uva306 - Cipher 置换群

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介绍了一种基于整数序列的置换编码方案，该方案使用特定的密钥进行消息编码，并通过多次迭代来增强安全性。文章详细解释了编码过程及其实现方法。

Cipher

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, , 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 , where 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 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

给你一种置换方法，a[i]=j就是经过一次置换，第i个位置的元素到j位置去了。问经过n次置换以后的字符串。

如果暴力会超时的，但是n只有200。。因为置换是不重复的，所以最多经过n次又换成了他自己。。也就是它有一个周期，周期最大是n。

比如第一个样例 1-4-7-1。。。2-5-2。。。3-3。。。6-8-6。。。9-10-9。。(147)(25)(3)(68)(9，10)这么循环，所以只要找到每个位置的周期就很快了不会超时。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cctype>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int N,M;
char str[210],ans[210],code[210];
int t[210],a[210];
void init(){
    int i,j,p;
    memset(t,0,sizeof(t));
    for(i=1;i<=N;i++){
        p=a[i];
        t[i]=1;
        while(p!=i){
            p=a[p];
            t[i]++;
        }
    }
}
void solve(){
    memset(ans,' ',sizeof(ans));
    int i,j;
    for(i=1;str[i];i++) ans[i]=str[i];
    for(i=1;i<=N;i++){
        int n=M%t[i],p=i;
        for(j=0;j<n;j++) p=a[p];
        code[p]=ans[i];
    }
    for(i=1;i<=N;i++) printf("%c",code[i]);
    puts("");
}
int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d",&N),N){
        int i;
        for(i=1;i<=N;i++) scanf("%d",&a[i]);
        init();
        while(scanf("%d",&M),M){
            gets(str);
            solve();
        }
        puts("");
    }
    return 0;
}