joj 1019解题报告

 

 1019: Do the Untwist

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	992	556	Standard

Cryptography deals with methods of secret communication that transform a message (theplaintext) 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. Twistingis 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, andplaincode 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 inplaincode 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 inciphercode 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'	'.'

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 keyk 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.)

 

Sample Input

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

 

Sample Output

cat
this_is_a_secret
beware._dogs_barking

。。。我一直在考虑这道题应该归在水题类还是数论类，但是曾经看到有人把它归为数论类，也就随了，其实我感觉就是一道模拟题，按照题目要求，先是求出ciptext的长度len，然后i=0tolen开始，将ciptext[i]其变为cipcode数字，然后求出(cipcode+i)%28,plaintext[k*i%len]=(cipcode+i)%28,最后将整型数组转化为字符型就好了。。。

代码：

语言：c++

#include<iostream> using namespace std; #include<cstring> char change[28]={'_','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','.'}; int main() { char ciptext[100],plaintext[100]; int cipcode,plaincode[100],k,m; int str2int(char c); void int2str(int plaincode[],char plaintext[],int len); while(cin>>k) { if(k==0) break; cin>>ciptext; int len=strlen(ciptext); for(int i=0;i<len;++i) { cipcode=str2int(ciptext[i]); m=(cipcode+i)%28; plaincode[k*i%len]=m; } int2str(plaincode,plaintext,len); cout<<plaintext<<endl; } return 0; } int str2int(char c) { int i; for(i=0;i<28;++i) if(c==change[i]) return i; } void int2str(int plaincode[],char plaintext[],int len) { for(int i=0;i<len;++i) plaintext[i]=change[plaincode[i]]; plaintext[len]='/0'; }