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
- SpringBoot读取资源目录下的文件
需要读取resources目录下的文件,那么方法如下: 假设在资源目录下的template目录下有一个文件a.txt,获取到文件流的方式 InputStream stream = this.getCl ...
- oracle can't kill session
oracle 在杀会话时,会出现杀不掉的情况. 原因是在回滚大事物 解决方法: alter system disconnect session 'sid, serial#' immediate; ...
- XSS跨站脚本攻击(1)
将跨站脚本攻击缩写为XSS,恶意攻击者往Web页面里插入恶意Script代码,当用户浏览该页面的时候,嵌入其中的Web里面的Script代码就会被执行,从而达到恶意攻击用户的目的. 反射型XSS 反射 ...
- Java基本概念:多态
一.简介 描述: 多态性是面向对象编程中的一个重要特性,主要是用来实现动态联编的.换句话说,就是程序的最终状态只有在执行过程中才被决定,而非在编译期间就决定了.这对于大型系统来说能提高系统的灵活性和扩 ...
- 使用lua-nginx模块实现请求解析与调度
系统版本及需求: OS:CentOS 7.7.1908 OpenResty:1.15.8.2 目录 描述 安装配置 安装openresty 使用示例 HTTP请求复制 HTTP报文解析 总结 描述 l ...
- JUC-ThreadLocalRandom
目录 Radndom类的局限性 ThreadLocalRandom 这个类是在JDK7中新增的随机数生成器,它弥补了Random类在多线程下的缺陷. Radndom类的局限性 在JDK7之前包括现在j ...
- PVE更新WEB管理地址
PVE也是一台Linux系统,如果PVE更换了网络环境,比如从家里拿到了办公室,那么就需要对其更新网络,才能让其它机器访问到它的8006管理地址. 具体做法是通过修改配置文件来更改IP. 更新网卡配置 ...
- 500GJava/Hadoop/Spark/机器学习...视频教程免费分享 百度云持续更新
参加工作这么长时间了,工作中遇到了不少技能都是看视频教程学习的,相比较看书而言看视频确实比较容易理解.分享一下自己看过的和收集的视频教程. 资源包括: 大数据方面的Hadoop(云帆,小象学院,八斗学 ...
- HDOJ-1043 Eight(八数码问题+双向bfs+高效记录路径+康拓展开)
bfs搜索加记录路径 HDOJ-1043 主要思路就是使用双向广度优先搜索,找最短路径.然后记录路径,找到结果是打印出来. 使用康拓序列来来实现状态的映射. 打印路径推荐使用vector最后需要使用a ...
- 简单的ssm练手联手项目
简单的ssm练手联手项目 这是一个简单的ssm整合项目 实现了汽车的品牌,价格,车型的添加 ,修改,删除,所有数据从数据库中拿取 使用到了jsp+mysql+Mybatis+spring+spring ...