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
- 全球扫货指南:面向中国消费者的全球奢侈品旅游导购服务 |PingWest中文网
全球扫货指南:面向中国消费者的全球奢侈品旅游导购服务 |PingWest中文网 用"全球扫货指南
- sizeof(long)
16位系统:long是4字节,int是2字节32位系统:long是4字节,int是4字节64位系统:long是8字节,int是4字节
- STM32F051 IAP源代码分享
STM32F051 IAP源代码分享 假设不懂IAP的请自己主动脑补或者怒戳这里 http://www.openedv.com/posts/list/11494.htm 然后STM32F051的IAP ...
- android EditText插入字符串到光标所在位置
EditText mTextInput=(EditText)findViewById(R.id.input);//EditText对象 int index = mTextInput.getSelect ...
- kvm 启动libvirtd市出现错误
kvm 启动libvirtd市出现错误: /etc/init.d/libvirtd start启动 libvirtd 守护进程:libvirtd: relocation error: libvirtd ...
- 自定义HttpHandler
1.创建自定义类型 2.继承IHttpHandler接口,并实现 3.配置Web.Config文件,注册类型 4.访问 public class QuickMsgSatisticsHandler : ...
- Android测试框架1(非原创)
1.继承AndroidTestCase :public class JunitTest3 extends AndroidTestCase {} 2.在AndroidManifest.xml清单文件中添 ...
- 《第一行代码》学习笔记2-Android开发特色
1.四大组件:活动(Activity),服务(Service),广播接收器(Broadcast Receiver),内容提供器(Content Provider). Activity:应用中看得到的东 ...
- .NET踩坑记录【不断更新】
NET 4.0 Tasks 使用 ThreadPool 可设置最大并发级别. 多个WebClient多线程下载受System.Net.ServicePointManager.DefaultConnec ...
- POJ 1001 Exponentiation 模拟小数幂
模拟小数幂 小数点位 pos 非零末位 e 长度 len 只有三种情况 pos > len pos < e e < pos < len #include <iostrea ...