# C++ Program to Count rotations divisible by 4

• Last Updated : 06 Jan, 2022

Given a large positive number as string, count all rotations of the given number which are divisible by 4.

Examples:

```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 ```

For large numbers it is difficult to rotate and divide each number by 4. Therefore, ‘divisibility by 4’ property is used which says that a number is divisible by 4 if the last 2 digits of the number is divisible by 4. Here we do not actually rotate the number and check last 2 digits for divisibility, instead we count consecutive pairs (in circular way) which are divisible by 4.

Illustration:

```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.```

Below is the implementation of the approach.

## 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;``}`

Output:

`Rotations: 2`

Time Complexity : O(n) where n is number of digits in input number.

Please refer complete article on Count rotations divisible by 4 for more details!

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