poj 1026 Cipher 【置换群】 【求一个字符串经过 k 次置换后 的新字符串】

Cipher

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20026		Accepted: 5376

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

题意就不说了吧。。。       

前提：对于置换的数组（我们用next数组代替），里面的每一个元素都会存在于（有且只有）一个置换子群里面。

     例如 题中数据: （4 -> 7 -> 1 -> 4）（5 -> 2 -> 3 -> 3） （8 -> 6 -> 8）（10 -> 9 -> 10）。   可找到4个子群。

思路：每个子群都有一定数目的各不相同的元素，用变量t记录子群元素个数，用son[]存储子群元素，就有son[0],son[1]...son[t-1]个元素。

      那么我们可以得到如下变换规则-> 新字符数组gain[son[(k + j) % t]] = 原字符数组str[son[j]] (0 <= j <= t-1)。

公式推导过程：一个子群 有t个元素说明它的循环节为t， 经过k次置换 那么肯定就有son[(k + j) % t] = son[j]； 

            例如 题目数据：4-1 5-1 3-1 7-1 2-1 8-1 1-1 6-1 10-1 9-1 （为了对应字符串的下标自减1） Hello+Bob+ 空格用+表示

            拿子群（3 -> 6 -> 0 -> 3【已经减一】）来说，其中t = 3,son[0] = 3, son[1] = 6, son[2] = 0。

            当k = 1时， { j = 0,son[(k + j) % t] = 6，gain[6] = str[3] = l

                         j = 1,son[(k + j) % t] = 0，gain[0] = str[6] = B

                         j = 2,son[(k + j) % t] = 3，gain[3] = str[0] = H

                       } 

代码实现：

#include 
#include 
#define MAX 200+66
using namespace std;
char str[MAX];//将要进行变换的字符串 
char gain[MAX];//变换k次后 得到的新字符串 
int next[MAX];//str[i] 置换 str[next[i]] 
int pos[MAX];//保存当前子群元素 也就是变换过程的位置 
int vis[MAX];//标记当前元素是否存在于某个子群中 (一个元素只能存在于一个置换子群中)  若可以为1否则为0  
int main()
{
	int n, k;
	int i, j;
	int t;//记录当前子群元素个数  
	while(scanf("%d", &n), n)//n个元素 
	{
		for(i = 0; i < n; i++)
		{
			scanf("%d", &next[i]);
			next[i]--;//自减一  与字符串下标协调 
		}
		while(scanf("%d", &k), k)//k次变换  注意输入后空格 
		{
			getchar();
			memset(str, '\0', sizeof(str)); 
			gets(str);
			for(i = strlen(str); i < n; i++) str[i] = ' ';//填充空格 
			memset(vis, 0, sizeof(vis));//初始化
			memset(gain, '\0', sizeof(gain)); 
			//printf("%s\n", str); 
			for(i = 0 ;i < n; i++)
			{
				j = i;//便于运算 
				t = 0;//子群元素个数清 0 
				if(!vis[j])//不存在于 已找到的所有子群中 
				{
					while(!vis[j])//找循环节
					{
						vis[j] = 1;//存在于当前子群
						pos[t++] = j;//为当前子群保存元素 
						j = next[j];//下一位置 
					} 
				}
				for(j = 0; j < t; j++)//更新 子群所有元素 
				{
					gain[pos[(j+k)%t]] = str[pos[j]];//位置为son[j] 经过k次变换 会到达位置son[(j+k)%t] 
				}
			}
			printf("%s\n", gain); 
		}
		printf("\n");
	} 
	return 0;
}