Compute n modulo d without division(/) and modulo(%) operators, where d is a power of 2 number.
Let ith bit from right is set in d. For getting n modulus d, we just need to return 0 to i-1 (from right) bits of n as they are and other bits as 0.
For example if n = 6 (00..110) and d = 4(00..100). Last set bit in d is at position 3 (from right side). So we need to return last two bits of n as they are and other bits as 0, i.e., 00..010.
Now doing it is so easy, guess it….
Yes, you have guessing it right. See the below program.
Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.
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Improved By : vt_m