如何使用循环运行 c 程序来查找两个数字的余数而不使用乘法、除法和模运算符？

Question

#include <stdio.h>

int main()
{
    int num1, num2;

    printf ("Input value for num1: ");
    scanf ("%d", &num1);
    printf ("Input value for num2: ");
    scanf ("%d", &num2);   

    int prod =0, i;
    for(i = 1; i <= num1; i++){
        prod += num2;
    }

    int quo = 0 , rem = 0;
    for(rem = num1 - num2; rem >= 0; rem = rem-num2) {
        if(rem < 0)             
            break;
        else
            quo++;
    }

    //The last part is that i need to find the remainder without  using multiplication, division and the modulo itself.

    printf ("The product of %d and %d is: %d\n", num1, num2, prod);
    printf ("The integer quotient of %d and %d is: %d\n", num1, num2, quo);

    return 0;
}

Answer 1

If you want a slow calculation of num1 % num2 (ie without multiplication/division) you can do:

// Calculate num1 % num2

unsigned rem(unsigned num1, unsigned num2)
{
    if (num2 == 0) {.... error handling ....}

    while (num1 >= num2) num1 -= num2;
    return num1;
}

int main(void)
{
    unsigned num1 = 42;
    unsigned num1 = 3;
    unsigned rem = rem(num1, num2);
    printf("%u", rem);
    return 0;
}

Answer 2

The simplest solution for calculating a mod b for positive integers a and b with only subtraction and addition is to subtract b from a until the result is smaller than a . However, this takes many iterations if b is much smaller than a .

A method with better worst-case performance is the following:

#include <stdio.h>

unsigned rem(unsigned a, unsigned b)
{
    if(b == 0) return 0;  // Error
    while(a >= b)
    {
        unsigned s = b;
        do
        {
            a = a - s;
            s = s + s;
        } while(a >= s);
    }
    return a;
}

int main(void)
{
    unsigned example = rem(32453, 3);
    printf("%u\n", example);
}

This method is based on the fact that to get closer to the result, we can subtract any multiple of b as long as it is smaller than a , so in each inner iteration we try to subtract twice the multiples of the last iteration until the subtractor becomes too large and we start over again with a single multiple of b .

Be aware that this will give wrong results if s = s + s; overflows the unsigned range. Hence, a should not be larger than half the upper limit of unsigned .