Проверка делимости с помощью побитовых операции C++

посмотри на это: Как получить остаток от деления через битовые операции?

#include <stdio.h>

#include <stdint.h>

int subtract(uint32_t x, uint32_t y) {
    // Iterate till there 
    // is no carry 
    while (y) {
        // borrow contains common  
        // set bits of y and unset 
        // bits of x 
        uint32_t borrow = (~x) & y;

        // Subtraction of bits of x  
        // and y where at least one 
        // of the bits is not set 
        x = x ^ y;

        // Borrow is shifted by one  
        // so that subtracting it from 
        // x gives the required sum 
        y = borrow << 1;
    }
    return x;
}

uint32_t inc(uint32_t i) {
    uint32_t mask = 1;
    while (i & mask) {
        i &= ~mask;
        mask <<= 1;
    }
    i |= mask;
    return i;
}

uint32_t divide(uint32_t n, uint32_t d) {
    // n is dividend, d is divisor
    // store the result in q: q = n / d
    uint32_t q = 0;

    // as long as the divisor fits into the remainder there is something to do
    while (n >= d) {
        uint32_t i = 0, d_t = d;
        // determine to which power of two the divisor still fits the dividend
        //
        // i.e.: we intend to subtract the divisor multiplied by powers of two
        // which in turn gives us a one in the binary representation 
        // of the result
        while (n >= (d_t << 1)) {
            i = inc(i);
            d_t <<= 1;
        }
        // set the corresponding bit in the result
        q |= 1 << i;
        // subtract the multiple of the divisor to be left with the remainder
        n = subtract(n, d_t);
        // repeat until the divisor does not fit into the remainder anymore
    }
    return q;
}

int add(uint32_t x, uint32_t y) {
    // Iterate till there is no carry  
    while (y) {
        // carry now contains common set bits of x and y
        uint32_t carry = x & y;

        // Sum of bits of x and y where at least one of the bits is not set
        x = x ^ y;

        // Carry is shifted by one so that adding it to x gives the required sum
        y = carry << 1;
    }
    return x;
}

uint32_t multiply(uint32_t a, uint32_t b) {
    uint32_t result = 0;
    while (b) // Iterate the loop till b==0
    {
        if (b & 01) // Bitwise &  of the value of b with 01
        {
            result = add(result, a); // Add a to result if b is odd .
        }
        a <<= 1; // Left shifting the value contained in 'a' by 1 
        // multiplies a by 2 for each loop
        b >>= 1; // Right shifting the value contained in 'b' by 1.
    }
    return result;
}

uint32_t mod(uint32_t a, uint32_t b) {
    uint32_t r = divide(a, b); // truncated division
    return subtract(a, multiply(r, b));
}

int main(void) {
    printf("%un", mod(33, 31));
    return 0;
}