prob014.c

/*
    Project Euler, Problem 14

    The following iterative sequence is defined for the set of positive
    integers:

            n -> n / 2    (n is even)
            n -> 3n + 1   (n is odd)

    Using the rule above and starting with 13, we generate the following
    sequence:

            13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1

    It can be seen that this sequence (starting at 13 and finishing at 1)
    contains 10 terms. Although it has not been proved yet (Collatz Problem),
    it is thought that all starting numbers finish at 1.

    Which starting number, under one million, produces the longest chain?

    NOTE: Once the chain starts the terms are allowed to go above one million.

    Lawrence Velazquez
    10 March 2012
*/

#include <stdio.h>

int main(int argc, char const *argv[])
{
    unsigned int ans, start, j;
    unsigned int max_len, len;
    size_t i;
    
    ans = max_len = 0;
    for (i = 1; i < 1000000; ++i) {
        start = j = i;
        len = 1;
        while (j != 1) {
            j = (j % 2 == 0) ? (j / 2) : (3 * j + 1);
            ++len;
        }
        if (len > max_len) {
            ans = start;
            max_len = len;
        }
    }
    
    printf("answer: %u, length: %u\n", ans, len);
    return 0;
}