POJ-2141 Message Decowding-奶牛密码

Message Decowding

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12242 Accepted: 6844

Description

The cows are thrilled because they've just learned about encrypting messages. They think they will be able to use secret messages to plot meetings with cows on other farms.

Cows are not known for their intelligence. Their encryption method is nothing like DES or BlowFish or any of those really good secret coding methods. No, they are using a simple substitution cipher.

The cows have a decryption key and a secret message. Help them decode it. The key looks like this:

        yrwhsoujgcxqbativndfezmlpk

Which means that an 'a' in the secret message really means 'y'; a 'b' in the secret message really means 'r'; a 'c' decrypts to 'w'; and so on. Blanks are not encrypted; they are simply kept in place.

Input text is in upper or lower case, both decrypt using the same decryption key, keeping the appropriate case, of course.

Input

* Line 1: 26 lower case characters representing the decryption key

* Line 2: As many as 80 characters that are the message to be decoded

Output

* Line 1: A single line that is the decoded message. It should have the same length as the second line of input.

Sample Input

eydbkmiqugjxlvtzpnwohracsfKifq oua zarxa suar bti yaagrj fa xtfgrj

Sample Output

Jump the fence when you seeing me coming

Source

USACO 2003 March Orange

/*    *Copyright (c)2015,烟台大学计算机与控制工程学院    *All rights reserved.      *作    者：单昕昕    *完成日期：2015年8月22日    *版 本 号：v1.0        */ #include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    char k[26],c;    int i;    for(i=0; i<26; ++i)        cin>>k[i];    getchar();    while((c=getchar()))//用cin不能接受到空格    {        if(c>='A'&&c<='Z')            cout<<char(k[c-'A']-'a'+'A');        else if(c>='a'&&c<='z')            cout<<k[c-'a'];        else if(c=='\n')            break;        else            cout<<c;    }    return 0;}

getchar（）每次用的时候都觉得好危险。。可能自己还是掌握的不够好