Algorithm writing 3 Instructions Algorithms are a generic way of describing a so

Question

Algorithm writing 3

Instructions Algorithms are a generic way of describing a solution to a computational problem without use of any specific programming language syntax. It cannot be executed on a real computer, but it can model and resemble real programming code. Natural language is used whenever details are unimportant or distracting. Your algorithm: Must be unambiguous (exact).are the steps numbered? ...do we always know where to go next or are we left hanging? .Must be correct (solves the problem) .are the requirements' being met? . Must come to an end (terminate)...how do we know when to end this? . Must be general (for all cases)....can we input any data & still solve the problem? Must be at the right level of detail.... will a computer (in general) understand what we are asking it to do? Requirements>Write a solution to the problem accepts one number from the user in the range of 1 to 10 and displays the Roman numeral version of that number. If the number is outside of the acceptable range, then display an error message (tell the user she entered something wrong). Number Roman Numeral 3 4 VI 7 8 03 10 VII TX

Explanation / Answer

#include <stdio.h>

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;
}

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