//----------------------------------------------------------------------------------
// enigma1352.c (c) 2005 by Charles Petzold (www.charlespetzold.com)
// 
// "Old MacDonald" by Adrian Somerfield, from New Scientist, 6 August 2005, page 50
//
// Four varieties of potato: Primes, Queens, Roosters, and Superb.
//
// A one-stone (14-pound) bag of Chippers contains an integral number of pounds of
// each of the four varieties in increasing weight:
//
// 	pc + qc + rc + sc = 14;
//
//	pc < qc < rc < sc
//
// The variable pc (for example) is the number of pounds of Primes in the bag of Chippers.
//
// A one-stone bag of Boilers contains an integral number of pounds of each of the
// four varieties in decreasing weight:
//
//	pb + qb + rb + sb = 14;
//
//	pb > qb > rb > sb
//
// Also,
//
//	pc != pb
//	qc != qb
//	rc != rb
//	sc != sb
//
// Although that fact may help in working out the problem manually, it's not 
// required in the program shown below.
//
// A hundred-weight (8 stone) sack of Commercial consists of 1 stone bags
// of Chippers and Boilers.  Thus,
//
//	c + b = 8
//
// So, the total weight of Primes in this sack (the solution to the problem)
// is:
//
//	c * pc + b * pb
//
// We are also told:
//
//	b * sb = c * sc
//
//	c * qc + b * qb = c * rc + b * rb	
//-----------------------------------------------------------------------------------


#include <stdio.h>

int main(void)
{
	int pc, qc, rc, sc;
	int pb, qb, rb, sb;
	int c, b;

	for (pc = 1; pc < 3; pc++)
	for (qc = pc + 1; qc < 4; qc++)
	for (rc = qc + 1; rc < 6; rc++)
	{
		sc = 14 - pc - qc - rc;

		if (sc <= rc)
			continue;

		for (sb = 1; sb < 3; sb++)
		for (rb = sb + 1; rb < 4; rb++)
		for (qb = rb + 1; qb < 6; qb++)
		{
			pb = 14 - sb - rb - qb;

			if (pb <= qb)
				continue;

			for (c = 1; c < 8; c++)
			{
				b = 8 - c;

				if (c * sc != b * sb)
					continue;

				if (c * qc + b * qb != c * rc + b * rb)
					continue;

				printf("pc=%i qc=%i rc=%i sc=%i\n", pc, qc, rc, sc);
				printf("sb=%i rb=%i qb=%i pb=%i\n", sb, rb, qb, pb);
				printf("c=%i b=%i\n", c, b); 
				printf("%i\n", c * pc + b * pb);
			}
		}
	}
	return 0;
}