# the 5th sharif acm contest: number 2:SMS lock decrypt program
"""
THE PROBLEM:
This year, ACM scientific committee members use emails to discuss
about the problems and edit the selected ones. They know that email is
not a secure way of communication, especially on such an important
topic. So they transfer password-protected compressed file among
themselves. In order to send the passwords, they use SMS. To increase
the security level, the encrypted passwords are sent by SMS. To do this, a
multi-tap SMS typing method is used.
Multi-tap is currently the most common text input method for mobile
phones. With this approach, the user presses each key one or more times
to obtain the wanted characters. For example, the key 2 is pressed once to
get character A, twice for B, and three times for C.
The encryption algorithm that is used is quite simple: to encrypt the ith
character of the password, the key used to obtain that character is tapped i
more times. For if the 4th
character of password is U, the key 8 is tapped 6
times, getting the character V. Note that to make the problem simple, we
have assumed that the keypad does not generate digits.
The scientific committee needs a program to decrypt the received
passwords. They are too busy to write this program and have asked you to help! Write a program to get a correct
encrypted text and print the original password.
INPUT OUTPUT
----------------------------------
BACE ABCD
GgaudQNS IhateSMS
__to imagine better use your phone keyboard
__there is two function: encrypt and decrypt the answer of problem is decrypt and encrypt do reversed decrypt.
# <><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>
code explanation:
__r: real number ,real character position in cellphone keyboard ,for example "E" is 1.
__v: virtual number (encrypted)
__n: character number in string
__notice that in python lists start with 0,
__for example to encrypt "ABCD", number of "A" in string is 0
"B" is 1,"C" is 2 and "D" is 1("DEF").
__to encrypt "A":r=0 , n:1 then v=r+n ,v=1 ,1 in keyboard is "B".
__to convert "B":r=1 , n=2 then v=r+n ,v=3 ,4 we have not 4 in keybord(ABC IN KEY "2")
but we cvan use "%" operator so we have 3%3=0 then we have "A". 3 is length of("ABC"),
see your cell phone if you not understand.
__to encrypt "C":r=2 , n=3 then v=r+n , 5%3=2 and 2 is "C" itself.
__to encrypt "D":r=0 , n=4 then v=4 , 4%3=1 and 1 is "E" .
__so we convert "ABCD" to "BACE".
"""
# <><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>
# <><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>
def det(c,num):
""" c:character and num :r or real number, takes a character
and a number ,then returns character that belongs to c's list(given character ' lsit) using num
for example we give it "a" and 2 then returns c """
flag=0
l2=['a','b','c']
gl2=['A','B','C']
l3=['d','e','f']
gl3=['D','E','F']
l4=['g','h','i']
gl4=['G','H','I']
l5=['j','k','l']
gl5=['J','K','L']
l6=['m','n','o']
gl6=['M','N','O']
l7=['p','q','r','s']
gl7=['P','Q','R','S']
l8=['t','u','v']
gl8=['T','U','V']
l9=['w','x','y','z']
gl9=['W','X','Y','Z']
all=[l2,l3,l4,l5,l6,l7,l8,l9,gl2,gl3,gl4,gl5,gl6,gl7,gl8,gl9]
retl=[] # return list
for i in all:
if c in i:
retl=i
return retl[num]
# ------------------------------------
def num(list,c):
" return character number in a given list "
if ord(c)>65 and ord(c)<91:
c=chr(ord(c)+32)
for i in range(len(list)):
if list[i]==c:
return i
return False
# ------------------------------------
# ------------------------------------
def what(c):
" determines that the given charater (c) belongs to witch list "
flag=0
l2=['a','b','c']
gl2=['A','B','C']
l3=['d','e','f']
gl3=['D','E','F']
l4=['g','h','i']
gl4=['G','H','I']
l5=['j','k','l']
gl5=['J','K','L']
l6=['m','n','o']
gl6=['M','N','O']
l7=['p','q','r','s']
gl7=['P','Q','R','S']
l8=['t','u','v']
gl8=['T','U','V']
l9=['w','x','y','z']
gl9=['W','X','Y','Z']
all=[l2,l3,l4,l5,l6,l7,l8,l9,gl2,gl3,gl4,gl5,gl6,gl7,gl8,gl9]
retl=[] # return list
if ord(c)<91 and ord(c)>64: # in all case we make characters in small case
flag=1
c=chr(ord(c)+32)
for i in all:
if c in i:
return i
# --------------------------------------------------
####################################################
#### # ENCRYPT Fun() ####
####################################################
def encrypt(string):
list=[]
retstr=''
snum=1 # character number in string
r=int() # real number usin num
v=int() # virtual number
for chr in string:
if chr==' ' or type(i)==type(2): # if you put number it pirnts space.
retstr+=' '
continue
list=what(chr)
r=num(list,chr) # real number
v=snum+r # virtual number
v=v%len(list)
retstr+=det(chr,v)
snum+=1
return retstr
# --------------------------------------------------------------
################################################################
##### DECRYPT fun() #####
################################################################
def decrypt(string):
r=0
v=0 # number in list usin num() , virtual
n=0 # number in string; num() can be used - * virtual
secchr='' # for returning
list=[]
count=0
for i in string:
if i==" " or type(i)==type(2): # if you put number it pirnts space.
secchr+=' '
continue
list=what(i)
v=num(list,i)
n=count+1
r=v-n # real number
r=r%len(list)
secchr+=det(i,r)
count+=1
return secchr
# RUNNING:
# ---------------------------------------------x
if __name__=="__main__":
a=raw_input('decrypt or encrypt ?( d/e) ')
word=raw_input('enter word ')
if a=='d':
print decrypt(word)
elif a=='e':
print encrypt(word)
