Clairewd’s message

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. Sii 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. 

Hint

Range of test data: 
T<= 100 ; 
n<= 100000; 

Output

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

Sample Input

2abcdefghijklmnopqrstuvwxyzabcdabqwertyuiopasdfghjklzxcvbnmqwertabcde

Sample Output

abcdabcd
qwertabcde
题目意思：题目看了好久，很难理解，看的莫名奇妙
          题目就是让你通过一个密文转换表，将给出得一个串“密文+明文”（密文是完整的，但明文
          未知）补全。
解题思路：理解了题目意思就好写了，一道kmp的题，其实密文和明文构成了一种映射的关系，在转换表
          中是一一对应的，于是可以先将整个串转换成“明文”，然后与原串比较，重合的部分的长度
          就是密文的长度，然后输出就行；
#include<iostream>#include<cstring>using namespace std;int const maxn=100005;char str[maxn],str1[maxn];int next1[maxn];int len,len1,mark,Max;void sign(char *str1,int next1[]){int i,j;next1[0]=-1;i=-1,j=0;int len1=strlen(str1);while(j<len1-1){    if(i==-1||str1[i]==str1[j])    {    i++;    j++;    next1[j]=i;}else{i=next1[i];}}}int kmp(char *str,char *str1){int m=0,n=0;sign(str1,next1);int len=strlen(str);int len1=strlen(str1);if(len%2==0){m=len/2;}else{m=len/2+1;}while((m<len)&&(n<len1)){if(n==-1||str[m]==str1[n]){m++;n++;}else{ n=next1[n];}}return n;}int main(){char s,ss,strm[maxn],strm1[maxn];int T;cin>>T;while(T--){scanf("%s",strm);    //转换表scanf("%s",str);    //密文+明文 for(int i=0;i<26;i++){s=strm[i];strm1[s]=i;         //转换表中每一字母与26个字母一一对应 }int lenn=strlen(str);for(int i=0;i<lenn;i++){ss=strm1[str[i]];   //密文的每一个字母在26个字母中的位置         str1[i]='a'+ss;      //ASCII转换为明文 }int num=kmp(str,str1);if(2*num==lenn){cout<<str<<endl;}else{cout<<str;for(int i=num;i<lenn-num;i++){cout<<str1[i];}cout<<endl;}}return 0;   }