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  • All in All 找子串 水题

    Problem Description
    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
    The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace.The length of s and t will no more than 100000.
     
    Output
    For each test case output "Yes", if s is a subsequence of t,otherwise output "No".
     
    Sample Input
    sequence subsequence person compression VERDI vivaVittorioEmanueleReDiItalia caseDoesMatter CaseDoesMatter
     
    Sample Output
    Yes No Yes No
    ***************************************************************************************************************************
     1 #include<iostream>
     2 #include<string>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<cstdio>
     6 #include<algorithm>
     7 using namespace std;
     8 char str1[100001],str2[100001];
     9 int i,j,k;
    10 int main()
    11 {
    12     while(scanf("%s %s",&str1,&str2)!=EOF)
    13     {
    14         int len1=strlen(str1);
    15         int len2=strlen(str2);
    16         if(len1>len2)
    17         {
    18             puts("No");
    19         }
    20         else
    21            if(len1==len2)
    22             {
    23                if(strcmp(str1,str2)==0)
    24                  puts("Yes");
    25                else
    26                  puts("No");
    27             }
    28             else
    29             {
    30                 j=0;
    31                 bool gs=0;
    32                 for(i=0;str2[i];i++)
    33                 {
    34                     if(str2[i]==str1[j])
    35                        j++;
    36                     if(j==len1)
    37                     {
    38                         gs=1;
    39                         break;
    40                     }
    41                 }
    42                 if(gs==1)
    43                   puts("Yes");
    44                 else
    45                   puts("No");
    46             }
    47     }
    48     return 0;
    49 }
    View Code
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  • 原文地址:https://www.cnblogs.com/sdau--codeants/p/3395442.html
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