用于计数可被 8 整除的旋转的Java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 8 整除的旋转。

例子：

Input: 8
Output: 1

Input: 40
Output: 1
Rotation: 40 is divisible by 8
          04 is not divisible by 8

Input : 13502
Output : 0
No rotation is divisible by 8

Input : 43262488612
Output : 4

方法：对于大数，很难将每个数字旋转并除以 8。因此，使用“被 8 整除”属性，即如果数字的最后 3 位数字可以被 8 整除，则该数字可以被 8 整除。这里我们实际上并没有旋转数字并检查最后 8 位数字的可分性，而是计算可被 8 整除的 3 位数字的连续序列（以循环方式）。

插图：

Consider a number 928160
Its rotations are 928160, 092816, 609281, 
160928, 816092, 281609.
Now form consecutive sequence of 3-digits from 
the original number 928160 as mentioned in the 
approach. 
3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6), 
(1, 6, 0),(6, 0, 9), (0, 9, 2)
We can observe that the 3-digit number formed by 
the these sets, i.e., 928, 281, 816, 160, 609, 092, 
are present in the last 3 digits of some rotation.
Thus, checking divisibility of these 3-digit numbers
gives the required number of rotations. 

Java

// Java program to count all 
// rotations divisible by 8
import java.io.*;
  
class GFG 
{
    // function to count of all 
    // rotations divisible by 8
    static int countRotationsDivBy8(String n)
    {
        int len = n.length();
        int count = 0;
      
        // For single digit number
        if (len == 1) {
            int oneDigit = n.charAt(0) - '0';
            if (oneDigit % 8 == 0)
                return 1;
            return 0;
        }
      
        // For two-digit numbers 
        // (considering all pairs)
        if (len == 2) {
      
            // first pair
            int first = (n.charAt(0) - '0') * 
                        10 + (n.charAt(1) - '0');
      
            // second pair
            int second = (n.charAt(1) - '0') * 
                         10 + (n.charAt(0) - '0');
      
            if (first % 8 == 0)
                count++;
            if (second % 8 == 0)
                count++;
            return count;
        }
      
        // considering all three-digit sequences
        int threeDigit;
        for (int i = 0; i < (len - 2); i++) 
        {
            threeDigit = (n.charAt(i) - '0') * 100 + 
                        (n.charAt(i + 1) - '0') * 10 + 
                        (n.charAt(i + 2) - '0');
            if (threeDigit % 8 == 0)
                count++;
        }
      
        // Considering the number formed by the 
        // last digit and the first two digits
        threeDigit = (n.charAt(len - 1) - '0') * 100 + 
                    (n.charAt(0) - '0') * 10 + 
                    (n.charAt(1) - '0');
      
        if (threeDigit % 8 == 0)
            count++;
      
        // Considering the number formed by the last 
        // two digits and the first digit
        threeDigit = (n.charAt(len - 2) - '0') * 100 +
                    (n.charAt(len - 1) - '0') * 10 + 
                    (n.charAt(0) - '0');
        if (threeDigit % 8 == 0)
            count++;
      
        // required count of rotations
        return count;
    }
      
    // Driver program 
    public static void main (String[] args)
    {
        String n = "43262488612";
        System.out.println( "Rotations: "
                       +countRotationsDivBy8(n));
          
    }
}
  
// This code is contributed by vt_m.

输出：

Rotations: 4

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 8 整除的计数旋转的完整文章！