POJ 3650 The Seven Percent Solution

来源:互联网 发布:工业大数据 智能制造 编辑:程序博客网 时间:2024/05/24 01:23
The Seven Percent Solution

Description

Uniform Resource Identifiers (or URIs) are strings like http://icpc.baylor.edu/icpc/mailto:foo@bar.orgftp://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.

CharacterEncoding" " (space)%20"!" (exclamation point)%21"$" (dollar sign)%24"%" (percent sign)%25"(" (left parenthesis)%28")" (right parenthesis)%29"*" (asterisk)%2a

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!http://icpc.baylor.edu/icpc/plain_vanilla(**)?the 7% solution#

Sample Output

Happy%20Joy%20Joy%21http://icpc.baylor.edu/icpc/plain_vanilla%28%2a%2a%29?the%207%25%20solution
这道题也是一道水题,直接遍历数组,遇到要求变化的字符就输出相应的字符。其他字符就原样输出。
AC代码:
#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;int main(){char a[1111];while(gets(a)){if(a[0]=='#')break;for(int i=0;a[i];i++){if(a[i]==' ')cout<<"%20";else if(a[i]=='!')cout<<"%21";else if(a[i]=='$')cout<<"%24";else if(a[i]=='%')cout<<"%25";else if(a[i]=='(')cout<<"%28";else if(a[i]==')')cout<<"%29";else if(a[i]=='*')cout<<"%2a";elsecout<<a[i];}cout<<endl;}return 0;}


 
原创粉丝点击