zju 1006 zoj 1006

Do the Untwist

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Cryptography deals with methods of secret communication thattransform a message (theplaintext) into a disguised form (theciphertext) so that no one seeing the ciphertext will be ableto figure out the plaintext except the intended recipient.Transforming the plaintext to the ciphertext isencryption;transforming the ciphertext to the plaintext isdecryption.Twisting is a simple encryption method that requires that thesender and recipient both agree on a secret keyk, which is apositive integer.

The twisting method uses four arrays: plaintext and ciphertextare arrays of characters, andplaincode and ciphercode arearrays of integers. All arrays are of lengthn, where n is thelength of the message to be encrypted. Arrays are origin zero, so theelements are numbered from 0 ton - 1. For this problem allmessages will contain only lowercase letters, the period, and theunderscore (representing a space).

The message to be encrypted is stored in plaintext. Given a keyk, the encryption method works as follows. First convert theletters inplaintext to integer codes in plaincode according tothe following rule: '_' = 0, 'a' = 1, 'b' = 2, ..., 'z' = 26,and '.' = 27. Next, convert each code inplaincode to anencrypted code in ciphercode according to the following formula: for alli from 0 ton - 1,

ciphercode[i] = (plaincode[ki modn] - i) mod 28.

(Here x mod y is the positive remainder when x is divided byy.For example, 3 mod 7 = 3, 22 mod 8 = 6, and -1 mod 28 = 27. Youcan use the C '%' operator or Pascal 'mod' operator to computethis as long as you addy if the result is negative.)Finally, convert the codes in ciphercode back to letters inciphertext according to the rule listed above. The final twistedmessage is inciphertext. Twisting the message cat using the key 5yields the following:

Array012plaintext'c''a''t'plaincode3120ciphercode31927ciphertext'c''s''.'

Your task is to write a program that can untwist messages,i.e., convert the ciphertext back to the originalplaintext given the keyk. For example, given the key 5 andciphertext 'cs.', your program must output the plaintext 'cat'.

The input file contains one or more test cases, followed by a linecontaining only the number 0 that signals the end of the file.Each test case is on a line by itself and consists of the keyk, aspace, and then a twisted message containing at least one and at most 70characters. The keyk will be a positive integer not greater than300. For each test case, output the untwisted message on a lineby itself.

Note: you can assume that untwisting a message always yieldsa unique result. (For those of you with some knowledge of basic numbertheory or abstract algebra, this will be the case provided that thegreatest common divisor of the keyk and length n is 1, which it will be for all test cases.)

Example input:

5 cs.101 thqqxw.lui.qswer3 b_ylxmhzjsys.virpbkr0

Example output:

catthis_is_a_secretbeware._dogs_barking

这题主要的问题在于如何求plaincode,由题中的公式可得plaincode[ki mod n] =(ciphercode[i]+i) mod 28，通过循环把每一个求出来即可

#include<iostream>
#include<string>
using namespace std;
char plaincode[28];
int plain[1000];
int main(){
    string ciphertext,plaintext;
    int key,n,ciphercode[1000];
    plaincode[0]='_';
    plaincode[27]='.';
    for(int i=1;i<27;i++){
        plaincode[i]=i+'a'-1;
    }
    while(cin>>key,key){
        cin>>ciphertext;
        n=ciphertext.size();
        for(int i=0;i<n;i++){
            for(int j=0;j<28;j++){
                if(plaincode[j]==ciphertext[i]) ciphercode[i]=j;
            }
        }
        plaintext="";
        for(int i=0;i<n;i++){
            plain[key*i%n]=(ciphercode[i]+i)%28;
        }
        for(int i=0;i<n;i++){
            plaintext+=plaincode[plain[i]];
        }
        cout<<plaintext<<endl;
    }
    return 0;
}