This script slowly cracks a cypher.

cyphcrack.py

import urllib2
def docrypt(password):
        # insert hash here
	orig = "sl/iECN%22ttp%3AP%20%3Chwxh.Iw3g/TR%22.Prhl/l%201//xhtTD/D1.%20r%20anionaOsTtdxa%22/%3EmhtmlLmY%3CmwDs%20%3D%22p%3A//Dh%20twxB3T.o1999nrH/h%3ETlC%22%3E%3Chea%22mtmmea%3CXmehttpot%20%20qTlvi%3D%22tentTC/nytC%22E%20cent%3DioLtes0/-ht%20chammn%3Bem/uMtf%20/%3Emt-/%22%3C0wtole9%2033g%3E/11im0t%20311%3C/m3m9ttt%3EWmmtyleamlsy%3Ed%3D1%22t/cssoe%20tm%20n%3C%21%21-mbodn-wm%7BdwmU%09bgroula/k-0mlDor00008%3Ad%23%3Bdmmr%7Dmody%2C1m-b%2C%3Ai%20m%7Bmcolo%20m-%09%20meF3FFmmm%7D%3CF%22%3Bm%3Et%3E.mmstylxmr/%3Cb%3De/admmmmm%3Emmoii%3Ewmmiv%20acm%20dgm1%22tcer%22%3Em%20nte%20%20ep-%3E133%200m1%20%201me3t3%20%3C/p%3Em9%221mpm%3Clp%3Esp%3B%3CF%26yb%3E%3Bmm.%20%20%26nbs%3C%3Cn%3E%3Cmo%3Etmmdiv%3Emm0/mtmbmodmm%3C/%3Cytmm%3Dhmummc/F-tmrm%20%3Cpppm9mp-//pte0%20%3Cxm11tmhd%3C/%20%3C%20%20%3Etmmn%23hynl%3Ellm"
        #urldeode:
	orig = urllib2.unquote(orig)
        # Here is the algorithm from the js
	passnum = len(orig) % len(str(password))
	i = 0
	while i >= 0:
		passnum -= 1
		if passnum <= -1 or passnum >= len(str(password)) - 1:
			passnum = len(str(password -1))
		pos1 = i
		pos2 = i + int(str(password)[passnum-1])
		if pos2 >= len(orig):
			continue
		char1 = orig[pos1]
		char2 = orig[pos2]

		orig = list(orig)

		orig[pos2] = char1
		orig[pos1] = char2
		i -= 1
	orig1 = ""
	i = 0
	while i < len(orig):
		orig1 = orig1 + orig[i]
		i += 1
	orig1 = orig1.replace('/mmm/g','\r\n')
	return orig1
if __name__ == "__main__":
# You can start at whatever number you want.
# as of writing I'm at about 50m, so there's
# no point searching under that if you trust
# me
	count = 1
# This simply counts up numbers and tries them
# as passwords. This works for text passwords
# too as the js just converts chars to their
# ASCII value anyway, so we're still going to
# decrypt it, even if we don't have the intended
# password
	while 1:
		output = docrypt(count)
		print count
		if '3301' in output or 'alhok' in output or 'http://' in output or 'https://' in output:
			print count + '\n\n\n' + output + '\n\n\n'
			f = open ('passwords.log','a')
			f.write(count + '\n' + output + '\n')
		count += 1

I think line 31 causes some problems as the JS seems strange, and I don't know how to implement that in Python.
From what I can see, it's only applying the cipher to the passwords length at the start of the ciphertext.