Find Sum of Digits of a Number using Recursion
#include
#include
int sum (int a);
int main()
{
int num, result;
printf("Enter the number: ");
scanf("%d", &num);
result = sum(num);
printf("Sum of digits in %d is %d\n", num, result);
return 0;
}
int sum (int num)
{
if (num != 0)
{
return (num % 10 + sum (num / 10));
}
else
{
return 0;
}
}
Output
Enter the number: 2345
Sum of digits in 2345 is 14
C Program to Convert Octal to Decimal
#include
#include
#include
int main()
{
long int octal, decimal = 0;
int i = 0;
printf("Enter any octal number: ");
scanf("%ld", &octal);
while (octal != 0)
{
decimal = decimal +(octal % 10)* pow(8, i++);
octal = octal / 10;
}
printf("Equivalent decimal value: %ld",decimal);
return 0;
}
Output
Enter any octal number: 67
Equivalent decimal value: 55
Find Multiplication of two Binary Numbers
#include
#include
int binaryproduct(int, int);
int main()
{
long binary1, binary2, multiply = 0;
int digit, factor = 1;
printf("Enter the first binary number: ");
scanf("%ld", &binary1);
printf("Enter the second binary number: ");
scanf("%ld", &binary2);
while (binary2 != 0)
{
digit = binary2 % 10;
if (digit == 1)
{
binary1 = binary1 * factor;
multiply = binaryproduct(binary1, multiply);
}
else
binary1 = binary1 * factor;
binary2 = binary2 / 10;
factor = 10;
}
printf("Product of two binary numbers: %ld", multiply);
return 0;
}
int binaryproduct(int binary1, int binary2)
{
int i = 0, remainder = 0, sum[20];
int binaryprod = 0;
while (binary1 != 0 || binary2 != 0)
{
sum[i++] =(binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder =(binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0)
sum[i++] = remainder;
--i;
while (i >= 0)
binaryprod = binaryprod * 10 + sum[i--];
return binaryprod;
}
Output
Enter the first binary number: 10010
Enter the second binary number: 10101
Product of two binary numbers: 101111010
Find the Sum of two Binary Numbers
#include
#include
int main()
{
long binary1, binary2;
int i = 0, remainder = 0, sum[20];
printf("Enter the first binary number: ");
scanf("%ld", &binary1);
printf("Enter the second binary number: ");
scanf("%ld", &binary2);
while (binary1 != 0 || binary2 != 0)
{
sum[i++] =(binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder =(binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0)
sum[i++] = remainder;
--i;
printf("Sum of two binary numbers: ");
while (i >= 0)
printf("%d", sum[i--]);
return 0;
}
Output
Enter the first binary number: 100000
Enter the second binary number: 101010
Sum of two binary numbers: 1001010
Convert Numbers to Roman Numerals
#include
#include
void predigit(char num1, char num2);
void postdigit(char c, int n);
char romanval[1000];
int i = 0;
int main()
{
int j;
long number;
printf("Enter the number: ");
scanf("%d", &number);
if (number <= 0)
{
printf("Invalid number");
return 0;
}
while (number != 0)
{
if (number >= 1000)
{
postdigit('M', number / 1000);
number = number - (number / 1000) * 1000;
}
else if (number >= 500)
{
if (number < (500 + 4 * 100))
{
postdigit('D', number / 500);
number = number - (number / 500) * 500;
}
else
{
predigit('C','M');
number = number - (1000-100);
}
}
else if (number >= 100)
{
if (number < (100 + 3 * 100))
{
postdigit('C', number / 100);
number = number - (number / 100) * 100;
}
else
{
predigit('L', 'D');
number = number - (500 - 100);
}
}
else if (number >= 50 )
{
if (number < (50 + 4 * 10))
{
postdigit('L', number / 50);
number = number - (number / 50) * 50;
}
else
{
predigit('X','C');
number = number - (100-10);
}
}
else if (number >= 10)
{
if (number < (10 + 3 * 10))
{
postdigit('X', number / 10);
number = number - (number / 10) * 10;
}
else
{
predigit('X','L');
number = number - (50 - 10);
}
}
else if (number >= 5)
{
if (number < (5 + 4 * 1))
{
postdigit('V', number / 5);
number = number - (number / 5) * 5;
}
else
{
predigit('I', 'X');
number = number - (10 - 1);
}
}
else if (number >= 1)
{
if (number < 4)
{
postdigit('I', number / 1);
number = number - (number / 1) * 1;
}
else
{
predigit('I', 'V');
number = number - (5 - 1);
}
}
}
printf("Roman number is: ");
for(j = 0; j < i; j++)
printf("%c", romanval[j]);
return 0;
}
void predigit(char num1, char num2)
{
romanval[i++] = num1;
romanval[i++] = num2;
}
void postdigit(char c, int n)
{
int j;
for (j = 0; j < n; j++)
romanval[i++] = c;
}
Output
Enter the number: 500
Roman number is be: D
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