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# C++ Program to Count rotations divisible by 8

• Last Updated : 09 Jun, 2022

Given a large positive number as string, count all rotations of the given number which are divisible by 8.

Examples:

```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```

Approach: For large numbers it is difficult to rotate and divide each number by 8. Therefore, ‘divisibility by 8’ property is used which says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not actually rotate the number and check last 8 digits for divisibility, instead we count consecutive sequence of 3 digits (in circular way) which are divisible by 8.

Illustration:

```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. ```

## C++

 `// C++ program to count all rotations divisible``// by 8``#include ``using` `namespace` `std;` `// function to count of all rotations divisible``// by 8``int` `countRotationsDivBy8(string n)``{``    ``int` `len = n.length();``    ``int` `count = 0;` `    ``// For single digit number``    ``if` `(len == 1) {``        ``int` `oneDigit = n - ``'0'``;``        ``if` `(oneDigit % 8 == 0)``            ``return` `1;``        ``return` `0;``    ``}` `    ``// For two-digit numbers (considering all``    ``// pairs)``    ``if` `(len == 2) {` `        ``// first pair``        ``int` `first = (n - ``'0'``) * 10 + (n - ``'0'``);` `        ``// second pair``        ``int` `second = (n - ``'0'``) * 10 + (n - ``'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[i] - ``'0'``) * 100 +``                     ``(n[i + 1] - ``'0'``) * 10 +``                     ``(n[i + 2] - ``'0'``);``        ``if` `(threeDigit % 8 == 0)``            ``count++;``    ``}` `    ``// Considering the number formed by the``    ``// last digit and the first two digits``    ``threeDigit = (n[len - 1] - ``'0'``) * 100 +``                 ``(n - ``'0'``) * 10 +``                 ``(n - ``'0'``);` `    ``if` `(threeDigit % 8 == 0)``        ``count++;` `    ``// Considering the number formed by the last``    ``// two digits and the first digit``    ``threeDigit = (n[len - 2] - ``'0'``) * 100 +``                 ``(n[len - 1] - ``'0'``) * 10 +``                 ``(n - ``'0'``);``    ``if` `(threeDigit % 8 == 0)``        ``count++;` `    ``// required count of rotations``    ``return` `count;``}` `// Driver program to test above``int` `main()``{``    ``string n = ``"43262488612"``;``    ``cout << ``"Rotations: "``         ``<< countRotationsDivBy8(n);``    ``return` `0;``}`

Output:

`Rotations: 4`

Time Complexity: O(n), where n is the number of digits in the input number.
Auxiliary Space: O(1)
Please refer complete article on Count rotations divisible by 8 for more details!

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