[CG] MadKnight's Fibonacci String Bit Sum Challenge Solution

#include <stdio.h>
#include <string.h>

int
fib_get(int n)
{
    static int fib[21] = { 0, 1 }; // fib(21) = 10946, fib(20) = 6765, max string length * bytesize = 8000
    static int fi = 2;

    if(n == 0)
        return 0;

    while(!fib[n])
        fib[fi++] = fib[fi - 2] + fib[fi - 3];

    return fib[n];
}

char *
ch_to_bin(char *buf, char ch)
{
    for(int i = 7; i >= 0; i--, buf++)
        *buf = ch & 1 << i ? '1' : '0';

    return buf;
}

void
build(char *buf, char *src)
{
    while(*src)
        buf = ch_to_bin(buf, *(src++));
}

int
is_fib(size_t n)
{
    for(int i = 0; i <= 21; i++)
        if(n == fib_get(i))
            return 1;

    return 0;
}

void
parse(char *src)
{
    int total = 0;

    for(int i = strlen(src) - 1; i >= 0; i--)
        if(is_fib(i))
            total += src[i] - '0';

    printf("%u\n", total);
}

int
main()
{
    char text[502],
         binary[8001] = ""; // max string length * bytesize = 8000

    fgets(text, 501, stdin);
    build(binary, text);
    parse(binary);
    return 0;
}

/*
FIB sequence = [0 1 2 3 5 8 ...]

"testcase" = 74 65 73 74 63 61 73 65

74 0111 0100
65 0110 0101
73 0111 0011
74 0111 0100
63 0110 0011
61 0110 0001
73 0111 0011
65 0110 0101

0111 0100
0110 0101
0111 0011
0111 0100
0110 0011
0110 0001
0111 0011
0110 0101
7654 3210

01 [1] 1 [0100]
01 [1] 0 [0101]
01 [1] 1 [0011]
01 [1] 1 [0100]
01 [1] 0 [0011]
01 [1] 0 [0001]
01 [1] 1 [0011]
01 [1] 0 [0101]
76  5  4  3210
    ^     ^^^^ < add all these together

expect: 7
got: 6 + 3 + 3 + 4 + 0 + 8 = 24 ... huh?

[0111 0100] 0 < 4
[0110 0101] 1 < 4
[0111 0011] 2 < 5
[0111 0100] 3 < 4
 0110 0011  4
[0110 0001] 5 < 3
 0111 0011  6
 0110 0101  7

also more than 7. Not sure what to do. The description is too vague. Lemme try both.

0111 0100  0 < 2 (2)
0110 0101  1 < 3 (5)
0111 0011  2 < 3 (8) !
0111 0100  3 < 4
0110 0011  4
0110 0001  5 < 3
0111 0011  6
0110 0101  7

7654 3210
  ^  ^^^^

Someone in the comments said something about a stream of bytes, so what if I...

0111 0100 0110 0101 0111 0011 0111 0100 0110 0011 0110 0001 0111 0011 0110 0101
          ^                         ^               ^         ^     ^   ^  ^^^^
1 + 0 + 0 + 1 + 0 + 1 + 1 + 1 + 1 + 1 + 0 = 7

A-ha. Now to implement...
*/