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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[4] = "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);
} else 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 string,
// an extra '\n' is needed
char o[2];
int on;
word = (char *) tens[n / 10];
printf("%s", word);
memcpy(o, &num[1], 2);
on = lengthOfName(o);
return (strlen(word) + on);
}
} else if(digits == 3) {
word = (char *) ones[n / 100];
printf("%shundred", word);
if(num[1] == '0' && num[2] == '0') {
printf("\n");
return strlen(word) + strlen("hundred");
} else {
char t[3];
int te;
strncpy(t, &num[1], 3);
printf("%s", nd);
te = lengthOfName(t);
// example: the user inputs, num = 121
// one hundred AND twenty one
return (strlen(word) + strlen("hundred") + strlen(nd) + te);
}
} else if(digits == 4) {
word = (char *) ones[n / 1000];
printf("%sthousand", word);
if(num[1] == '0' && num[2] == '0' && num[3] == '0') {
printf("\n");
return strlen(word) + strlen("thousand");
} /*else {
char h[4];
int hu;
strncpy(h, &num[1], 3);
hu = lengthOfName(h);
return (strlen(word) + strlen("thousand") + hu);
}*/
}
// error condition for debugging, i.e. in case I messed up, return something
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 = 0;
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");
// Test code to make sure the function works
//out = lengthOfName(n);
//printf("%d Characters... wow you'll hurt your hands typing that!\n", out);
for(int i = 1; i <= atoi(n); i++){
char curr[5];
sprintf(curr, "%d\n", i);
out += lengthOfName(curr);
}
printf("Sum of charachters: %d\n", out);
return 0;
}
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