All in All-poj 1936 (substring)

String subsequence lookup problem, set two pointers, one point to the subsequence, the other point to the sequence to be looked up, find a string to judge once.

1#include <iostream>2#include <string.h>3 using namespacestd;4 Chars[100002];5 Chart[100002];6 intMain () {7 8      while(cin>>s>>t) {9         intlength1=strlen (s);Ten         intLength2=strlen (t); One         intI=0, j=0; A          for(i=0, j=0;i<length1&&j<length2;) { -             if(s[i]==T[j]) { -i++; theJ + +; -}Else{ -J + +; -             } +         } -         if(i==length1) +cout<<"Yes"<<Endl; A         Else atcout<<"No"<<Endl; -     } -     return 0; -}

Report:

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 29592		Accepted: 12269

Description

You are devised a new encryption technique which encodes a message by inserting between its characters randomly generated Strings in a clever. Because of pending patent issues we won't discuss in detail how the strings is generated and inserted into the Origina L message. To validate your method, however, it's necessary to write a program that checks if the message was really encoded in the F Inal string.

Given-strings s and T, you had to decide whether s are a subsequence of T, i.e. if you can remove characters from T su CH, the concatenation of the remaining characters is S.

Input

The input contains several testcases. Each are specified by and strings s, T of alphanumeric ASCII characters separated by whitespace. The length of S and T would no more than 100000.

Output

For each test, the output "Yes", if S is a subsequence of t,otherwise output "No".

Sample Input

Sequence Subsequenceperson Compressionverdi vivavittorioemanuelerediitaliacasedoesmatter CaseDoesMatter

Sample Output

Yesnoyesno

All in All-poj 1936 (substring)