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.
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
解题思路：
读了半天才看懂题意。题目不难，但要静下心去理解。
题目大意是先给一串数字， 再给一串信息。信息中的每一个字符与给的数字相相应，而相应的数字表示的是该字符应该出如今第几个位置。
依据数字调整好字符的位置后相当于编码一次。然后让你求编码k次后的信息是如何的。
这题必需要找出循环的节点。
字符在编码m次后一定会回到原来的位置，这就是一个循环。记录一个循环中每一个点的位置。编码次数会给的非常大。
假设不用循环来做。是会超时的。
注意：这题每组数据输出之间要空一行。
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
int n, k, a[205], m[205][205], huan[205]; //m记录一个循环中每一个节点。huan记录每一个点多久循环一次
char mes[205], ans[205];
while(scanf("%d", &n) && n)
{
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
while(scanf("%d", &k) && k)
{
gets(mes);
for(int i = strlen(mes); i <= n; i++)
mes[i] = ' ';
memset(huan, 0, sizeof(huan));
for(int i = 1; i <= n; i++)
{
int temp, j;
temp = a[i];
j = 1;
m[i][j] = temp;
while(a[temp] != a[i])
{
temp = a[temp];
m[i][++j] = temp; // 记录循环中点的位置
huan[i]++; //仅仅要没有回到原来位置。循环长度加1
}
huan[i]++;
}
for(int i = 1; i <= n; i++)
{
int p, q;
p = k % huan[i]; // 计算循环后的位置
if(p)
q = m[i][p];
else
q = m[i][huan[i]]; //若能整除说明循环到最后一个
ans[q - 1] = mes[i];
}
ans[n] = ' ';
printf("%s
", ans);
}
printf("
"); //每组数据之间要空一行
}
return 0;
}