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  • UVaOJ 10340

    All in All

    Input: standard input

    Output: standard output

    Time Limit: 2 seconds

    Memory Limit: 32 MB

    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
    
    
    本题需要注意:
    数组s和t要开的足够大,而且要放在main()外面;
    其他地方还是较为简单。
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    
    char s[102400],t[102400];
    
    int main()
    {
    
        int len_s,len_t;
        int i,j,count,temp;
        while(scanf("%s",s)!=EOF)
        {
            scanf("%s",t);
            len_s = strlen(s);
            len_t = strlen(t);
            count=0;
            temp = 0;
            for(i=0; i<len_s; i++)
            {
                for(j=temp; j<len_t; j++)
                {
                    if(s[i] == t[j])
                    {
                        temp = j+1;
                        count++;
                        break;
                    }
                }
            }
            if(count == len_s)
                printf("Yes
    ");
            else
                printf("No
    ");
        }
    
        return 0;
    }
    


    
       
    
    
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  • 原文地址:https://www.cnblogs.com/kunsoft/p/5312809.html
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