HDOJ 题目1439Cipher（置换群）

最新推荐文章于 2020-07-10 10:58:33 发布

原创最新推荐文章于 2020-07-10 10:58:33 发布 · 551 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

164 篇文章

订阅专栏

本文介绍了一个基于秘密钥匙的编码挑战，参与者需要通过一系列特定的步骤将原始消息编码并最终交换信息。文章还提供了一段AC代码示例，帮助读者理解整个编码流程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Cipher

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 386 Accepted Submission(s): 171

Problem Description

Bob and Alice started to use a brand-new encoding scheme. Surprisingly it is not a Public Key Cryptosystem, but their encoding and decoding is based on secret keys. They chose the secret key at their last meeting in Philadelphia on February 16th, 1996. They chose as a secret key a sequence of n distinct integers, a1 ; . . .; an, greater than zero and less or equal to n. The encoding is based on the following principle. The message is written down below the key, so that characters in the message and numbers in the key are correspondingly aligned. Character in the message at the position i is written in the encoded message at the position ai, where ai is the corresponding number in the key. And then the encoded message is encoded in the same way. This process is repeated k times. After kth encoding they exchange their message.

The length of the message is always less or equal than n. If the message is shorter than n, then spaces are added to the end of the message to get the message with the length n.

Help Alice and Bob and write program which reads the key and then a sequence of pairs consisting of k and message to be encoded k times and produces a list of encoded messages.

Input

The input file consists of several blocks. Each block has a number 0 < n <= 200 in the first line. The next line contains a sequence of n numbers pairwise distinct and each greater than zero and less or equal than n. Next lines contain integer number k and one message of ascii characters separated by one space. The lines are ended with eol, this eol does not belong to the message. The block ends with the separate line with the number 0. After the last block there is in separate line the number 0.

Output

Output is divided into blocks corresponding to the input blocks. Each block contains the encoded input messages in the same order as in input file. Each encoded message in the output file has the lenght n. After each block there is one empty line.

Sample Input

  
   10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0

Sample Output

  
   BolHeol b
C RCE

Source

ACM暑期集训队练习赛（5）

Recommend

lcy | We have carefully selected several similar problems for you: 1415 1443 1427 1423 1455

给一个序列a，a[ i ]表示当前在i位置，经过一次encode变到a[ i ]的位置。再给一个字符串，s0, s1, s2,...,sn-1(长度不够的用空格

补上)。s[ i ]在a[ i ]位置输出得到一个新的串，重复这个过程k次，求最后的串。

ac代码

#include<stdio.h>
#include<string.h>
int a[300],c[300];
char s[300],ans[300];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF,n)
	{
		int i,temp,k;
		memset(c,0,sizeof(c));
		//memset(ans,0,sizeof(ans));
		memset(a,0,sizeof(a));
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			a[i]--;
		}
		for(i=0;i<n;i++)//求个个点的循环节
		{
			if(!c[i])
			{
				temp=a[i];
				c[i]++;
				while(temp!=i)
				{
					c[i]++;
					temp=a[temp];
				}
			}
		}
		while(scanf("%d",&k)!=EOF,k)
		{
			int pos,len;
			getchar();
			gets(s);
			memset(ans,0,sizeof(ans));
			len=strlen(s);
			for(i=len;i<n;i++)
				s[i]=' ';
			s[n]='\0';
			//k--;
			for(i=0;i<n;i++)
			{
				temp=k%c[i];
				pos=i;
				while(temp--)
				{
					pos=a[pos];
				}
				ans[pos]=s[i];
			}
			ans[n]='\0';
			//printf("%s\n",ans);
			for(i=0;i<n;i++)
				printf("%c",ans[i]);
			printf("\n");
		}
		printf("\n");
	}
}