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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// 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;
}
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