To count number of quadruplets with sum = 0

marek.rocki at wp.pl marek.rocki at wp.pl
Fri Mar 16 21:05:14 CET 2007


My attempt uses a different approach: create two sorted arrays, n^2
elements each; and then iterate over them looking for matching
elements (only one pass is required). I managed to get 58,2250612857 s
on my 1,7 MHz machine. It requires numpy for decent performance,
though.

import numpy
import time

def parse_input():
	al, bl, cl, dl = [], [], [], []
	for i in xrange(int(raw_input())):
		a, b, c, d = map(int, raw_input().split())
		al.append(a)
		bl.append(b)
		cl.append(c)
		dl.append(d)
	return al, bl, cl, dl

def count_zero_sums(al, bl, cl, dl):
	n = len(al) # Assume others are equal

	# Construct al extended (every element is repeated n times)
	ale = numpy.array(al).repeat(n)
	del al
	# Construct bl extended (whole array is repeated n times)
	ble = numpy.zeros((n*n,), int)
	for i in xrange(n): ble[i*n:(i+1)*n] = bl
	del bl
	# Construct abl - sorted list of all sums of a, b for a, b in al, bl
	abl = numpy.sort(ale + ble)
	del ale, ble

	# Construct cl extended (every element is repeated n times)
	cle = numpy.array(cl).repeat(n)
	del cl
	# Construct dl extended (whole array is repeated n times)
	dle = numpy.zeros((n*n,), int)
	for i in xrange(n): dle[i*n:(i+1)*n] = dl
	del dl
	# Construct cdl - sorted list of all negated sums of a, b for a, b in
cl, dl
	cdl = numpy.sort(-(cle + dle))
	del cle, dle

	# Iterate over arrays, count matching elements
	result = 0
	i, j = 0, 0
	n = n*n
	try:
		while True:
			while abl[i] < cdl[j]:
				i += 1
			while abl[i] > cdl[j]:
				j += 1
			if abl[i] == cdl[j]:
				# Found matching sequences
				ii = i + 1
				while ii < n and abl[ii] == abl[i]: ii += 1
				jj = j + 1
				while jj < n and cdl[jj] == cdl[j]: jj += 1
				result += (ii - i)*(jj - j)
				i, j = ii, jj
	except IndexError:
		pass

	return result

t = time.clock()
print count_zero_sums(*parse_input())
print time.clock() - t




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