My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       
#include <iostream>
       
 
       
// memoize intermediate results

       
unsigned long long cache[80*2*3] = { 0 };
       
 
       
// recursively evaluate a day:
       
// days   - number of days left
       
// absent - number of consecutive absent day (only consider up to three past days)
       
// late   - total number of late days

       
unsigned long long count(unsigned int days, unsigned int absent = 0, unsigned int late = 0)
       
{
       
  // too many consecutive absent days ?
       
  if (absent == 3)
       
    return 0;
       
  // too late to often ?
       
  if (late > 1)
       
    return 0;
       
 
       
  // all days passed ? => collect prize
       
  if (days == 0)
       
    return 1;
       
 
       
  // unique ID of current parameter set
       
  unsigned int hash = days * 2 * 3 + absent * 2 + late;
       
  // already computed ?
       
  if (cache[hash] != 0)
       
    return cache[hash];
       
 
       
  unsigned long long result;
       
  // assume pupil is today neither late nor absent
       
  result  = count(days - 1, 0, late);
       
  // assume pupil is absent today
       
  result += count(days - 1, absent + 1, late);
       
  // assume pupil is late today
       
  result += count(days - 1, 0, late + 1);
       
 
       
  // store result
       
  cache[hash] = result;
       
  return result;
       
}
       
 
       
int main()
       
{
       
  unsigned int days;
       
  std::cin  >> days;
       
  std::cout << count(days) << std::endl;
       
  return 0;
       
}

This solution contains 7 empty lines, 14 comments and 1 preprocessor command.