zoukankan      html  css  js  c++  java
  • NUC1657 All in All【字符串匹配】

    All in All

    时间限制: 1000ms 内存限制: 30000KB

    通过次数: 1总提交次数: 1

    问题描述
    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.
    输入描述
    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.
    输出描述
    For each test case output "Yes", if s is a subsequence of t,otherwise output "No".
    样例输入
    sequence subsequence
    person compression
    VERDI vivaVittorioEmanueleReDiItalia
    caseDoesMatter CaseDoesMatter
    
    样例输出
    Yes
    No
    Yes
    No
    
    来源
    Ulm Local 2002


    问题分析:(略)

    这个问题和《UVA10340 POJ1936 ZOJ1970 All in All【字符串匹配】》是同一个问题,代码拿过来用就AC了。

    程序说明:参见参考链接。

    参考链接:UVA10340 POJ1936 ZOJ1970 All in All【字符串匹配】

    题记:程序做多了,不定哪天遇见似曾相识的。



    AC的C++程序如下:

    /* UVA10340 POJ1936 ZOJ1970 All in All */
    
    #include <stdio.h>
    #include <string.h>
    
    #define MAXN 110000
    
    char s[MAXN], t[MAXN];
    
    int delstrcmp(char *s, char *t)
    {
        int i, j, slen, tlen;
    
        slen = strlen(s);
        tlen = strlen(t);
    
        for(i=0, j=0; i<slen && j<tlen;) {
            if(s[i] == t[j]) {
                i++;
                j++;
            } else
                j++;
        }
    
        return i == slen;
    }
    
    int main(void)
    {
        while(scanf("%s%s", s, t) != EOF)
            printf("%s
    ", delstrcmp(s, t) ? "Yes" : "No");
    
        return 0;
    }




  • 相关阅读:
    P、NP、NPC、NPH问题介绍
    过河卒 bfs搜索
    对迪杰斯特拉算法的理解
    第七周
    周作业
    月考一
    第四周
    第三周
    第二周作业
    46期第一次作业
  • 原文地址:https://www.cnblogs.com/tigerisland/p/7563677.html
Copyright © 2011-2022 走看看