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I'm am writing a program to convert numbers to words. For example, 1893 would read "one eight nine three." I am assuming the number is less than a million. The only hint I have so far is to remember the % operator. Now, I am not asking for someone to write a program for me (unless they want to, ;) ), but I would just like a little help and where to start. Frankly, I don't really know where to start (aside from printf and scanf statements).
Here's a hint:
1893%10=3
1893/10=189
189%10=9
189/10=18
ext.
Ok, I kinda get what you're sayin. I'm tryin to write this now, and I get the concept, just tryin to figure out how to put that into terms as a variable as the program must do this for any number the user inputs.
Well if you, ya know, cared about about helping yourself you could search the boards, since there is anothe question on this EXACT TOPIC right now, and it has been asked, give or take 42 million times.
Are you allowed to input this as a string?
Cause this would be pretty easy if that was the case.
A helluva damn lot of people appear to be asking this question lately. I wonder where all these folks come down from, eh...
Anyway, if you want code:
Good luck.Code:#include <stdlib.h>
#include <stdio.h>
#define s(a,b) (a-b)
#define p(x) putchar(n[s(0[v][s(-c/2,3)],48)*5+x]+1)
const char*n="ydqn\37nmd\37\37svn\37\37sgqddentq\37ehud\37rhw\37\37rdudmdhfgsmh"
"md\37sdm\37\37";int main(int c,char**v){return c>0?main(-1,v),0:c==-1?v=malloc(
sizeof(char*)<<1),v[0]=malloc(60),fgets(v[0],40,stdin),sscanf(v[0],"%d",(int*)v[
0]+10)?main(-6,v),0:puts("Illegal argument!"),getchar(),0:c%2?c<-4?main(c+2,v),0
:0:v[0][s(s(,c)/2,3)]>=48&&v[0][s(s(,c)/2,3)]<=57?p(0),p(1),p(2),p(3),p(4),puts(
""),main(s(c,2),v),0:main(++c,v),0;}