ZOJ Problem Set - 1006 Do the Untwist

Do the Untwist

Time Limit: 2 Seconds Memory Limit: 65536 KB

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 messagecat 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'	'.'

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

Source: Zhejiang University Local Contest 2001

分析：
题意：编码译码问题。给定一个暗码，要求按照一定规则转换为明码。

读懂题意就可以，另外注意输出的坑。

ac代码：
#include <iostream>

#include<cstdio>

#include<cstring>

#include<string>

using namespace std;

char a[75],b[75];

int c[75],d[75];

int main()

{

int k,i,j,t;

while(scanf("%d",&k)&&k)

{

scanf("%s",a);

int n=strlen(a);

for(i=0;i<n;i++)

{

if(a[i]>='a'&&a[i]<='z')

c[i]=a[i]-96;

else if(a[i]=='_')

c[i]=0;

else

c[i]=27;

}

for(i=0;i<n;i++)

{

t=i*k%n;

d[t]=(c[i]+i)%28;

}

for(i=0;i<n;i++)

{

if(d[i]>0&&d[i]<27)

b[i]=d[i]+96;

else if(d[i]==0)

b[i]='_';

else

b[i]='.';

}

// printf("%s\n",b);//这样输出的字符数为75，有多余的字符（没有赋值的字符）。是个坑。

for(i=0;i<n;i++)

cout<<b[i];

cout<<endl;

}

return 0;

}