J - Keyboard Layouts

CodeForces - 831B

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.

Example

Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017

Output
HelloVKCup2017

Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7

Output
7uduGUDUUDUgudu7

题意：有两个键盘，键盘上字母序列不同，已知两种键盘的字母排列方式，和一种键盘的输入内容，求另一种键盘同样键序列输出的字母序列。

解法：按字母顺序存第一个键盘键的位置，按位置存第二个键盘每个位置的字母，然后转换得到第二个键盘输出的序列。

代码：

#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int a[28];
char b[28];
int main()
{
    char x;
    for(int i=1;i<27;i++){
        cin>>x;
        a[x-'a'+1]=i;
    }
    for(int i=1;i<27;i++){
        cin>>b[i];
    }
    char c[1000];
    cin>>c;
    int n=strlen(c);
    for(int i=0;i<n;i++){
        if(c[i]>='a'&&c[i]<='z'){
            cout<<b[a[c[i]-'a'+1]];
        }
        else if(c[i]>='A'&&c[i]<='Z'){
            cout<<char(b[a[c[i]+32-'a'+1]]-32);
        }
        else
            cout<<c[i];
    }
    cout<<endl;
}