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the answer of an acm problem to further explantion look at source code.

Python, 191 lines
 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191``` ```# 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) ```

this problem is interesting and creative but yet rather simple.

#### 1 comment

amir naghavi (author) 13 years, 1 month ago

if you run it please comment.

 Created by amir naghavi on Mon, 14 Mar 2011 (MIT)

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