Project Euler, Problem #14, C

#include <stdio.h>

/* 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. */

void solution_one(void);

int main()
{
     solution_one();
     /* The number is 837799.
      * real    0m1,664s
      * user    0m1,642s
      * sys 0m0,016s */

     return(0);
}

void solution_one(void)
{
     int number;
     long int item;
     int sequence_length;
     int current_max_sequence_length = 4;
     int the_number = 1;

     for(number = 2; number < 1000000; number++)
     {
          item = number;
          sequence_length = 2;

          while(item > 1)
          {
               if(item % 2 == 0)
                    item /= 2;
               else
                    item = item * 3 + 1;

               sequence_length++;
          }

          if(current_max_sequence_length < sequence_length)
          {
               current_max_sequence_length = sequence_length;
               the_number = number;
          }
     }

     printf("The number is %d.\n", the_number);
}