ACM: polya模拟 数论题 poj 1026

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该博客介绍了一种Bob和Alice使用的基于密钥的编码方式，利用一系列不重复的正整数作为秘密钥匙对消息进行多次编码。文章详细阐述了编码原理，并给出了一段C++代码实现对给定序列的编码过程。最后，提供了样例输入和输出，帮助理解编码算法的工作方式。

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Cipher

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

4 5 3 7 2 8 1 6 10 9

1 Hello Bob

1995 CERC

Sample Output

BolHeol b

C RCE

题意: 给你置换的编码表格, 再给你重复编码的次数和序列 , 求出新序列.

解题思路:

1. 4 5 3 7 2 8 1 6 10 9

1 2 3 4 5 6 7 8 9 10

循环有: (4 1 7) (5 2) (3) (8 6) (10 9)

循环节是: 编码次数 % 循环长度.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 205

int n, k;
char str1[MAX], str2[MAX];
int a[MAX];
int cir[MAX];

int main()
{
//   freopen("input.txt","r",stdin);
   while(scanf("%d",&n) != EOF && n != 0)
   {
       for(int i = 1; i <= n; ++i)
           scanf("%d",&a[i]);
       for(int i = 1; i <= n; ++i)
       {
           int j = i;
           int t = 1;
           while(a[j] != i)
           {
               t++;
               j = a[j];
           }
           cir[i] = t;
       }

       while(scanf("%d",&k) != EOF && k != 0)
       {
           getchar();
           gets(str1+1);
           int max = -1;
           int len = strlen(str1+1);
           memset(str2,' ',sizeof(str2));

           for(int i = 1; i <= len; ++i)
           {
               int x = k % cir[i];
               int j = i;
               while(x--)
                   j = a[j];
               str2[j] = str1[i];
               if(j > max) max = j;
           }
           for(int i = 1; i <= max; ++i)
               str1[i] = str2[i];
           str1[max+1] = '\0';
           printf("%s",str1+1);
           for(int i = n-max; i >= 0; --i)
               printf(" ");
           printf("\n");
       }
       printf("\n");
   }
   return 0;
}