旋转次数的 C++ 程序

给定一个大的正数作为字符串，计算给定数的所有可被 4 整除的旋转。

例子：

Input: 8
Output: 1

Input: 20
Output: 1
Rotation: 20 is divisible by 4
          02 is not divisible by 4

Input : 13502
Output : 0
No rotation is divisible by 4

Input : 43292816
Output : 5
5 rotations are : 43292816, 16432928, 81643292
                  92816432, 32928164 

对于大数字，很难将每个数字旋转并除以 4。因此，使用“被 4 整除”属性，即如果数字的最后 2 位数字可以被 4 整除，则该数字可以被 4 整除。这里我们做实际上并没有旋转数字并检查最后 2 位数字的可分性，而是我们计算可被 4 整除的连续对（以循环方式）。

插图：

Consider a number 928160
Its rotations are 928160, 092816, 609281, 160928, 
    816092, 281609.
Now form pairs from the original number 928160
as mentioned in the approach.
Pairs: (9,2), (2,8), (8,1), (1,6), 
         (6,0), (0,9)
We can observe that the 2-digit number formed by the these 
pairs, i.e., 92, 28, 81, 16, 60, 09, are present in the last
2 digits of some rotation.
Thus, checking divisibility of these pairs gives the required
number of rotations. 

Note: A single digit number can directly
be checked for divisibility.

下面是该方法的实现。

C++

// C++ program to count all rotation divisible
// by 4.
#include 
using namespace std;
  
// Returns count of all rotations divisible
// by 4
int countRotations(string n)
{
    int len = n.length();
  
    // For single digit number
    if (len == 1)
    {
        int oneDigit = n.at(0)-'0';
        if (oneDigit%4 == 0)
            return 1;
        return 0;
    }
  
    // At-least 2 digit number (considering all
    // pairs)
    int twoDigit, count = 0;
    for (int i=0; i<(len-1); i++)
    {
        twoDigit = (n.at(i)-'0')*10 + (n.at(i+1)-'0');
        if (twoDigit%4 == 0)
            count++;
    }
  
    // Considering the number formed by the pair of
    // last digit and 1st digit
    twoDigit = (n.at(len-1)-'0')*10 + (n.at(0)-'0');
    if (twoDigit%4 == 0)
        count++;
  
    return count;
}
  
//Driver program
int main()
{
    string n = "4834";
    cout << "Rotations: " << countRotations(n) << endl;
    return 0;
}

输出：

Rotations: 2

时间复杂度： O(n)，其中 n 是输入数字的位数。

有关详细信息，请参阅有关可被 4 整除的计数旋转的完整文章！