/******* Problem 4--Clock Angles **************************************
 One important fact about these programming contests is that the
 problems are not ordered by difficulty.  Before the contest, they
 announced that one of the problems was essentially a "Hello World"
 problem, meaning that it was easy.  This was that problem.  Most teams
 were able to get this one.  The computation was straightforward.
 **********************************************************************/

#include <fstream>
#include <cstdlib>
#include <cmath>
using namespace std;

int main (int argc, char **argv);
double angle (int h, int m);

ifstream in ("prob4.in");
ofstream out ("prob4.out");

/******* main ********************************************************
 The program reads a series of hour-minute pairs from the input file
 and prints the angle between the clock hands correct to the nearest
 tenth of a degree.
 *********************************************************************/

int main (int argc, char **argv) {

  int h, m;   // The hour-minute pair

 while (1) {               // read in hour-minute pairs and print out
  in >> h >> m; // the angle between them until hour and
  if (h==0 && m==0) break; // minute are both 0.
  out << "At " << h << ":";
  out.width(2); out.fill ('0'); out << m << " the angle is ";
  int a = (int) (10*angle (h,m)+0.5);
  out.width(0); out << a/10 << "." << a%10 << " degrees." << endl;
 }
 return EXIT_SUCCESS;
}

/******** angle *******************************************************
 angle accepts two integers, the hour and minute, and returns a double
 containing the angle between the two hands of the clock.  The angle
 must be between 0 and 180 degrees.
 **********************************************************************/

double angle (int h, int m) {

  double ha, ma,   // The angle of the hour and minute hands
         a;        // The angle (to be returned) between the hands

 ma = m * 6;   // The difference between minutes is 6 degrees each
 ha = (h%12) * 30 + ma/12;  // The difference between hours is 30
                            // degrees each, plus each time the minute
                            // hand is advanced, the hour hand is
                            // advanced by one-twelfth as much. */
 a = fabs (ma-ha);  // The distance between the hands in degrees
 return a > 180 ? 360-a : a;  // Making sure we return the smaller angle
                              // between the hands
}