Codeforces 831 B. Keyboard Layouts

本文介绍了一种键盘布局转换的方法,通过给定两种不同的键盘布局映射,实现从一种键盘输入到另一种键盘输出的转换,同时保留大小写和非字母字符不变。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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
Copy
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
output
Copy
HelloVKCup2017
input
Copy
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
output
Copy
7uduGUDUUDUgudu7


思路:给你两个长度为26的字符串,表示两种a-z的映射(也就是键盘改了键),现在给出你用第一种键盘打出的字符,让你求用第二种键盘打出的结果

Ac代码如下:

#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
	char str1[27],str2[27],str3[1001],str4[1001];
	cin>>str1>>str2>>str3;
	int l=strlen(str3);
	int i,j;
	for(i=0;i<l;i++)
	{
		if(str3[i]>=97&&str3[i]<=122)
		{
			for(j=0;j<26;j++)
			{
				if(str3[i]==str1[j])
				  break;
			}
			cout<<str2[j];
		}
		else if(str3[i]>=65&&str3[i]<=90)
		{    
			for(j=0;j<26;j++)
			{
				if(str3[i]+32==str1[j])
				 break;
			}
			printf("%c",str2[j]-32);
		}
		else
		   cout<<str3[i];
	}
	cout<<endl;
	return 0;
} 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值