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 [cpp] /* 題意為：給定一個翻譯表，即第i個字母用哪個字母表示 再給一個串，裡面前面為密文，後面為明文，密文一定是完整的，但明文不完整或可能沒有 求這個完整的前面密文後面明文的串 把給定的串全部按表翻譯為密文，這樣例如樣例 qwertabcde 翻譯之後為： jvtkzqwert 可發現翻譯後的後半部分與原傳相同，把原串作為T串，翻譯後的為S串，求B 題目要求求最短長度， 從B的一半開始， 當找到第一個滿足 B[i]== len- i條件的，i即為第一個明文的位置 */ #include <iostream> #include <cstdio> #include <string.h> using namespace std; int next[100005]; char str[27]; char s1[100005],s2[100005]; void getnext(char *t) { int i = 0,j = -1; next[0] = -1; while(t[i]) { if(j == -1 || t[i] == t[j]) { i++; j++; next[i] = j; } else j = next[j]; } } int kmp(char *s,char *t) { int i = 0,j = 0; int slen =strlen(s),tlen = strlen(t); getnext(t); while(i<slen && j<tlen) { if(j == -1 || s[i] == t[j]) { i++; j++; if(i == slen) return j; } else j = next[j]; } return 0; } int main() { int t; cin >> t; while(t--) { scanf("%s",str); scanf("%s",s1); int len = strlen(s1); strcpy(s2,s1+(len+1)/2); printf("%s",s1); for(int i = 0; s1[i]; i++) { for(int j = 0; j<26; j++) { if(s1[i] == str[j]) { s1[i] = 'a'+j; break; } } } int flag = kmp(s2,s1); for(int i = flag; i<len-flag; i++) { printf("%c",s1[i]); } cout << endl; } return 0; }
