//----------------------------------------------------------------------
// enigma1366.c (c) 2005 by Charles Petzold
//
// "Done in Half the Time" by Susan Denham, New Scientist, 12 November 2005, page 62
//
// You look at your watch 9 times over a two-day period, always at an
// exact number of minutes past the hour. Each time, the angle between
// the hour hand and the minute hand is half what it was previously.
//
// What's the last time you look at your watch?
//
// Measuring angles in degrees clockwise (of course) from the stroke of 12, 
//
//	Angle of Minute hand = 6 * Minutes
//	Angle of Hour hand = 30 * Hour + Minutes / 2
//
// The difference is always a multiple of 0.5 degrees. Is 0.5 degrees
// actually the angle between the two hands seen on the 9th time the 
// watch is observed?
//
// If so, the other eight angles are (working backwards) 1, 2, 4, 8, 16,
// 32, 64, and 128. The next angle would be 256, but that would probably
// be interpreted as 104, so we have the 9 angles.
//
// The hour and minute hand are separated by each of these 9 angles only
// twice during every 12-hour period, as shown in the following table:
//
// 	Angle	    Times
//	-----	-------------
//	 128	 5:04 &  6:56
//	  64	 3:28 &  8:32
//	  32	 4:16 &  7:44
//	  16	 2:08 &  9:52
//	   8	 1:04 & 10:56
//	   4	 5:28 &  6:32
//	   2	 3:16 &  8:44
//	   1	 4:22 &  7:38
//	 0.5	 2:11 &  9:49
//
// If you're looking at your watch within a two-day period (Saturday and 
// Sunday are mentioned in the text of the puzzle) then the last time 
// must be 9:49 PM, which is the solution.
//
// The code shown below solves the problem as well. All angles are 
// doubled so they can be represented as integers.
//----------------------------------------------------------------------

#include <stdio.h>

// This function calculates the angle between the hour hand and the
// minute hand for a particular time. Actually, as the name of the
// function implies, it calculates TWICE that angle.

int TwiceAngleDifference(int hour, int min)
{
	int angleMin = 12 * min;
	int angleHour = 60 * hour + min;
	int angleDiff = angleHour - angleMin;

	// Reflect around 0
	if (angleDiff < 0)
		angleDiff = -angleDiff;

	// NOrmalize to between 0 and 720
	while (angleDiff > 720)
		angleDiff -= 720;

	// Reflect around 360
	if (angleDiff > 360)
		angleDiff = 720 - angleDiff;
	
	return angleDiff;
}

// This function determines the time when twice the angle between 
// the hour hand and minute hand equals the third argument.
//
// Notice that the arguments are pointers. Thus, the values on 
// entry to the function indicate a start time, and the values
// when the function returns indicate the determined time.
//
// The hour can go up to 48. Values above 12 are PM. Values above
// 24 are the second day. 
//
// The function returns 0 if it can't find a time, but I don't use that.

int FindTimeAtAngle(int * pHour, int * pMin, int angle)
{
	for ( ; *pHour < 48; (*pHour)++)
	{
		for ( ; *pMin < 60; (*pMin)++)
		{
			if (angle == TwiceAngleDifference(*pHour, *pMin))	
				return 1;
		}
		*pMin = 0;
	}
	return 0;
}

// Entry point

int main(void)
{
	int hour = 0, min = 0, angle = 256;	// actually twice the angle

	while (angle > 0)
	{
		// Find the time at the specified angle
		FindTimeAtAngle(&hour, &min, angle);

		// Print the angle and the time
		printf("%3i/2: %2i:%02i %s\n", angle, 1 + (hour + 11) % 12, min, 
			hour % 24 < 12 ? "AM" : "PM");

		// Half the angle for the next time through
		angle /= 2;
	}
	return 0;
}