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  • OpenJudge/Poj 1936 All in All

    1.链接地址:

    http://poj.org/problem?id=1936

    http://bailian.openjudge.cn/practice/1936

    2.题目:

    All in All
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 26651   Accepted: 10862

    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
    

    Source

    3.思路:

    4.代码:

     1 #include "stdio.h"
     2 //#include "stdlib.h"
     3 #include "string.h"
     4 #define NUM 100002
     5 char s[NUM],t[NUM];
     6 int main()
     7 {
     8     int i,j;
     9     int len_s,len_t;
    10     while(scanf("%s%s",s,t) != EOF)
    11     {
    12         len_s=strlen(s);
    13         len_t=strlen(t);
    14         i=0;j=0;
    15         while(i<len_s && j<len_t)
    16         {
    17            if(s[i]==t[j]) i++;
    18            j++;
    19         }
    20         //printf("%s%s
    ",s,t);
    21         if(i>=len_s) printf("Yes
    ");
    22         else printf("No
    ");
    23     }
    24     //system("pause");
    25     return 0;
    26 }
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  • 原文地址:https://www.cnblogs.com/mobileliker/p/3576774.html
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