Method 1
Thanks to Raj for suggesting this method.
1. Convert both i/p base 14 numbers to base 10.
2. Add numbers.
3. Convert the result back to base 14.
Method 2
Just add the numbers in base 14 in same way we add in base 10. Add numerals of both numbers one by one from right to left. If there is a carry while adding two numerals, consider the carry for adding next numerals.
Let us consider the presentation of base 14 numbers same as hexadecimal numbers
A --> 10
B --> 11
C --> 12
D --> 13
Example:
num1 = 1 2 A
num2 = C D 3
1. Add A and 3, we get 13(D). Since 13 is smaller than
14, carry becomes 0 and resultant numeral becomes D
2. Add 2, D and carry(0). we get 15. Since 15 is greater
than 13, carry becomes 1 and resultant numeral is 15 - 14 = 1
3. Add 1, C and carry(1). we get 14. Since 14 is greater
than 13, carry becomes 1 and resultant numeral is 14 - 14 = 0
Finally, there is a carry, so 1 is added as leftmost numeral and the result becomes
101D
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Implementation of Method 2:
C
# include <stdio.h>
# include <stdlib.h>
# define bool int
int getNumeralValue(char );
char getNumeral(int );
/* Function to add two numbers in base 14 */
char *sumBase14(char *num1, char *num2)
{
int l1 = strlen(num1);
int l2 = strlen(num2);
char *res;
int i;
int nml1, nml2, res_nml;
bool carry = 0;
if(l1 != l2)
{
printf("Function doesn't support numbers of different"
" lengths. If you want to add such numbers then"
" prefix smaller number with required no. of zeroes");
getchar();
assert(0);
}
/* Note the size of the allocated memory is one
more than i/p lenghts for the cases where we
have carry at the last like adding D1 and A1 */
res = (char *)malloc(sizeof(char)*(l1 + 1));
/* Add all numerals from right to left */
for(i = l1-1; i >= 0; i--)
{
/* Get decimal values of the numerals of
i/p numbers*/
nml1 = getNumeralValue(num1[i]);
nml2 = getNumeralValue(num2[i]);
/* Add decimal values of numerals and carry */
res_nml = carry + nml1 + nml2;
/* Check if we have carry for next addition
of numerals */
if(res_nml >= 14)
{
carry = 1;
res_nml -= 14;
}
else
{
carry = 0;
}
res[i+1] = getNumeral(res_nml);
}
/* if there is no carry after last iteration
then result should not include 0th character
of the resultant string */
if(carry == 0)
return (res + 1);
/* if we have carry after last iteration then
result should include 0th character */
res[0] = '1';
return res;
}
/* Function to get value of a numeral
For example it returns 10 for input 'A'
1 for '1', etc */
int getNumeralValue(char num)
{
if( num >= '0' && num <= '9')
return (num - '0');
if( num >= 'A' && num <= 'D')
return (num - 'A' + 10);
/* If we reach this line caller is giving
invalid character so we assert and fail*/
assert(0);
}
/* Function to get numeral for a value.
For example it returns 'A' for input 10
'1' for 1, etc */
char getNumeral(int val)
{
if( val >= 0 && val <= 9)
return (val + '0');
if( val >= 10 && val <= 14)
return (val + 'A' - 10);
/* If we reach this line caller is giving
invalid no. so we assert and fail*/
assert(0);
}
/*Driver program to test above functions*/
int main()
{
char *num1 = "DC2";
char *num2 = "0A3";
printf("Result is %s", sumBase14(num1, num2));
getchar();
return 0;
}