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

  1. uva--1339 - Ancient Cipher(模拟水体系列)

    1339 - Ancient Cipher Ancient Roman empire had a strong government system with various departments, ...

  2. 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, ...

  3. 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 ...

  4. 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 ...

  5. poj 2159 D - Ancient Cipher 文件加密

    Ancient Cipher Description Ancient Roman empire had a strong government system with various departme ...

  6. POJ2159 Ancient Cipher

    POJ2159 Ancient Cipher Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38430   Accepted ...

  7. 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 ...

  8. 2159 -- Ancient Cipher

    Ancient Cipher Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 36074   Accepted: 11765 ...

  9. 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

  1. You don't have permission to access / on this server for debian_8

    Forbidden You don't have permission to access / on this server. Apache/2.4.10 (Debian) Server at www ...

  2. js如何判断字符串是否进行过window.btoa()转码

    window.btoa()是基于Base64算法的.window.btoa()只能将ASCII字符进行转码 因此我们需要了解Base64的原理及主要特征:Base64的原理在这里就不多说了,网上很多讲 ...

  3. poj 2392 Space Elevator(多重背包+先排序)

    Description The cows are going to space! They plan to achieve orbit by building a sort of space elev ...

  4. JAVA模拟表单提交

    这是我网上搜的,自己使用也蛮方便,所以上传供大家分享. package wzh.Http;   import java.io.BufferedReader; import java.io.IOExce ...

  5. 细说php(六) 数组

    一.数组概述 1.1 数组是复合类型 1.2 数组中能够存储随意长度的数据, 也能够存储随意类型的数据 二.数组的类型 2.1 索引数组: 下标是顺序整数作为索引 <?php $user[0] ...

  6. PowerDesigner Mysql 主键自增、初始值、字符集

    自增 在你所要设为自增型的键上(比如你的id)双击,弹出一个Column Properties对话框,右下角有一个Identify的选择框,选中它OK,就可以了. 再去查看Preview,就能看到AU ...

  7. arry()数组的理解及api的使用(一)

    我们想要了解数组,首先就要先要了解到什么是数据结构,所谓的数据结构就是把数据与数据见的关系按照特定的结构来保存.设计合理的数据结构是解决问题的前提.了解了数据结构后我们下面来数组的定义:数组(arra ...

  8. java运算

    (一) 截图: 程序: import javax.swing.JOptionPane; public class Addition { public static void main (String ...

  9. Render和template?

    Template是一个模板. render = web.template.render('templates/') 这会告诉web.py到你的模板目录中去查找模板.然后把 index.GET改成: 告 ...

  10. UWP开发小记

    针对个人的上一篇文章中提到的遇到的几个问题,做一下个人解答 DLL部署的问题,可以将DLL添加到工程中,属性中设置content为true,这样,部署目录下就会有这个文件. 需要说明的是,这个文件确实 ...