Logic:
- input two integers
- print all Armstrong numbers between these two integers
- Using a for loop check numbers in this range, in the loop, call function check_armstrong which returns ‘1’ if a number is Armstrong and ‘0’ otherwise.
#include <stdio.h>
#include<conio.h>
int check_armstrong(int);
int power(int, int);
void main ()
{
int c, a, b;
clrscr();
printf("Input two integers\n");
scanf("%d%d", &a, &b);
for (c = a; c <= b; c++)
{
if (check_armstrong(c) == 1)
printf("%d\n", c);
}
getch();
}
int check_armstrong(int n)
{
long long sum = 0, temp;
int remainder, digits = 0;
temp = n;
while (temp != 0)
{
digits++;
temp = temp/10;
}
temp = n;
while (temp != 0)
{
remainder = temp%10;
sum = sum + power(remainder, digits);
temp = temp/10;
}
if (n == sum)
return 1;
else
return 0;
}
int power(int n, int r)
{
int c , p = 1;
for (c = 1; c <= r; c++)
p = p*n;
return p;
}
