hdu4300 Clairewd’s message

Problem 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.

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

2 abcdefghijklmnopqrstuvwxyz abcdab qwertyuiopasdfghjklzxcvbnm qwertabcde

 

Sample Output

abcdabcd qwertabcde

这道题花了很长时间啊。。。题意也很长时间才看懂，题意是这样的：给你一个26字母表，这些代表的是原来顺序的字母表变成密文后所对应的字母，如样例中的qwertyuiopasdfghjklzxcvbnm，那么原来的'a'就变成了'q'，'b'变成了'w'.然后再给你一个字符串，里面包含了密文和原文，其中顺序一定是先密文，后原文，但原文不一定写全，可能有缺少，问你把密文和明文补充完整输出。

思路：因为密文一定比明文长或者相等，所以(len-1)/2个字符一定是密文，先把它们转化为明文，然后和后面的明文进行匹配，看最后从哪个字符开始后面的所有字符可以和密文匹配，一直到最后，那么这个字符就是明文的开始。然后就可以先输出密文，再输出明文了。

#include<stdio.h>
#include<string.h>
char s1[100006],s2[100006],s[100006],s3[100],s4[100006];
int len1,len2,next[100006];
void nextt()
{
	int i,j;
	i=0;j=-1;
	memset(next,-1,sizeof(next));
	while(i<len2){
		if(j==-1 || s2[i]==s2[j]){
			i++;j++;next[i]=j;
		}
		else j=next[j];
	}
}

int kmp()
{
	int i,j;
	i=0;j=0;
	while(i<len1 && j<len2){
		if(j==-1 || s1[i]==s2[j]){
			i++;j++;
		}
		else j=next[j];
	}
	return j;
}

int main()
{
	int n,m,i,j,T,len,t,k;
	scanf("%d",&T);
	while(T--){
		memset(s1,0,sizeof(s1));memset(s2,0,sizeof(s2));
		scanf("%s%s",s3,s4);
		len=strlen(s4);
		for(i=0;i<len;i++){
			for(k=0;k<26;k++){
				if(s3[k]==s4[i]){
					s[i]='a'+k;break;
				}
			}
			if(i<=(len-1)/2)s2[i]=s[i];
		}
		for(i=(len-1)/2+1;i<len;i++){
			s1[i-(len-1)/2-1]=s4[i];
		}
		//printf("%s %s\n",s1,s2);
		nextt();
		len1=strlen(s1);len2=strlen(s2);
		t=len-kmp();
		//printf("%d\n",kmp());
		for(i=0;i<t;i++)printf("%c",s4[i]);
		
		for(i=0;i<t;i++){
			for(k=0;k<26;k++){
				if(s3[k]==s4[i]){
					printf("%c",'a'+k);
				}
			}
		}
		printf("\n");
	}
	return 0;
}