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

HINT

  这道题目的映射是指的一个字母可以映射对应一个字母。映射的方式也不一定是按照一定的规律的。因此,只需要看看每一个字符串出现的次数是不是一样的就可以,要看是不是一样的就用到了排序算法,这里使用的是快排函数

Accepted

#include<stdio.h>
#include<stdlib.h>
#include<string.h> int cmp(const void* a, const void* b)
{
return *(int*)a - *(int*)b;
}
int main()
{
char arr[105];
char arr1[105]; while (scanf("%s", arr) != EOF && scanf("%s", arr1) != EOF)
{
int a[30] = { 0 };
int b[30] = { 0 };
for (int i = 0;i < strlen(arr); i++)
a[arr[i] - 'A']++;
for (int i = 0;i < strlen(arr1);i++)
b[arr1[i] - 'A']++;
qsort(a, 30, sizeof(int), cmp);
qsort(b, 30, sizeof(int), cmp);
int flag = 0;
for(int i=0;i<30;i++)
if(a[i]!=b[i])
{
flag = 1;
break;
}
if (flag)printf("NO\n");
else printf("YES\n");
}
}

Ancient Cipher UVA - 1339的更多相关文章

  1. 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, ...

  2. UVa1399.Ancient Cipher

    题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

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

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

  4. 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 ...

  5. 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 ...

  6. poj 2159 D - Ancient Cipher 文件加密

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

  7. POJ2159 Ancient Cipher

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

  8. 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 ...

  9. 2159 -- Ancient Cipher

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

随机推荐

  1. js---it笔记

    typeof a返回的是字符串 vscode scss安装的easy scss中的配置settingjson文件中的css编译生成路径是根目录下的

  2. 开源OA办公平台搭建教程:O2OA+Arduino实现物联网应用(一)

    O2OA平台是一个企业办公类系统的低代码开发平台,更够方便的开发和部署协同办公.流程管理等应用,但它能做的远不止这些,今天这个案例就为大家介绍一下,O2OA可以做的更多. 最近对养鱼产生了浓厚的兴趣, ...

  3. 力扣832. 翻转图像-C语言实现-简单题

    题目 传送门 文本 给定一个二进制矩阵 A,我们想先水平翻转图像,然后反转图像并返回结果. 水平翻转图片就是将图片的每一行都进行翻转,即逆序.例如,水平翻转 [1, 1, 0] 的结果是 [0, 1, ...

  4. CVer想知道的都在这里了,一起分析下《中国计算机视觉人才调研报告》吧!

    最近闲来无事,老潘以一名普通算法工程师的角度,结合自身以及周围人的情况,理性也感性地分析一下极市平台前些天发布的2020年度中国计算机视觉人才调研报告. 以下的"计算机视觉人才"简 ...

  5. idea快捷键:查找类中所有方法的快捷键

    查找类中所有方法的快捷键 第一种:ctal+f12,如下图 第二种:alt+7,如下图

  6. .NET测试断言工具Shouldly

    .NET测试断言工具Shouldly .NET测试 Shouldly在GitHub的开源地址:https://github.com/shouldly/shouldly Shouldly的官方文档:ht ...

  7. 腾讯一面问我SQL语句中where条件为什么写上1=1

    目录 where后面加"1=1″还是不加 不用where 1=1 在多条件查询的困惑 使用where 1=1 的好处 使用where 1=1 的坏处 where后面加"1=1″还是 ...

  8. SpineRuntime-Presentation - 基于 spine-libgdx 实现在 AndroidPresentation 上展示 Spine 动画

    SpineRuntime-Presentation 基于 spine-libgdx 实现在 AndroidPresentation 上展示 Spine 动画 Github地址 效果 可以在 Andro ...

  9. 40. 组合总和 II + 递归 + 回溯 + 记录路径

    40. 组合总和 II LeetCode_40 题目描述 题解分析 此题和 39. 组合总和 + 递归 + 回溯 + 存储路径很像,只不过题目修改了一下. 题解的关键是首先将候选数组进行排序,然后记录 ...

  10. pytorch(04)简单的线性回归

    线性回归 线性回归是分析一个变量与另外一个变量之间关系的方法 因变量:y 自变量:x 关系:线性 y = wx+b 分析:求解w,b 求解步骤: 确定模型,Model:y = wx+b 选择损失函数, ...