POJ 1026-Cipher（置换群-K次置换取模循环节长度）

迷南。 2022-06-17 05:53 131阅读 0赞

Cipher

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;21967</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;6059</td> 
  </tr> 
 </tbody> 
</table>

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

[Central Europe 1995][]

## 题目意思： ##

给出N个数构成的置换群，长度不大于N的字符串，输出K次置换之后的串。

##    解题思路： ##

据（巨巨）说：

发现每个字母经过一定次数变换  
后一定会回到原来的位置，且这个变换次数肯定不会大于N。

所以我们可以找出每个循环节的长度，用K对其取模之后再计算置换的情况。

#include<iostream>
    #include<cstdio>
    #include<iomanip>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<map>
    #include<algorithm>
    #include<vector>
    #include<queue>
    using namespace std;
    #define INF 0xfffffff
    #define MAXN 300
    int a[MAXN],cir[MAXN];
    char c[MAXN],ans[MAXN];
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        int n;
        while(~scanf("%d",&n))
        {
            if(n==0) break;
            memset(c,'\0',sizeof(c));
            memset(ans,'\0',sizeof(ans));
            memset(cir,0,sizeof(cir));
            for(int i=0; i<n; ++i)
            {
                scanf("%d",&a[i]);
                --a[i];
            }
            int cnt=0,res=1;//注意本身就算是1个
            for(int i=0; i<n; ++i)
            {
                int temp=a[i];
                while(1)//求置换群中每个循环节长度
                {
                    if(i==temp)
                    {
                        cir[cnt++]=res;
                        res=1;
                        break;
                    }
                    ++res;
                    temp=a[temp];
                }
            }
            int k;
            while(~scanf("%d",&k))
            {
                if(k==0) break;
                char ch=getchar();//先吃掉k后面的空格
                gets(c);
                int len=strlen(c);
                while(len<=n) c[len++]=' ';//长度不足要用空格补齐
                c[len]='\0';
                //cout<<"/"<<c<<"/"<<endl;
                for(int i=0; i<n; i++)
                {
                    int temp=k%cir[i];
                    int t=i;
                    while(temp--)//交换
                        t=a[t];
                    ans[t]=c[i];
                }
                //cout<<strlen(ans)<<endl;
                printf("%s\n",ans);
            }
            printf("\n");//PE的罪恶之源…哭/(ㄒoㄒ)/~~
        }
        return 0;
    }

[Central Europe 1995]: http://poj.org/searchproblem?field=source&key=Central+Europe+1995