Problem 109: Friday the Thirteenth

Today is Friday the thirtieth (30^th)!

Not that hard, but I somehow misread 1900 as 1990 once...

Simply find the day of week of the 13^th day of the next month.
Then count the days that pass with each month mod 7 for the day of the week for the "zero"th day of the next month.

/*
ID: victor4
PROG: friday
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>

using namespace std;

// const int daysinmonth[12] = {  };
int daysinmonth(int month, int year){
 // 1 = January
 if(month == 1)
  // year is offset from 1900
  // 1900 % 4 = 0 = 1900 % 100 = 0, 1900 % 400 = 300
  /*
  (leap year) iff (divisible by 4) and (not (divisible by 100) or (divisible by 400))
  not (leap year) iff not( (divisible by 4) and (not (divisible by 100) or (divisible by 400)))
  not (leap year) iff (not (divisible by 4)) or (not (not (divisible by 100) or (divisible by 400)))
  not (leap year) iff (not (divisible by 4)) or ((divisible by 100) and not (divisible by 400))
  */
  return ((year % 4) || (!(year % 100) && ((year+300) % 400))) ? 28 : 29;
 // September, April, June, and November
 switch(month){
  case 8:
  case 3:
  case 5:
  case 10:
   return 30;
   // break;
  default:
   return 31;
   // break;
 }
}

int main() {
 ofstream fout ("friday.out");
 ifstream fin ("friday.in");
 
 // sanitize input
 int N;
 fin >> N;
 if(N < 0) N = 0;
 if(N > 400) N = 400;
 
 // failed once due to forgetting the = {0}
 int result[7] = {0};
 
 int dow = 0, month = 0;
 for(int i = 0; i < N; ++i)
  for(int m = 0; m < 12; ++m){
   // (a + 13) mod 7 = (a + 6) mod 7
   ++result[((dow + 6) % 7)];
   dow = (dow + daysinmonth(m, i)) % 7;
  }
 
 // output!
 for(int i = 0; i < 7; ++i){
  // failreason: Saturday first!
  fout << result[(i + 6) % 7];
  if(i == 6) fout << endl;
  else fout << " ";
 }
 return 0;
}