1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
|
#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";
int lengthOfName(char *num){
char *word;
// minus 1 for null terminator
int digits = strlen(num) - 1;
int n = atoi(num);
if(digits == 1){
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{
// since the input to the function expects a null terminator, an extra '\n' is needed
/// substring would be a better way to do this
char o[2];
int on;
word = (char *) tens[n / 10];
printf("%s", word);
o[0] = num[1];
o[1] = '\n';
on = lengthOfName(o);
//return (strlen(word) + lengthOfName(o));
return (strlen(word) + on);
}
}
if(digits == 3){
word = (char *) ones[n / 100]
int t = lengthOfName(
}
// error condition
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;
}
|
