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

1. ## 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;
}```

2. 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.

3. Originally Posted by Salem
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.
Thank you very much your answer put me definitely in the right direction I think I know how to do it now!