B. Keyboard Layouts
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.

You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.

You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.

Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.

Input

The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.

The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.

The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.

Output

Print the text if the same keys were pressed in the second layout.

Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
map构建映射就可以了
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
map<char,char>m;
map<int,char>p;
char a[],b[];
int main()
{
scanf("%s",&a);
for(int i=;i<;i++)
{
p[i+]=a[i];
}
scanf("%s",&a);
for(int i=;i<;i++)
{
m[p[i+]]=a[i];
}
scanf("%s",&b);
for(int i=;b[i]!='\0';i++)
{
if(b[i]>='' && b[i]<='') printf("%c",b[i]);
else
{
if(b[i]>='A' && b[i]<='Z')
printf("%c",toupper(m[tolower(b[i])]));
else printf("%c",m[b[i]]);
}
}
printf("\n");
}

Codefroces 831B Keyboard Layouts的更多相关文章

  1. codeforces 831B. Keyboard Layouts 解题报告

    题目链接:http://codeforces.com/contest/831/problem/B 题目意思:给出两个长度为26,由小写字母组成的字符串s1和s2,对于给出的第三个字符串s3,写出对应s ...

  2. Codeforces831B Keyboard Layouts

    B. Keyboard Layouts time limit per test 1 second memory limit per test 256 megabytes input standard ...

  3. Codeforces Round #424 B. Keyboard Layouts(字符串,匹配,map)

    #include <stdio.h> #include <string.h> ][]; ]; ]; int main(){ scanf(]); scanf(]); scanf( ...

  4. 【Codeforces Round #424 (Div. 2) B】Keyboard Layouts

    [Link]:http://codeforces.com/contest/831/problem/B [Description] 两个键盘的字母的位置不一样; 数字键的位置一样; 告诉你第一个键盘按某 ...

  5. CF831B Keyboard Layouts 题解

    Content 给你 \(26\) 个字母的映射(都是小写,大写的映射方式相同),再给你一个字符串 \(s\),求它的映射结果(如果有非字母的字符保持不变). 数据范围:\(1\leqslant |s ...

  6. Fedora 22中的Locale and Keyboard Configuration

    Introduction The system locale specifies the language settings of system services and user interface ...

  7. OS X: Keyboard shortcuts

    Using keyboard shortcuts To use a keyboard shortcut, press a modifier key at the same time as a char ...

  8. Educational Codeforces Round 82 C. Perfect Keyboard

    Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him ...

  9. Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem A - B

    Array of integers is unimodal, if: it is strictly increasing in the beginning; after that it is cons ...

随机推荐

  1. rac重新启动遭遇ORA-01078、ORA-01565、ORA-17503、ORA-12547

    今天測试环境server重新启动导致一个节点集群无法重新启动,遭遇ORA-12547错误.详细例如以下: server重新启动后,rac1集群无法启动,rac2正常启动: [root@rac1 ~]# ...

  2. 杭电1425 sort

    Problem Description 给你n个整数.请按从大到小的顺序输出当中前m大的数.   Input 每组測试数据有两行,第一行有两个数n,m(0<n,m<1000000).第二行 ...

  3. Centos7.4 modsecurity with nginx 安装

    1.准备: 系统环境:Centos7.4 软件及版本: nginx:OpenResty1.13.6.1 ModSecurity:ModSecurity v3.0.0rc1 (Linux) modsec ...

  4. USACO2002 Open:雄伟的山峦

    简要题意: 奶牛们在落基山下避暑,从它们的房子向外望去,可以看到N 座山峰构成的山峦,奶牛发现每座山峰都是等腰三角形,底边长度恰好是高度的两倍.所以山峰的顶点坐标可由两个底部端点求出.设i 座第山峰的 ...

  5. VMware中CentOS6.5启动出现An error occurred during the file system check

     

  6. mybatis :实现mybatis分页

    上一篇文章里已经讲到了mybatis与spring MVC的集成,并且做了一个列表展示,显示出所有article 列表,但没有用到分页,在实际的项目中,分页是肯定需要的.而且是物理分页,不是内存分页. ...

  7. lvs为何不能完全替代DNS轮询--转

    原文地址:http://mp.weixin.qq.com/s?__biz=MjM5ODYxMDA5OQ==&mid=2651959595&idx=1&sn=5f0633afd2 ...

  8. Sql Server新手学习入门

    Sql Server新手学习入门 http://www.tudou.com/home/_117459337

  9. xcode 条件调试

    添加条件 有时候我们可能会在某个循环中创建断点,但一次又一次地点击 continue 直到我们想要的条件出现,显然是一种非常低效的方式.好在 Xcode 为我们提供了条件断点. 首先在下列代码中插入一 ...

  10. 把华为交换机设置成时钟源服务器(NTP)

    把华为交换机设置成时钟源服务器(NTP),提供给下面客户端Linux服务器使用, 1,先设置交换机的时区,和正确时间 # 假设地理位置在中国北京,设置本地时区名称为BJ. 如果系统默认的UTC是伦敦时 ...