1002: Clairewd’s message

1002: Clairewd’s message

Time Limit: 1 Sec  Memory Limit: 128 MB

Description

Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was preparing for sending it to ykwd. They had agreed that each letter of these messages would be transfered to another one according to a conversion table.

    Unfortunately, GFW(someone's name, not what you just think about) has detected their action. He also got their conversion table by some unknown methods before. Clairewd was so clever and vigilant that when she realized that somebody was monitoring their action, she just stopped transmitting messages.

    But GFW knows that Clairewd would always firstly send the ciphertext and then plaintext(Note that they won't overlap each other). But he doesn't know how to separate the text because he has no idea about the whole message. However, he thinks that recovering the shortest possible text is not a hard task for you.

    Now GFW will give you the intercepted text and the conversion table. You should help him work out this problem.

Input

The first line contains only one integer T, which is the number of test cases.

Each test case contains two lines. The first line of each test case is the conversion table S. S[i] is the ith latin letter's cryptographic letter. The second line is the intercepted text which has n letters that you should recover. It is possible that the text is complete.

T<= 100 ;
n<= 100000;

Output

For each test case, output one line contains the shorest possible complete text.

Sample Input

2

abcdefghijklmnopqrstuvwxyz

abcdab

qwertyuiopasdfghjklzxcvbnm

qwertabcde

Sample Output

abcdabcdqwertabcde

HINT

Source

介个题目的意思就是给你一个26个字母的对照表，就是例如qwertyuiopasdfghjklzxcvbnm 表示q对应的a，w对应的b，e对应的c.....类推m对应的z。然后再给你一个字符串，密码包括密文+明文（这个概念我理解了好久-。-）qwert就是密文，然后qwert按对照表出来的明文就是abcde，然鹅，这个字符串可能是残缺的，它并不一定是密码！只有密文是一定全部都在的。所以就要从中间开始找啦。你要做的就是找到密文和明文的分界点，输出密文+明文。具体怎么找 ↓↓↓

#include <stdio.h>
#include <string.h>
int main()
{
    int n;
    scanf("%d",&n);

    while(n--)
    {
        char a[100]={0},b[100101]={0},dic[200]={0};
        scanf("%s%s",a,b);
        int n=strlen(b),m,w=97;
        m=(1+n)/2;   //一开始用了n/2 结果一直错，后来发现有点小问题-0-  n为奇数的时候可能会出现中间字母恰好和第一个和最后一个对应
                     //例一来说 abbabba 输出本应该是abbabbabbabb  但是n/2 就会是abbabb
        int i=0,j;
        for(j=0;j<strlen(a);j++)dic[a[j]]=w++;
        while((m+i)<n)
        {
            if(dic[b[i]]==b[i+m])i++;
            else {m++;i=0;}
        }
            for(i=0;i<m;i++)printf("%c",b[i]);
            for(i=0;i<m;i++)printf("%c",dic[b[i]]);
            printf("\n");
    }
    return 0;
}