Thread: How to get rid of repeated if statements and make a loop.

Hi I have written a small program in which I am trying to assign values to variables a and b (1000-5000 with increment 1000 and 100-500 with increment of 100) dependent to the counter interval of 20. So that on each counter interval all possible combinations are assigned to these two variable (a and b). I apologize for my English and hope that my description makes some sense to the reader.
The program works as expected! But I want to make it look simpler and get rid of if repetitions. I have tried a while and for loop but due to lack of experience I could not simplify it. I hope some people can show me the way how to do it. And get my coding level further.
Thank you

Here is the code:


insert

Code:

#include <stdio.h>
#include <stdlib.h>


int main()
{
    int xinterval = 20;
    int y=5;
    int x=5;


    int count;
    int states = x*y;


    int a=0;
    int b=0;






    for(count=0;count<(xinterval*states);count++)
    {
        printf("%i\t%i\t%i\n", count,a,b);
        if(count>0){a=1000;b=100;}
        if(count>20){a=1000;b=200;}
        if(count>40){a=1000;b=300;}
        if(count>60){a=1000;b=400;}
        if(count>80){a=1000;b=500;}


        if(count>100){a=2000;b=100;}
        if(count>120){a=2000;b=200;}
        if(count>140){a=2000;b=300;}
        if(count>160){a=2000;b=400;}
        if(count>180){a=2000;b=500;}


        if(count>200){a=3000;b=100;}
        if(count>220){a=3000;b=200;}
        if(count>240){a=3000;b=300;}
        if(count>260){a=3000;b=400;}
        if(count>280){a=3000;b=500;}


        if(count>300){a=4000;b=100;}
        if(count>320){a=4000;b=200;}
        if(count>340){a=4000;b=300;}
        if(count>360){a=4000;b=400;}
        if(count>380){a=4000;b=500;}


        if(count>400){a=5000;b=100;}
        if(count>420){a=5000;b=200;}
        if(count>440){a=5000;b=300;}
        if(count>460){a=5000;b=400;}
        if(count>480){a=5000;b=500;}
    }




    return 0;
}

Well
a = 1000 + (count/100) * 1000;
b = ((count % 100) / 20 + 1) * 100;

Play around with ideas like this, there is clearly an arithmetic relationship between the values.
Figure out what it is.