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
- 手游与App测试如何快速转型? —— 过来人科普手游与App测试四大区别
随着智能设备的普及和移动互联网的兴起,各家互联网巨头纷纷在往移动端布局和转型,同时初创的移动互联网公司也都盯着这个市场希望分一杯羹.在这个大环境下,互联网的重心已经慢慢从Web端转向了移动端,而移动端 ...
- C# 使用Sqlite 如何返回统计行数
Visual 2010 with Sqlite 需要这样Query 数据: select count(*) from tblOrder where OrderStartTime >= '2013 ...
- poj 3104 Drying(二分搜索之最大化最小值)
Description It is very hard to wash and especially to dry clothes in winter. But Jane is a very smar ...
- pyqt例子搜索文本
#!/usr/bin/env python #-*- coding:utf-8 -*- import sip sip.setapi('QString', 2) sip.setapi('QVariant ...
- 下载配置MySql,高速启动MySql批处理,MySQLclient软件SQL-Front的配置---ShinePans
首先,下载 sql 绿色版,: http://yunpan.cn/cgERHhTtV8XGh 提取码 85bc 然后解压就可以用, 安装文件夹下有bin文件夹,从里面的命令中启动服务 例如以下: ...
- Leetcode_num3_Same Tree
题目: Given two binary trees, write a function to check if they are equal or not. Two binary trees are ...
- PCM文件格式简单介绍
PCM文件格式简单介绍 PCM文件:模拟音频信号经模数转换(A/D变换)直接形成的二进制序列,该文件没有附加的文件头和文件结束标志.Windows的Convert工具能够把PCM音频格式的文件转换成M ...
- unbantu相关笔记
很多项目使用的系统是centos或者redhat,最近有一个项目使用的系统竟然是阿里云unbantu,不知道他们负责人怎么想的,明明有centos,非要用unbantu.抱怨到此,unbantu的学习 ...
- Js字面变量,定义问题
Js字面变量.浏览器的版本问题:
- Eclipse MAT: Understand Incoming and Outgoing References
引用:http://xmlandmore.blogspot.hk/2014/01/eclipse-mat-understand-incoming-and.html?utm_source=tuicool ...