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
- Spring初学(一)
Spring核心机制:依赖注入 依赖注入简单的理解就是 由Spring负责对model进行设置,而非由controller直接设置. 通过依赖注入,javaEE各种组件可以解耦. 依赖注入(Depen ...
- Entify Framewrok - 学习链接
http://blogs.msdn.com/b/adonet/archive/2011/01/31/using-dbcontext-in-ef-feature-ctp5-part-6-loading- ...
- python入门第一天,循环与判断
学习一门新的语言最重要的就是练习. 一.脚本需求: 编写登陆接口 输入用户名密码 认证成功后显示欢迎信息 输错三次后锁定 二.脚本流程图: 写代码之前画个流程图总是好的,可以让你理清思路,避免写着写着 ...
- WebService-相关概念介绍
WebService学习总结(二)——WebService相关概念介绍 一.WebService是什么? 1. 基于Web的服务:服务器端整出一些资源让客户端应用访问(获取数据) 2. 一个跨语言.跨 ...
- Android + eclipse +ADT安装完全教程
最近几天没事做,网上看来看去突然就想弄个android来学学... 首先,是要下载android SDK,在http://developer.android.com/sdk/index.html这个 ...
- 初学者学Java设计模式(一)------单例设计模式
单例设计模式 单例设计模式是指一个类只会生成一个对象,优点是他可以确保所有对象都访问唯一实例. 具体实现代码如下: public class A { public static void main(S ...
- [Redux] Passing the Store Down Implicitly via Context
We have to write a lot of boiler plate code to pass this chore down as a prop. But there is another ...
- IPMITOOL常用操作指令
一.开关机,重启 1. 查看开关机状态: ipmitool -H (BMC的管理IP地址) -I lanplus -U (BMC登录用户名) -P (BMC 登录用户名的密码) power statu ...
- oracle监听服务开启
输入命令netca即可开启oracle的监听服务 弹出对话框 选择监听服务配置,单击下一步 选择增加监听,单击下一步 监听的名字,默认即可,下一步 监听链接的协议,默认TCP协议即可,下一步 监听默认 ...
- jquery之onblur事件
onblur事件与html结合 function discountCheck(){ //var checkVal=$('input:text[name="Fee1"]').val( ...