zoukankan      html  css  js  c++  java
  • POJ 1936 All in All

    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
    题目要求就是在s2中找字串s1!

    #include <iostream>
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    using namespace std;
    char a[100005],b[100005];
    int main()
    {
        while(~scanf("%s",a)){
            scanf("%s",b);
            int a_length = strlen(a);
            int b_length = strlen(b);
            int a1,b1=0;
            int i;
            for(i=0;i<a_length;i++){
                    bool flag=false;
                for(int j=b1;j<b_length;j++){
                    if(a[i]==b[j]){
                        b1=j+1;
                        flag=true;
                        break;
                    }
                }
                if(!flag)
                    break;
            }
            if(i==a_length)
                printf("Yes
    ");
            else
                printf("No
    ");
        }
        return 0;
    }
  • 相关阅读:
    rdb 和 aof
    nginx 遇见问题与解决问题
    linux 每天一个命令
    Consul 集群搭建
    Consul 安装的与启动
    hession RMI 远程调用
    3、使用Lucene实现千度搜索
    1、什么是Lucene,Lucene能干什么
    Tengine笔记2:通过IP、域名、端口实现虚拟主机
    Tengine笔记3:Nginx的反向代理和健康状态检查
  • 原文地址:https://www.cnblogs.com/webmen/p/5739603.html
Copyright © 2011-2022 走看看