Ancient Cipher UVA - 1339
Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents were sent between provinces and the capital in encrypted form to prevent eavesdropping. The most popular ciphers in those times were so called substitution cipher and permutation cipher.
Substitution cipher changes all occurrences of each letter to some other letter. Substitutes for all letters must be different. For some letters substitute letter may coincide with the original letter. For example, applying substitution cipher that changes all letters from ‘A’ to ‘Y’ to the next ones in the alphabet, and changes ‘Z’ to ‘A’, to the message “VICTORIOUS” one gets the message “WJDUPSJPVT”.
Permutation cipher applies some permutation to the letters of the message. For example, applying the permutation ⟨2, 1, 5, 4, 3, 7, 6, 10, 9, 8⟩ to the message “VICTORIOUS” one gets the message “IVOTCIRSUO”.
It was quickly noticed that being applied separately, both substitution cipher and permutation cipher were rather weak. But when being combined, they were strong enough for those times. Thus, the most important messages were first encrypted using substitution cipher, and then the result was encrypted using permutation cipher. Encrypting the message “VICTORIOUS” with the combination of the ciphers described above one gets the message “JWPUDJSTVP”.
Archeologists have recently found the message engraved on a stone plate. At the first glance it seemed completely meaningless, so it was suggested that the message was encrypted with some substitution and permutation ciphers. They have conjectured the possible text of the original message that was encrypted, and now they want to check their conjecture. They need a computer program to do it, so you have to write one.
Input
Input file contains several test cases. Each of them consists of two lines. The first line contains the message engraved on the plate. Before encrypting, all spaces and punctuation marks were removed, so the encrypted message contains only capital letters of the English alphabet. The second line contains the original message that is conjectured to be encrypted in the message on the first line. It also contains only capital letters of the English alphabet.
The lengths of both lines of the input file are equal and do not exceed 100.
Output
For each test case, print one output line. Output ‘YES’ if the message on the first line of the input file could be the result of encrypting the message on the second line, or ‘NO’ in the other case.
Sample Input
JWPUDJSTVP
VICTORIOUS
MAMA
ROME
HAHA
HEHE
AAA
AAA
NEERCISTHEBEST
SECRETMESSAGES
Sample Output
YES
NO
YES
YES
NO
HINT
这道题目的映射是指的一个字母可以映射对应一个字母。映射的方式也不一定是按照一定的规律的。因此,只需要看看每一个字符串出现的次数是不是一样的就可以,要看是不是一样的就用到了排序算法,这里使用的是快排函数。
Accepted
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int cmp(const void* a, const void* b)
{
return *(int*)a - *(int*)b;
}
int main()
{
char arr[105];
char arr1[105];
while (scanf("%s", arr) != EOF && scanf("%s", arr1) != EOF)
{
int a[30] = { 0 };
int b[30] = { 0 };
for (int i = 0;i < strlen(arr); i++)
a[arr[i] - 'A']++;
for (int i = 0;i < strlen(arr1);i++)
b[arr1[i] - 'A']++;
qsort(a, 30, sizeof(int), cmp);
qsort(b, 30, sizeof(int), cmp);
int flag = 0;
for(int i=0;i<30;i++)
if(a[i]!=b[i])
{
flag = 1;
break;
}
if (flag)printf("NO\n");
else printf("YES\n");
}
}
Ancient Cipher UVA - 1339的更多相关文章
- UVa 1339 Ancient Cipher --- 水题
UVa 1339 题目大意:给定两个长度相同且不超过100个字符的字符串,判断能否把其中一个字符串重排后,然后对26个字母一一做一个映射,使得两个字符串相同 解题思路:字母可以重排,那么次序便不重要, ...
- UVa1399.Ancient Cipher
题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...
- uva--1339 - Ancient Cipher(模拟水体系列)
1339 - Ancient Cipher Ancient Roman empire had a strong government system with various departments, ...
- Poj 2159 / OpenJudge 2159 Ancient Cipher
1.链接地址: http://poj.org/problem?id=2159 http://bailian.openjudge.cn/practice/2159 2.题目: Ancient Ciphe ...
- Ancient Cipher UVa1339
这题就真的想刘汝佳说的那样,真的需要想象力,一开始还不明白一一映射是什么意思,到底是有顺序的映射?还是没顺序的映射? 答案是没顺序的映射,只要与26个字母一一映射就行 下面给出代码 //Uva1339 ...
- poj 2159 D - Ancient Cipher 文件加密
Ancient Cipher Description Ancient Roman empire had a strong government system with various departme ...
- POJ2159 Ancient Cipher
POJ2159 Ancient Cipher Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 38430 Accepted ...
- POJ2159 ancient cipher - 思维题
2017-08-31 20:11:39 writer:pprp 一开始说好这个是个水题,就按照水题的想法来看,唉~ 最后还是懵逼了,感觉太复杂了,一开始想要排序两串字符,然后移动之类的,但是看了看 好 ...
- 2159 -- Ancient Cipher
Ancient Cipher Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 36074 Accepted: 11765 ...
随机推荐
- 手把手教你Centos7 部署 gitlab社区版
一.前置说明: 操作系统:Centos 7 物理内存:>=2G 本人亲测,如果安装低版本的gitlab,比如我这里所使用的v8.17.0,物理内存1G,swap 2G虚拟内存即可部署.高版本的所 ...
- 微信小程序:如何实现两个按钮在最右侧并排
要实现的效果: wxml端代码: <view class="prepare_param"> <view clas ...
- 微信小程序登录流程解析
小程序可以通过微信官方提供的登录能力方便地获取微信提供的用户身份标识openid,快速建立小程序内的用户体系. 登录流程时序: 1.首先,调用 wx.login获取code ,判断用户是否授权读取用户 ...
- 导出----用Excel导出数据库表
根据条件导出表格: 前端 <el-form-item label=""> <el-button type="warning" icon=&qu ...
- SpringBoot(四): SpringBoot web开发 SpringBoot使用jsp
1.在SpringBoot中使用jsp,需要在pom.xml文件中添加依赖 <!--引入Spring Boot内嵌的Tomcat对JSP的解析包--> <dependency> ...
- 一个基于 Vue3 的开源项目,3个月时间 star 终于破千!
本文主要是对如何做开源项目的一些思考. 前文回顾: <Vue3 来了,Vue3 开源商城项目重构计划正式启动!> <一个基于 Vue 3 + Vant 3 的开源商城项目> 关 ...
- LNMP配置——PHP安装
一.下载 #cd /usr/local/src //软件包都放在这里方便管理 #wget http://cn2.php.net/distributions/php-5.6.30.tar.gz 二.解压 ...
- C语言入门-mingw64安装+配置
OK,大家好,结合上期所说,本期让我们来配置编译器吧! 首先先下载mingw64离线包,官网下载慢,可以去群里下载,*.7z格式(有些同学可能没有解压软件,为了照顾这部分同学,笔者提供*.exe格式的 ...
- [笔记] 扩展Lucas定理
[笔记] 扩展\(Lucas\)定理 \(Lucas\)定理:\(\binom{n}{m} \equiv \binom{n/P}{m/P} \binom{n \% P}{m \% P}\pmod{P} ...
- Shtml、html、xhtml、htm以及SSI的了解与认识(转载)
Shtml.html.xhtml.htm以及SSI的了解与认识(转载) 一.htm.html.shtml网页区别(博客园) 文章链接:https://www.cnblogs.com/Renyi-Fan ...