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 : Bob和Alice想用一种新的加密方式,他们编码和译码都是依靠秘密钥匙,秘密钥匙选用一串数字a1,……an(0<ai<=n),编码要求:将信息(长度不大于n,如果小于n请自己补上空格)中i位置上的字符放到ai位置上
。这个过程经过K遍之后就成了加密后的信息。现在要求你写一个程序帮他们实现编码。 思路: 因为K会很大, 暴力会超时, 将其看作一个置换群, 求出每个循环的长度, 可知一个循环的长度即他的周期; 考虑求出每个循环的长度L, 对K取模, 然后再暴力求解;
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn= ; int main()
{
int n;
while(scanf("%d", &n) && n)
{
int hay[maxn], pos[maxn];
for(int i = ;i < n; i++)
{
scanf("%d",&hay[i]);
hay[i] -= ;
}
for(int i =;i < n;i++)
{
pos[i] = ;
int tmp = hay[i];
while(tmp!=i)
{
pos[i]++;
tmp = hay[tmp];
}
}
int k;
char s1[maxn], s2[maxn];
while(scanf("%d", &k) && k)
{
getchar();
gets(s1);
if(strlen(s1) < n)
{
for(int i = strlen(s1); i<n; i++)
s1[i] = ' ';
s1[n] = '\0';
}
for(int i = ; i < n; i++)
{
int tim = k%pos[i];
int t = i;
while(tim--)
{
t = hay[t];
}
s2[t] = s1[i];
}
s2[n] = '\0';
puts(s2);
}
printf("\n");
}
return ; }
												

Cipher的更多相关文章

  1. JAVA实现AES 解密报错Input length must be multiple of 16 when decrypting with padded cipher

    加密代码 /**解密 * @param content 待解密内容 * @param password 解密密钥 * @return */ public static byte[] decrypt(b ...

  2. POJ1026 Cipher(置换的幂运算)

    链接:http://poj.org/problem?id=1026 Cipher Time Limit: 1000MS   Memory Limit: 10000K Total Submissions ...

  3. BZOJ 1031: [JSOI2007]字符加密Cipher 后缀数组

    1031: [JSOI2007]字符加密Cipher Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 6014  Solved: 2503[Submit ...

  4. freeCodeCamp:Caesars Cipher

    让上帝的归上帝,凯撒的归凯撒. 下面我们来介绍风靡全球的凯撒密码Caesar cipher,又叫移位密码. 移位密码也就是密码中的字母会按照指定的数量来做移位. 一个常见的案例就是ROT13密码,字母 ...

  5. ural Cipher Message

    Cipher Message Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Desc ...

  6. 紫书例题-Ancient Cipher

    Ancient Roman empire had a strong government system with various departments, including a secret ser ...

  7. 【BZOJ-1031】字符加密Cipher 后缀数组

    1031: [JSOI2007]字符加密Cipher Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 5504  Solved: 2277[Submit ...

  8. uva-1339Ancient Cipher

    Ancient Roman empire had a strong government system with various departments, including a secret ser ...

  9. scp使用加密算法报错unknown cipher type

    为了提高scp的传输速度指定了scp的加密算法为arcfour $ scp -c arcfour localFile userName@remoteIP:remoteFile 得到报错unknown ...

  10. POJ - 1107 W's Cipher

    POJ - 1107 W's Cipher Time Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %I64d & %I64u De ...

随机推荐

  1. python循环字符转换

    pyhon函数传参的时候穿的是引用,而不是实际值,这样可以节省内存 变量名要求:最好是以字母下划线作为变量名,不能和py关键字重复 import getpass提供了平台无关的在命令行下输入密码的方法 ...

  2. elasticsearch安装IK分词插件

    一 打开网页:https://github.com/medcl/elasticsearch-analysis-ik/releases 这个是ik相关的包,找到你想下载的版本,下载对应的zip包 二 然 ...

  3. jquery.axios无刷新机制删除

    思路:无刷新机制就是不用的刷新动作 ,用前端html语法删除和后端的数据库删,同时删除达到效果 除操作,来实现无刷洗的方法

  4. Windows7安装两个jdk配置

    win7 配置两个jdk,之前装了jdk1.6,项目需要安装1.8. 首先去oracle官网下载一个和你eclipse版本一致的jdk(我的是32位). 网址:https://www.oracle.c ...

  5. zabbix server源码安装

    一.准备工作 yum -y install net-snmp-devel php-bcmath php-ctype php-xml php-xmlreader php-xmlwriter php-se ...

  6. Oracle 11g服务详细介绍及哪些服务是必须开启的

    成功安装Oracle 11g后,共有7个服务,这七个服务的含义分别为: 1. Oracle ORCL VSS Writer Service: Oracle卷映射拷贝写入服务,VSS(Volume Sh ...

  7. what's the python之python介绍

    其实这一篇文章的大部分都是啰嗦话,大部分在百度百科中都有详尽的叙述.既然决定学python了就要风雨兼程,你不用洞悉python到底是什么,你只要知道这是一门编程语言,跟Java.C++等语言一样都是 ...

  8. grub启动流程和配置

    grub stage  1 MBR中前 446个字节,如果把这里面的内容损坏,那么系统会认为当前磁盘没有启动引导功能,会尝试从光盘或者网络启动系统 grub  stage  1.5 存放识别/boot ...

  9. eos 空投列表网址 及 工具网站列表

    https://eosdrops.io https://www.shensi.com/#/eos eos 区块链浏览器: https://eostracker.io/ https://eospark. ...

  10. OpenVPN简介及架构详解

    OpenVPN简介 1 简介    VPN(Virtual Private Network)直译就是虚拟专用通道,是提供给企业之间或者个人与公司之间安全数据传输的隧道. OpenVPN无疑是Linux ...