The Seven Percent Solution
|
Problem Description
Uniform Resource Identifiers (or URIs) are strings like http://icpc.baylor.edu/icpc/, mailto:foo@bar.org, ftp://127.0.0.1/pub/linux, or even just readme.txt that are used to identify a resource, usually on the Internet or a local computer. Certain characters are reserved within URIs, and if a reserved character is part of an identifier then it must be percent-encoded by replacing it with a percent sign followed by two hexadecimal digits representing the ASCII code of the character. A table of seven reserved characters and their encodings is shown below. Your job is to write a program that can percent-encode a string of characters.
Character Encoding |
|
Input
The input consists of one or more strings, each 1–79 characters
long and on a line by itself, followed by a line containing only "#" that signals the end of the input. The character "#" is used only as an end-of-input marker and will not appear anywhere else in the input. A string may contain spaces, but not at the beginning or end of the string, and there will never be two or more consecutive spaces. |
|
Output
For each input string, replace every occurrence of a reserved
character in the table above by its percent-encoding, exactly as shown, and output the resulting string on a line by itself. Note that the percent-encoding for an asterisk is %2a (with a lowercase "a") rather than %2A (with an uppercase "A"). |
|
Sample Input
Happy Joy Joy! |
|
Sample Output
Happy%20Joy%20Joy%21 |
#include<stdio.h>
void main()
{
char a[1000];
int n;
while(1)
{
gets(a);
//getchar();
if(a[0]=='#')
return;
for(int j=0;a[j]!='\0';j++);
j++; n=j;
for(int i=0;a[i]!='\0';i++)
{
switch(a[i])
{
case ' ' :
for(;j>i;j--)
a[j+2]=a[j];
a[i]='%';
a[i+1]='2';
a[i+2]='0';
i=i+2;
j=n=n+2;
break;
case '!':
for(;j>i;j--)
a[j+2]=a[j];
a[i]='%';
a[i+1]='2';
a[i+2]='1';
i=i+2;
j=n=n+2;
break;
case '$':
for(;j>i;j--)
a[j+2]=a[j];
a[i]='%';
a[i+1]='2';
a[i+2]='4';
i=i+2;
j=n=n+2;
break;
case '%':
for(;j>i;j--)
a[j+2]=a[j];
a[i]='%';
a[i+1]='2';
a[i+2]='5';
i=i+2;
j=n=n+2;
break;
case '(':
for(;j>i;j--)
a[j+2]=a[j];
a[i]='%';
a[i+1]='2';
a[i+2]='8';
i=i+2;
j=n=n+2;
break;
case ')':
for(;j>i;j--)
a[j+2]=a[j];
a[i]='%';
a[i+1]='2';
a[i+2]='9';
i=i+2;
j=n=n+2;
break;
case '*':
for(;j>i;j--)
a[j+2]=a[j];
a[i]='%';
a[i+1]='2';
a[i+2]='a';
i=i+2;
j=n=n+2;
break;
}
}
printf("%s\n",a);
}
}
The Seven Percent Solution的更多相关文章
- HDUOJ-------2719The Seven Percent Solution
The Seven Percent Solution Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Ja ...
- zoj 2932 The Seven Percent Solution
The Seven Percent Solution Time Limit: 2 Seconds Memory Limit: 65536 KB Uniform Resource Identi ...
- POJ 3650:The Seven Percent Solution
The Seven Percent Solution Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7684 Accep ...
- HDU 2719 The Seven Percent Solution
#include <cstdio> #include <cstring> int main() { ]; ]!='#') { ; while (i<strlen(s)) ...
- HDU 2719 The Seven Percent Solution (水题。。。)
题意:把字符串中的一些特殊符号用给定的字符串代替. 析:没的说. 代码如下: #include <iostream> #include <cstdio> #include &l ...
- HDU题解索引
HDU 1000 A + B Problem I/O HDU 1001 Sum Problem 数学 HDU 1002 A + B Problem II 高精度加法 HDU 1003 Maxsu ...
- Enterprise Solution 3.1 企业应用开发框架 .NET ERP/CRM/MIS 开发框架,C/S架构,SQL Server + ORM(LLBL Gen Pro) + Infragistics WinForms
行业:基于数据库的制造行业管理软件,包含ERP.MRP.CRM.MIS.MES等企业管理软件 数据库平台:SQL Server 2005或以上 系统架构:C/S 开发技术 序号 领域 技术 1 数据库 ...
- Enterprise Solution 开源项目资源汇总 Visual Studio Online 源代码托管 企业管理软件开发框架
Enterprise Solution 是一套管理软件开发框架,在这个框架基础上开发出一套企业资源计划系统Enterprise Edition. 现将Enterprise Solution开发过程中遇 ...
- Windows 10 部署Enterprise Solution 5.5
Windows 10正式版发布以后,新操作系统带来了许多的变化.现在新购买的电脑安装的系统应该是Windows 10.与当初用户不习惯Windows 7,购买新电脑后第一个想做的事情就是重装成XP,估 ...
随机推荐
- 10.16JS日记
1.parseint() 2.parsefloat() 这两个单词运行的时候遇到第一个非数字就结束了 3.var a="hello word" a这个变量为字符串,每一个字母为字 ...
- html与css关系
1.html是网页的内容的载体 内容是作者发在页面想让用户浏览的信息,包括文字,图片,视频等 2.css样式是表现 像网页的外衣,比如标题字体的变化,颜色的变化,背景颜色,边框等,所有这些都是用来改变 ...
- mybatis入门--mybatis和hibernate比较
mybatis和hibernate的比较 Mybatis和hibernate不同,它不完全是一个ORM框架,因为MyBatis需要程序员自己编写Sql语句,不过mybatis可以通过XML或注解方式灵 ...
- Gulp应用场景
转自:Gulp教程之:Gulp能做什么,前端装逼为何要用它 我们先说说 平时web开发遇到的一些场景 和 苦恼无奈的情况: JavaScript和CSS的版本问题 我们都知道 JavaScript ...
- Window10系统的安装
关于系统的安装网上有许多的教程,本文的教程并没有什么特别的.只是将自己在安装过程中遇到的问题记录下来,方便以后观看. 1.下载系统镜像 首先从MSDN上下载windows10镜像.在操作系统Windo ...
- C++11并发编程实战 免费书籍
C++11 博客http://www.cnblogs.com/haippy/p/3284540.html 网上推荐的C++多线程基本都是C++ Concurrency in Action 英文版的,中 ...
- unwind
unwind:可以将一个列表展开为一个行的序列1.列表 unwind[1,2,3]as x return x2.创建唯一列表with[1,2,3,3]as coll unwind coll as x ...
- 带token的get和post方法
GET和POST传值失败,多半是传输的字符串和URL的事 public static string ExcuteGetToken(string serviceUrl, string ReqInfo, ...
- 【C#】详解C#委托
目录结构: contents structure [+] 委托语法 泛型委托 委托链 lambda表达式 揭秘委托 类库中的委托 委托和反射 1.委托语法 本文会详细阐述委托的使用,以及实现,想必读者 ...
- Seaborn图形可视化库
一.绘图 1)快速生成图 import numpy as np import matplotlib.pyplot as plt def sinplot(filp=): x = np.linspace( ...