this problem  is from PAT, which website is http://pat.zju.edu.cn/contests/pat-a-practise/1050.

firstly i think i can use double circulation to solve it ,however the result of two examples is

proofed to be running time out. So as the problem said, it is not that simple to make it fast.

then i search the problem , there are many people using the search table. i think it is a very

good idea. so i changed the algorithm.

#include<iostream>
#include<string>
#include<vector>
using namespace std; int MAX=; int main()
{
string s1,s2;
getline(cin,s1);
getline(cin,s2);
vector<bool> isExisted(MAX,false);
for (int i = ; i < s2.size(); i++)
{
isExisted[s2[i]] = true;
}
for (int i = ; i < s1.size(); i++)
{
if (!isExisted[s1[i]]) cout<<s1[i];
}
cout<<endl;
}

so this time the Time complexity is O(n) ,which is much faster than the previous one.

1050. String Subtraction (20)的更多相关文章

  1. PAT 解题报告 1050. String Subtraction (20)

    1050. String Subtraction (20) Given two strings S1 and S2, S = S1 - S2 is defined to be the remainin ...

  2. PAT 甲级 1050 String Subtraction (20 分) (简单送分,getline(cin,s)的使用)

    1050 String Subtraction (20 分)   Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be t ...

  3. PAT (Advanced Level) 1050. String Subtraction (20)

    简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> ...

  4. PAT甲题题解-1050. String Subtraction (20)-水题

    #include <iostream> #include <cstdio> #include <string.h> #include <algorithm&g ...

  5. PAT Advanced 1050 String Subtraction (20 分)

    Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking ...

  6. PAT Advanced 1050 String Subtraction (20) [Hash散列]

    题目 Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string afer taking all th ...

  7. 1050 String Subtraction (20分)

    Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking ...

  8. PAT练习--1050 String Subtraction (20 分)

    题⽬⼤意:给出两个字符串,在第⼀个字符串中删除第⼆个字符串中出现过的所有字符并输出. 这道题的思路:将哈希表里关于字符串s2的所有字符都置为true,再对s1的每个字符进行判断,若Hash[s1[i] ...

  9. 【PAT甲级】1050 String Subtraction (20 分)

    题意: 输入两个串,长度小于10000,输出第一个串去掉第二个串含有的字符的余串. trick: ascii码为0的是NULL,减去'0','a','A',均会导致可能减成负数. AAAAAccept ...

随机推荐

  1. Linux(CentOs)下安装Phantomjs + Casperjs

    Linux(CentOs)下安装Phantomjs + Casperjs 是参照cnMiss's Blog http://ju.outofmemory.cn/entry/70691的博客进行安装的 1 ...

  2. linux常用命令之--用户与用户组管理命令

    linux的用户与用户组管理命令 1.用户和群组 groupadd:用于添加新的组群 其命令格式如下: groupadd [-option] 群组名 常用参数: -g GID:指定创建群组的GID(G ...

  3. 全栈工程师眼中的HTTP

    HTTP,是Web工程师每天打交道最多的一个基本协议.很多工作流程.性能优化都围绕HTTP协议来进行,但是我们对HTTP的理解是否全面呢?如果前端工程师和后台工程师坐在一起玩捉鬼游戏,他们对HTTP的 ...

  4. MVC3.0在各个版本IIS中的部署

    概述: 最近在做一个MVC 3的项目,在部署服务器时破费了一番功夫,特将过程整理下来,希望可以帮到大家! 本文主要介绍在IIS5.1.IIS6.0.IIS7.5中安装配置MVC 3的具体办法! 正文: ...

  5. 三道JS试题(遍历、创建对象、URL解析)

    最近在网上看到了三道不错的JS试题,还是很基础(一直认为学好前端基本功很重要...),现在记录如下: 原帖地址:http://www.w3cfuns.com/forum.php?mod=viewthr ...

  6. lighttpd为什么要accept多次呢

    在lighttpd网络模型里面我们可以看到以下代码 /* accept()s at most 100 connections directly * * we jump out after 100 to ...

  7. [Hive - LanguageManual] Select base use

    Select Syntax WHERE Clause ALL and DISTINCT Clauses Partition Based Queries HAVING Clause LIMIT Clau ...

  8. 【恒天云技术分享系列10】OpenStack块存储技术

    原文:http://www.hengtianyun.com/download-show-id-101.html 块存储,简单来说就是提供了块设备存储的接口.用户需要把块存储卷附加到虚拟机(或者裸机)上 ...

  9. shell下解码url

    http://aaronw.me/static/779.html 封装了一下,有需要的拿走 function url_decode() { local url=$ echo $url | awk 'B ...

  10. UVA 10054 The Necklace(欧拉回路,打印路径)

    题目链接: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...