program to display all Armstrong numbers between 0 and 100 solution here we are taking Armstrong number with 3 digit. but more digit also can be taken

//armstrong numbers lying in given range 0 and 1000
//we are taking numbers with three digits
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
clrscr();
int i=1,rem,k,sum=0;
while(i<=1000)
{
     k=i;
     sum=0;
while(k!=0)
 {
   rem=k%10;
   sum=sum+pow(rem,3);
   k=k/10;
 }

if(sum==i)
{
      printf("%d\n",i);

}
sum=0;
 i++;

}
getch();
}
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logics in mind:-
-------------------
->we have to display Armstrong numbers in given range 0 and 1000.Armstrong can be for 3/4 digits numbers.
   Here we taking 3 digits only.

->First  we take first number (k) then we apply following method        

                  ->For this, 

                 ->we have to divide by 10 to get last digit as remainder 

                 ->we find its cube and then go for sum

                -> to get second digit, we get first 12 and for this , we use 123/10.It is done to get integer                           part  only.

                  ->then we compare the number 'i' with that sum. If they are same then we display that.

->We repeat this until the value reaches 0 and for all numbers and individually. We use loop for this as shown above in the code.     

->as the first number is checked, we , once again, assign value '0' to sum to check next value.And this goes on.