Given a decimal number n, we have to convert this into Roman numeral. The value n lies in the range 1 to 4000. These are some Roman Numerals.
Number | Numeral |
---|---|
1 | I |
4 | IV |
5 | V |
9 | IX |
10 | X |
40 | XL |
50 | L |
90 | XC |
100 | C |
400 | CD |
500 | D |
900 | CM |
1000 | M |
4000 | MMMM |
So if the number n = 859, its Roman Numeral will be DCCCLIX
To solve this, we will follow these steps
- Define an array to store numeral and corresponding values for the given list. That is called nume array
- we are using a recursive approach, the function decToRom() is used. this is taking nume array and the number num.
- The decToRom() will be like
- if num is not 0, then
- max := find maximum value from nume array that is not larger than num
- append the value for that max into the result string
- num := num – max value
- decToRom(nume, num)
Example
Let us see the following implementation to get a better understanding −
#include<stdio.h> typedef struct{ char *sym; int val; }numeral; int maxNume(numeral *nu, int num){ int i, index; for(i = 0; i<15; i++){//15 numerals in array if(nu[i].val <= num) index = i; } //gretest value numeral index, not greater than number return index; } void decToRoman(numeral *nu, int num){ int max; if(num != 0){ max = maxNume(nu, num); printf("%s", nu[max].sym); num -= nu[max].val;//decrease number decToRoman(nu, num);//recursively print numerals } } main(){ int number; numeral nume[15] = {{"I",1},{"IV",4},{"V",5},{"IX",9}, {"X",10},{"XL",40},{"L",50},{"XC",90}, {"C",100},{"CD",400},{"D",500},{"CM",900},{"M",1000},{"MMMM",4000},{"V'",5000}}; printf("Enter a decimal number: "); scanf("%d", &number); if(number >0 && number <= 5000){//checking input number printf("The Roman equivalent of %d is ", number); decToRoman(nume, number); } else{ printf("Invalid Input"); } printf("\n"); }
Input
570 3574
Output
DLXX MMMDLXXIV