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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 " " (space) %20 "!" (exclamation point) %21 "$" (dollar sign) %24 "%" (percent sign) %25 "(" (left parenthesis) %28 ")" (right parenthesis) %29 "*" (asterisk) %2a |
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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.
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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").
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Sample Input
Happy Joy Joy!
http://icpc.baylor.edu/icpc/
plain_vanilla
(**)
?
the 7% solution
#
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Sample Output
Happy%20Joy%20Joy%21
http://icpc.baylor.edu/icpc/
plain_vanilla
%28%2a%2a%29
?
the%207%25%20solution
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本题核心：C++的字符串输入、输出等操作问题
1.基本输入
char buf[50];
cin>>buf;
这种方法输入会忽略最初的空白字符（换行符、空格符、制表符），而且会在下一个空格字符或换行符结束。
2.getline() 不定长输入字符串
string buf; //在C++中则把字符串封装成了一种数据类型string，可以直接声明变量并进行赋值等字符串操作。
getline(cin,buf);
对于getline()，读入整行数据，使用回车键的换行符来确定输入结尾。
3.getline()定长输入字符串
char buf[255];
cin.getline(buf,255);
读入整行数据，使用回车键的换行符来确定输入结尾，不保存换行符，在存储字符串时，用空值字符来替换换行符。
比如：
cin.getline(buf,5)
输入：abcde
输出：abcd （由于buf长度为5，输出时，最后一个为空值字符结尾，所以无法打印e）
4.string类和c字符串之间的转换。
可以将C字符串存储在string类型的变量中。
char a[]="Hello";
string b;
b=a;
但是string对象不能自动的转换为C字符串，需要进行显式的类型转换，需要用到string类的成员函数c_str()
例如：strcpy(a,b,c_str());
5.字符串到数字的转换
atoi函数获取一个C字符串参数，返回对应的int值。如果参数不与一个int值对应，atoi就会返回0.atoi函数在文件为cstdlib的库中。如果数字太大，不能转换为int类型的值，可以使用atoi将字符串转换为long类型的值
atoi("1234");//返回整数1234
atoi("#123");//返回0
6.string类的大小
string buf;
auto len ＝ buf.size();
len的数据类型不确定是int型还是无符号类型，所以使用auto。（还待深究....）
本题代码：
#include <iostream> #include <string> using namespace std; int main() { string str; while (getline(cin,str) && str !="#") { auto len=str.size(); for (int i=0; i<len; i++) { switch (str[i]) { case ‘ ‘:cout<<"%20";break; case ‘!‘:cout<<"%21";break; case ‘$‘:cout<<"%24";break; case ‘%‘:cout<<"%25";break; case ‘(‘:cout<<"%28";break; case ‘)‘:cout<<"%29";break; case ‘*‘:cout<<"%2a";break; default: cout<<str[i];break; } } cout<<"\n"; } return 0; }
题目1.2.4 The Seven Percent Solution（C++）
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原文地址：http://www.cnblogs.com/weekend/p/4954046.html
