首页 > ACM题库 > HDU-杭电 > HDU 1129 Do the Untwist-字符串[解题报告] C++
2013
11-28

HDU 1129 Do the Untwist-字符串[解题报告] C++

Do the Untwist

问题描述 :

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’ ‘.’

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.)

样例输入:

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

样例输出:

cat
this_is_a_secret
beware._dogs_barking

 

http://acm.hdu.edu.cn/showproblem.php?pid=1129

 

#include <iostream>
#define   N 75
using namespace std;
char ct[N],pt[N];
int cc[N],pc[N];
int k,len;
void ctTocc(){
for (int i=0;i<len;i++)
{
   if (ct[i]=='_')
    cc[i] = 0;
   else if(ct[i]=='.')
    cc[i] = 27;
   else{
    cc[i] = ct[i]-'a'+1;
   }
}
}
void ccTopc(){
int j;
for (int i=0;i<len;i++)
{
   j = (cc[i] +i+28)%28;
   pc[(k*i)%len]=j;
}
}
void pcTopt(){
int i,j;
for (i=0;i<len;i++)
{
   if (pc[i]==0)
    pt[i] = '_';
   else if (pc[i]==27)
    pt[i]='.';
   else
    pt[i]='a'+pc[i]-1;
}
pt[len]='\0';
}
int main()
{
while (scanf("%d",&k)&&k!=0)
{
   scanf("%s",&ct);
   len = strlen(ct);
   ctTocc();
   ccTopc();
   pcTopt();
   printf("%s\n",pt);
}
return 0;
}

 

解题思路:分析给的转换函数,然后逆推函数即可转换成功。

 

 

#include <stdio.h>
#include <string.h>
#define size 201

char PlainText[size],CipherText[size];//明码,密码
int PlainCode[size],CipherCode[size];//明码相应数值,密码相应数值
int len;

int StrToDig(char ch)
{
	if(ch == '_')
		return 0;
	else if(ch == '.')
		return 27;
	else
		return ch-'a'+1;
}

char DigToStr(int dig)
{
	if(dig == 0)
		return '_';
	else if(dig == 27)
		return '.';
	else
		return dig-1+'a';
}

/*密码数字转换成明码数字*/
void CipherCodeToPlainCode(int key)
{
	int i,tem;
	for (i=0;i<len;i++)
	{
		tem = CipherCode[i]+i;
		while (tem<0||tem>27)/*限定密码值的范围在0-27之间*/
		{
			if(tem<0)
				tem+=28;
			else if(tem>27)
				tem-=28;
		}
		PlainCode[(key*i)%len] = tem;
	}
};

int main()
{
	int key,i;
	while (scanf("%d",&key)&&key)
	{
		scanf("%s",CipherText);
		len = strlen(CipherText);
		for (i=0;i<len;i++)
			CipherCode[i] = StrToDig(CipherText[i]);
		CipherCodeToPlainCode(key);
		for(i=0;i<len;i++)
			PlainText[i] = DigToStr(PlainCode[i]);
		PlainText[len] = '/0';
		printf("%s/n",PlainText);
	}
	return 0;
}

  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮