C language, I created this code but I want to make it more efficient(reduce the runtime)

[Liam mallka]

<

#include <stdio.h>
#include <math.h>

void math(int a, int b);

int main()
{
    int a, b;
    printf("Enter 2 numbers: ");
    scanf("%d %d", &a, &b);
    math(a, b);
    return 0; 
}

void math(int a,int b)
{
    int i = 0;
    for (i = a; i >= b; i--)
    {
        if (i % b == 0)
        {
            printf("i = %d b = %d\n", i, b);
            printf("%d \n", i);
        }
    }
}> 

 

[AbydosOne]

17 minutes ago, Liam mallka said:

snip

What is the program supposed to do?

[Liam mallka]

6 minutes ago, AbydosOne said:

What is the program supposed to do?

checks if b can fit in a with 0 reminder  until b is 1

[AbydosOne]

10 minutes ago, Liam mallka said:

checks if a can fit in b with 0 reminder 

I'm not sure the reasoning behind your for loop, but the simplest way to get that answer is...

void math(int a, int b) {
  if (b % a == 0) {
    // print things here
  }
}

EDIT: you edited your previous response so my answer makes less sense...

Edited June 11, 2021 by AbydosOne

[badreg]

Why is a loop necessary? You just need to do the modulo operation once, and subtract the modulo from a to get your desired answer.

 

Edit: if you are trying to find every factor of b that is ≤ a, then you just need to find the first factor as I stated above, and then decrement that factor by b until it is ≤ 0. There's no need to decrement by 1 if b is > 1.

[wanderingfool2]

So to reduce the runtime, as @badreg said, you can reduce the problem so the comparison loop is done less.  @badreg is right.  If order isn't important, then it's actually most efficient to start from b and continue upwards adding b to i until you hit a (that way you only process the known values, instead of guessing)

 

To put a real world example

a = 10; b=3;

Your method

i = 10,9 (prints),8,7,6(prints),5,4,3(prints) (7 iterations of i%3 == 0)

 

badreg

i = 10,9 (prints) [i-b], 6 (prints), [i-b], 3 (prints) (2 iterations of i%3 == 0, then 2 subtract by b's...so only 4 real actions)

 

my altered method

i = b

while (i <= a)

i = i + b

 

i = 3 (prints), 6(prints), 9 (prints) (the optimal solution, only 3 real operations)

 

To put in a larger example

a = 1 999 999 999, b = 1000000

Your method

i = 1999999999,.... (so ~1.999 billion iterations, printing only 1999 times)

 

badreg's method

i = 1999999999, 1999999998..1999000000, 1998000000, 1997000000, .... (1 million initial operations, followed by 1998 operations so 2000x faster approximately) *yes, I intentionally did a contrived example*

 

altered method (where you don't care)

i = b while(i<=a) i = i + b

i = 1000000, 2000000, 3000000,... (1999 total additions, so about 1 million times faster...in real world numbers maybe more, since you only are doing 1999 adds, and <= comparisons...vs 1 billion subtracts, % operators, and >= comparisons)

[straight_stewie]

If I'm following this correctly, that just prints all numbers b <= n <= a where n % b = 0. In plain English, you are looking to find all multiples of b between b and a inclusive.

 

What you should do is find the largest multiple of b less than a. We can do this by writing:

int largestMultiple = div(a, b) * b;

Then you would just continually subtract b from largestMultiple until you are equal to or less than b, printing as you go along.

This is about as efficient as it gets, because it only requires as many iterations as there are multiples of b such that  b <= multiple <= a.

[mariushm]

could probably do something like  

while  a % b != 0  a = a -1;

This gets you to the maximum number that can be a multiple on b and lower than a. 

Now you can just do a loop and print number, substract b, print ..