zoukankan      html  css  js  c++  java
  • UVA10340 POJ1936 ZOJ1970 All in All【字符串匹配】

    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 33318   Accepted: 13901

    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



    问题链接UVA10340 POJ1936 ZOJ1970 All in All入门练习题,用C语言编写程序。

    题意简述:输入两个字符串s和t,看s是否是t的子串。t中的字符可以任意删除。

    问题分析:顺序匹配字符串即可。

    程序说明:(略)



    AC的C语言程序如下:

    /* 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:
    圣诞节快乐 | 圣诞特效来了!!
    前端特效demo | 值得收藏的6个 HTML5 Canvas 实用案例
    前端特效demo | 一起围观 10 种创意时钟
    即学即用,轻松搞定这些选择器!(下)
    架构师究竟要不要写代码?
    偷懒秘诀之变量篇
    弹幕,是怎样练成的?
    [C++]模板类和模板函数
    [C++]typedef用法
    [面试]CVTE 2019提前批 Windows应用开发一面
  • 原文地址:https://www.cnblogs.com/tigerisland/p/7564510.html
Copyright © 2011-2022 走看看