20、Do the Untwist

Cryptography deals with methods of secret communication that transform a message (the plaintext) into a disguised form (the ciphertext) so that no one seeing the ciphertext will be able to figure out the plaintext except the intended recipient. Transforming the plaintext to the ciphertext is encryption; transforming the ciphertext to the plaintext is decryption. Twisting is a simple encryption method that requires that the sender and recipient both agree on a secret key k, which is a positive integer.

The twisting method uses four arrays: plaintext and ciphertext are arrays of characters, and plaincode and ciphercode are arrays of integers. All arrays are of length n, where n is the length of the message to be encrypted. Arrays are origin zero, so the elements are numbered from 0 to n - 1. For this problem all messages will contain only lowercase letters, the period, and the underscore (representing a space).

The message to be encrypted is stored in plaintext. Given a key k, the encryption method works as follows. First convert the letters in plaintext to integer codes in plaincode according to the following rule: '_' = 0, 'a' = 1, 'b' = 2, ..., 'z' = 26, and '.' = 27. Next, convert each code in plaincode to an encrypted code in ciphercode according to the following formula: for all i from 0 to n - 1,

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

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

Array	0	1	2
plaintext	'c'	'a'	't'
plaincode	3	1	20
ciphercode	3	19	27
ciphertext	'c'	's'	'.'

5

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

The input file contains one or more test cases, followed by a line containing only the number 0 that signals the end of the file. Each test case is on a line by itself and consists of the key k, a space, and then a twisted message containing at least one and at most 70 characters. The key k will be a positive integer not greater than 300. For each test case, output the untwisted message on a line by itself.

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

Example input:

5 cs.
101 thqqxw.lui.qswer
3 b_ylxmhzjsys.virpbkr
0

Example output:

cat
this_is_a_secret

beware._dogs_barking

 

解题思路：这道题是有关明文与密文之间的转换问题，明文由“_”,a到z，“_”转换为0到27的明码，再把明码按照题目所给的式子ciphercode[i] = (plaincode[ki mod n] - i) mod 28. 转换为密码，最后把密码有数字按照与明文与明码之间相同的方式转换为密文。本题要求根据密文反求明文，ciphercode[i]为0到27之间的数，plaincode[ki mod n]也为0到27之间的数，又题目规定所有数无论正负对28取模均为非负，所以若plaincode[ki mod n] - i 大于0，则plaincode[ki mod n]=ciphercode[i]+i ;若plaincode[ki mod n] - i 小于0，对28取模之后为正数，AmodB=C, A<0 , B>0 , C>0，则A=C-mB(m为正整数），所以根据这样的关系，plaincode[ki mod n]=ciphercode[i]-m*28+i,m的取值根据plaincode[ki mod n]的取值区间[0,27]定。

具体代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

char transform[] = { '_', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k',
'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', '.' };

void convert(int key, char ctext[]){
     int n = strlen(ctext);
     int *ccode =(int*) malloc(n*sizeof(int));
     int *pcode = (int*)malloc(n*sizeof(int));
     char *ptext = (char*)malloc((n + 1)*sizeof(int));
     int i, j, k;
     for (i = 0; i < n; i++){
           if (ctext[i] >= 'a'&&ctext[i] <= 'z'){
                ccode[i] = ctext[i] - 96;
           }
           else if (ctext[i] == '_'){
                ccode[i] = 0;
           }
           else{
                ccode[i] = 27;
           }
     }
     for (j = 0; j < n; j++){
           int temp = ccode[j] + j;
           if (temp >= 28){
                while (temp >= 28){
                     temp -= 28;
                }
           }
           pcode[(key*j) % n] = temp;
     }
     for (k = 0; k < n; k++){
           ptext[k] = transform[pcode[k]];
     }
     ptext[n] = '';
    printf("%s
", ptext);
}

int main(){
     int number;
     char ciphertext[100];
     scanf("%d", &number);
     while (number){
           scanf("%s", ciphertext);
           convert(number, ciphertext);
           scanf("%d", &number);
     }
     return 0;
}

做题感悟：整个五月在考试，六月刚结束考试又要进行课程设计，不管怎样还是要每天抽出时间坚持。ZOJ对C99同样也不支持，所以for的格式，只能如下所示。

int i;
for(i=0;i<n;i++){...}