Question

7) Write a C program to swap first and last digits of a number.​

Answer 1

Answer:

#include <stdio.h>

#include <math.h>

int main()

{

   int n,no_of_digits, fst_digit,lst_digit,temp_a,temp_b,swapped_no

   printf("\n Please Enter any Number: ");

   scanf("%d", & n);

   no_of_digits = log10(n)

   fst_digit = n/pow(10,no_of_digits)

   lst_digit = n%10

   

   temp_a = fst_digit * pow(10,no_of_digits)

   temp_b = n%temp_a

   n = temp_b/10

   

   swapped_no = lst_digit *(pow(10,no_of_digits)) + (n*10 + fst_digit)

   printf("The Number after Swapping First Digit and Last Digit = %d", swapped_no);

   

   return 0;

}

Explanation:

Here, we are using log10(logarithm of base 10) to count the number of digits of a given number.

Digit count of number is always equal to the upper bound of log10 of the number. i.e

digit_count = upper_bound of log10(Number)

NOTE:  logarithm is define only for positive numbers, not defined for negative numbers.

Explanation of above code is given below by taking the example:

Suppose user entered the Number = 1234

no_of_digits = log10(n) = 3

fst_digit = 1234 / pow(10, 3) = 1234 / 1000 = 1.234 = 1

lst_digit = 1234 % 10 = 4

temp_a = fst_digit * (pow(10, no_of_digits))

temp_a = 1 * pow(10, 3)  = 1 * 1000 = 1000

temp_b = n % temp_a = 1234 % 1000 = 234

n = 234 / 10 = 23.4 = 23

swapped_no = lst_digit * (pow(10, no_of_digits)) + (n * 10 + fst_digit)

swapped_no = 4 * pow(10, 3) + (23 * 10 + 1)

swapped_no = 4000 + (230 + 1) = 4231