PAT_A1050#String Subtraction
Source:
Description:
Given two strings S1 and S2, S=S1−S2 is defined to be the remaining string after taking all the characters in S2 from S1. Your task is simply to calculate S1−S2 for any given strings. However, it might not be that simple to do it fast.
Input Specification:
Each input file contains one test case. Each case consists of two lines which gives S1 and S2, respectively. The string lengths of both strings are no more than 1. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.
Output Specification:
For each test case, print S1−S2 in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.
Keys:
Code:
/*
Data: 2019-07-21 19:47:24
Problem: PAT_A1050#String Subtraction
AC: 08:34 题目大意:
给定字符串S和T,把T中的字符从S中删去
*/ #include<cstdio>
#include<string>
#include<iostream>
using namespace std;
char mp[]={}; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif string s,t;
getline(cin,s);
getline(cin,t);
for(int i=; i<t.size(); i++)
mp[t[i]]=;
for(int i=; i<s.size(); i++)
if(mp[s[i]]==)
printf("%c", s[i]); return ;
}
Given two strings S1 and S2, S=S1−S2 is defined to be the remaining string after taking all the characters in S2 from S1. Your task is simply to calculate S1−S2 for any given strings. However, it might not be that simple to do it fast.
Input Specification:
Each input file contains one test case. Each case consists of two lines which gives S1 and S2, respectively. The string lengths of both strings are no more than 1. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.
Output Specification:
For each test case, print S1−S2 in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.
PAT_A1050#String Subtraction的更多相关文章
- 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 ...
- PAT 1050 String Subtraction
1050 String Subtraction (20 分) Given two strings S1 and S2, S=S1−S2 is defined to be t ...
- 1050 String Subtraction (20 分)
1050 String Subtraction (20 分) Given two strings S1 and S2, S=S1−S2 is defined to be the ...
- pat1050. String Subtraction (20)
1050. String Subtraction (20) 时间限制 10 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Giv ...
- pat 1050 String Subtraction(20 分)
1050 String Subtraction(20 分) Given two strings S1 and S2, S=S1−S2 is defined to be the ...
- 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 S1 and S2, S=S1−S2 is defined to be t ...
- PAT甲级——1050 String Subtraction
1050 String Subtraction Given two strings S1 and S2, S=S1−S2 is defined to be the remain ...
- 1050. String Subtraction (20)
this problem is from PAT, which website is http://pat.zju.edu.cn/contests/pat-a-practise/1050. firs ...
- String Subtraction
Given two strings S1 and S2, S = S1 - S2 is defined to be the remaining string after taking all the ...
随机推荐
- 转:Linux设备树(Device Tree)机制
目录 1. 设备树(Device Tree)基本概念及作用 2. 设备树的组成和使用 2.1. DTS和DTSI 2.2. DTC 2.3. DTB 2.4. Bootloader 3. 设备树中d ...
- java并发编程笔记(一)——并发编程简介
java并发编程笔记(一)--简介 线程不安全的类示例 public class CountExample1 { // 请求总数 public static int clientTotal = 500 ...
- 61、Queueable接口
public with sharing class QueueableSample implements Queueable{ private List<String> Name{get; ...
- datetime timestamp使用
#coding=utf-8 import time import datetime def yes_time(): #获取当前时间 now_time = datetime.datetime.now() ...
- vue 项目配置 element 出错
解决方法: 把main.js文件中的 import 'element-ui/lib/theme-default/index.css' 改为 import 'element-ui/lib/theme-c ...
- 深入理解finally关键字,Finally到底是在return前面执行还是在return后面执行
一:2种finally不会执行的情况 a.在try语句之前就return了 b.try语句中有System.exit();语句 二:finally语句在return执行之后,return返回之前执行 ...
- PAT甲级——A1130 Infix Expression【25】
Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with pa ...
- 20140922 tcpip3次握手 分段 分页 spooling 位示图
tcpip3次握手 http://www.cnblogs.com/CBDoctor/archive/2012/10/17/2727073.html 操作系统:http://blog.csdn.ne ...
- zabbix监控linux内存
通过free -m查看当前内存 可用内存:Available memory=free+buffers+cached,即31068=759+66+30243 已用内存:Used memory=used- ...
- bootstrap学习(五)代码
内联代码: <code>public static</code>void main 用户输入: to edit setting,press <kbd><kbd ...