#include #include #include // Problem 17 - Number Letter Counts // So I found the string approach to be a pain in the ass in C (see Python solution) // But then I realized, hey, I just need the lengths, so you don't have to pass the string! const char *ones[10] = {"","one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; const char *teens[10] = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"}; const char *tens[10] = {"","", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"}; const char *nd = "AND"; ///starting to think recursion is wrong way to go about this... int lengthOfName(char *num){ char *word; // minus 1 for null terminator int digits = strlen(num) - 1; int n = atoi(num); if(digits == 1){ printf("here"); word = (char *) ones[n]; printf("%s\n", word); return strlen(word); } if(digits == 2){ // if the first digit, ie. the tens column is 1 if(num[0] == '1'){ word = (char *) teens[n % 10]; printf("%s\n", word); return strlen(word); } // assuming the first digit is not one, if the second digit is 0 else if(num[1] == '0'){ word = (char *) tens[n / 10]; printf("%s\n", word); return strlen(word); } else{ char o[2]; int on; word = (char *) tens[n / 10]; printf("%s", word); o[0] = num[1]; printf("%s\n", o); on = lengthOfName(o); return (strlen(word) + on); } } return -1; } int main(){ // 4 is the max for this example, +1 for null terminator or if user tries to cheat char n[5]; int out; printf("Number from 1 to 1000: "); fgets(n, 10, stdin); if(atoi(n) > 1000) return printf("only up to 1000 in this example, sorry :(\n"); out = lengthOfName(n); printf("%d Characters... wow you'll hurt your hands typing that!\n", out); return 0; }