UVa 10340：All in All

連結：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=1281

原題：

You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input Specification

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace. Input is terminated by EOF.

Output Specification

For each test case output, if s is a subsequence of t.

Sample Input

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia

caseDoesMatter CaseDoesMatter

Sample Output

Yes

No

Yes

No

分析與總結：

更多精彩內容：http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

簡單題，直接判斷一個字串是否是另一個字串的子串。

代碼：

/*  *  UVa: 10340 - All in All  *  Time: 0.008s  *  Author: D_Double  */#include<iostream>  #include<cstdio>  #include<cstring>  using namespace std;  char str1[1000000], str2[1000000];       bool judge(){      int pos=0, i;      for(i=0; i<strlen(str1); ++i){          bool flag=false;          while(pos < strlen(str2)){              if(str2[pos]==str1[i])                  break;              ++pos;          }          if(pos==strlen(str2)) return false;          ++pos;      }      if(i==strlen(str1)) return true;      return false;  }       int main(){      while(scanf("%s %s",str1,str2)!=EOF){          if(judge()) printf("Yes\n");          else printf("No\n");      }      return 0;  }

作者：csdn部落格 shuangde800