February 2014 Bronze Problem 3: Secret Code

Result: 10/10 **********

This one took more time than Problem 1 and Problem 2, but at least I actually finished this Problem 3, unlike the last few ones.

scode.cpp

#include <iostream>
#include <fstream>

#define loopi(x) loop_start(i,x,0)
#define loopj(x) loop_start(j,x,0)
#define loop_start(i,x,s) for(int i = (s); i < (x); ++i)
#define loop_rev(a,x,stop) for(int a = (x); a >= (stop); --a)

using namespace std;

inline bool is_repeated(const char *str1, const char *str2, size_t len)
{
 // There are no NULL characters to worry about
 return !strncmp(str1, str2, len);
 /*
 // old code:
 while(len--)
  if(*str1++ != *str2++)
   return false;
 return true;
 */
}

/*
(s = F_L) may become:

Drop L: F_ + s or s + F_
Drop F: _L + s or s + _L

F_F_L
F_LF_
_LF_L
F_L_L

But F_ or _L cannot be distinguished, so
A_A_L
A_LA_
A_FA_
FA_A_

F and L cannot be distinguished either, so
A_A_B
BA_A_
A_BA_ (x2)
*/
int sources(char *start, size_t sz)
{
 /*
 1. Any string of length L will be encrypted to length (2L-1),
    which is always (even-1), which is odd.
 2. A string of length two is in minimal form.
 */
 if((sz & 1) == 0 || sz <= 2)
  return 0;
 int ret = 0;
 const int sublen = (sz - 1) >> 1; // length of A_
 // A_A_B
 if(is_repeated(start, start + sublen, sublen))
  // A_B plus anything formed by that
  ret += 1 + sources(start + sublen, sublen + 1);
 // BA_A_
 if(is_repeated(start + 1, start + 1 + sublen, sublen))
  // BA_ and A_B should be equivalent
  ret += 1 + sources(start, sublen + 1);
 // A_BA_ (x2)
 if(is_repeated(start, start + 1 + sublen, sublen))
  // 2 * (A_B + anything formed by that)
  ret += (1 + sources(start, sublen + 1)) << 1;
 return ret;
}

int main()
{
 ofstream fout ("scode.out");
 ifstream fin ("scode.in");
 // Read input
 static char str[100+1];
 fin >> str;
 // Calculate output
 int ret = sources(str, strlen(str));
 fout << ret << endl;
 return 0;
}