Skip to main content

Convert numbers to roman numerals



Convert numbers to roman numerals  

 

 

#include<stdio.h>
#include<conio.h>



void predigits(char c1,char c2);
void postdigits(char c,int n);

char roman_Number[1000];
int i=0;

int main()
{

    int j;
    long int number;
  
    printf("Enter any natural number: ");
    scanf("%d",&number);
  
    if(number <= 0){
         printf("Invalid number");
         return 0;
    }

    while(number != 0){

         if(number >= 1000){
             postdigits('M',number/1000);
             number = number - (number/1000) * 1000;
         }
         else if(number >=500){
             if(number < (500 + 4 * 100)){
                 postdigits('D',number/500);
                 number = number - (number/500) * 500;
             }
             else{
                 predigits('C','M');
                 number = number - (1000-100);
             }
         }
         else if(number >=100){
             if(number < (100 + 3 * 100)){
                 postdigits('C',number/100);
                 number = number - (number/100) * 100;
             }
             else{
                 predigits('L','D');
                 number = number - (500-100);
             }
         }
         else if(number >=50){
             if(number < (50 + 4 * 10)){

                 postdigits('L',number/50);
                 number = number - (number/50) * 50;
             }
             else{
                 predigits('X','C');
                 number = number - (100-10);
             }
         }
         else if(number >=10){
             if(number < (10 + 3 * 10)){
                 postdigits('X',number/10);
                 number = number - (number/10) * 10;
             }
             else{
                 predigits('X','L');
                 number = number - (50-10);
             }
         }
         else if(number >=5){
             if(number < (5 + 4 * 1)){
                 postdigits('V',number/5);
                 number = number - (number/5) * 5;
             }
             else{
                 predigits('I','X');
                 number = number - (10-1);

             }
         }
         else if(number >=1){
             if(number < 4){
                 postdigits('I',number/1);
                 number = number - (number/1) * 1;
             }
             else{
                 predigits('I','V');
                 number = number - (5-1);
             }
         }
    }

    printf("Roman number will be: ");
    for(j=0;j<i;j++)
         printf("%c",roman_Number[j]);

    return 0;

}

void predigits(char c1,char c2){

    roman_Number[i++] = c1;
    roman_Number[i++] = c2;
}

void postdigits(char c,int n){
    int j;
    for(j=0;j<n;j++)
         roman_Number[i++] = c;
  
}



Sample output:

Enter any natural number: 87
Roman number will be: LXXXVII

Comments

Popular posts from this blog

C Program to print the pattern

C Program to print the pattern X X Y X Y Z ........................................................................................ #include<stdio.h> #include<conio.h> void main () { int i , j ; for ( i= 88 ; i< 91 ;i ++) { for ( j = 88 ;j <= i ; j ++) { printf ( "%c" , j );                    }           printf ( "\n" ); } getch (); } ........................................................................................ OUTPUT

To check if given string is palindrome or not.

To check if given string is palindrome or not. #include <stdio.h> #include <string.h> #include <conio.h>   int main()   {   char a[100], b[100]; printf ( "Enter the string to check if it is a palindrome\n" );   gets (a);   strcpy (b,a);   strrev (b);   if ( strcmp(a,b) == 0 )   printf ( "Entered string is a palindrome.\n" );   else   printf ( "Entered string is not a palindrome.\n" );   return 0;   }

To Find Factorial of a Number

To Find Factorial of a Number   Factorial of a Number   For any positive number n , its factorial is given by: factorial = 1 * 2 * 3 * 4. ... n If a number is negative, factorial does not exist and factorial of 0 is 1. This program takes an integer from a user. If user enters negative integer, this program will display error message and if user enters non-negative integer, this program will display the factorial of that number. Source Code /* C program to display factorial of an integer if user enters non-negative integer. */ #include <stdio.h> int main () { int n , count ; unsigned long long int factorial = 1 ; /* you can only write int */ printf ( "\nEnter an integer: \n" ); scanf ( "%d" ,& n ); if ( n < 0 ) printf ( "\nError!!! Factorial of negative number doesn't e...