UVa1399.Ancient Cipher
题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4085
| 13855995 | 1339 | Ancient Cipher | Accepted | C++ | 0.012 | 2014-07-09 12:35:33 |
Ancient Cipher
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
解题思路:读了好长时间,没想到就是一道排序的水题。题目大意就是将一个字符串中的字符顺序改变后再不确定的一一对应是否能得到另外一个字符串。
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <numeric>
using namespace std;
int main() {
string str1, str2;
int cnt1[], cnt2[];
while (cin >> str1 >> str2) {
memset(cnt1, , sizeof(cnt1));
memset(cnt2, , sizeof(cnt2)); for(int i = ; i < str1.size(); i++) {
cnt1[str1[i] - 'A'] ++;
cnt2[str2[i] - 'A'] ++;
} sort(cnt1, cnt1 + );
sort(cnt2, cnt2 + ); bool flag = true; for(int i = ; i < ; i++) {
if(cnt1[i] != cnt2[i]) {
flag = false;
break;
}
} if(flag) cout << "YES" << endl;
else cout << "NO" << endl; }
return ;
}
UVa1399.Ancient Cipher的更多相关文章
- uva--1339 - Ancient Cipher(模拟水体系列)
1339 - Ancient Cipher Ancient Roman empire had a strong government system with various departments, ...
- 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, ...
- 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 ...
- 紫书例题-Ancient Cipher
Ancient Roman empire had a strong government system with various departments, including a secret ser ...
随机推荐
- 深入理解linux网络技术内幕读书笔记(四)--通知链
Table of Contents 1 概述 2 定义链 3 链注册 4 链上的通知事件 5 网络子系统的通知链 5.1 包裹函数 5.2 范例 6 测试实例 概述 [注意] 通知链只在内核子系统之间 ...
- 又一编辑神器-百度编辑器-Ueditor
(Lionden<hsdlionden@gmail.com> 转载说明) 前段时间发表过一篇关于“KindEditor在JSP中使用”的博文.这几天在沈阳东软进行JavaWeb方面的实习工 ...
- WHY IE AGAIN? - string.charAt(x) or string[x]?
近期今天在写一个"删除字符串中反复字符串"的函数,代码例如以下: 开门见山,重点 string.charAt(index) 取代 string[index] function re ...
- 光盘自动运行HTML页,Autorun文件写法
1.把你的网页放在一个根目录下面,起名为index.html 2.在目录新建一个autorun.inf的文件,打开后编辑为以下内容: 代码如下: [autorun]icon=***.ico(加图标) ...
- js获取get值
//获取get值 function getPar(par) { //获取当前URL var local_url = document.location.href; //获取要取得的get参数位置 va ...
- C# 将对象序列化为Json格式
public static string JsonSerializer<T>(T t) { DataContractJsonSerializer ser = new DataContrac ...
- oracle 存储过程返回结果集 (转载)
好久没上来了, 难道今天工作时间稍有空闲, 研究了一下oracle存储过程返回结果集. 配合oracle临时表, 使用存储过程来返回结果集的数据读取方式可以解决海量数据表与其他表的连接问题. 在存储过 ...
- 创建存储过程和函数【weber出品必属精品】
一.什么是存储过程和函数 1. 是被命名的pl/sql块 2. 被称之为pl/sql子程序 3. 与匿名块类似,有块结构: 声明部分是可选的(没有declare关键字) 必须有执行部分 可选的异常处理 ...
- 不用派生CTreeCtrl不用繁琐的过程 教你如何让CTreeCtrl的每一项有ToolTip提示
最近工作中需要让CTreeCtrl控件的每一项都有提示信息,于是谷歌百度,爬山涉水,结果是………….在CodeProject里找到一篇文章是把CTreeCtrl派生出新类,重载一些函数自定义内容.使用 ...
- 三个重要的游标sp_cursoropen
請問這三個存諸過程的作用是什么﹖ sp_cursoropen, sp_cursorfetch, sp_cursorclose API 服务器游标实现 SQL Server OLE DB 提供程序. ...