POJ 3650:the Seven Percent solution__pku

The Seven Percent Solution

Time Limit: 1000MS		Memory Limit: 65536K
Total submissions: 7684		accepted: 5159

Description

Uniform Resource Identifiers (or URIs) are strings ,  ftp://127.0.0.1/pub/linux, or even just readme.txt that are used to identify a resource, usually on the I Nternet or a local computer. Certain characters are reserved within URIs, and if a reserved character is part of identifier then it must Cent-encoded by replacing it with a percent sign followed by two hexadecimal digits representing the ASCII code of TH e 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

Input

The input consists of one or more strings, each 1–79 characters long and in a line by itself, followed by a line containin G only "#" which signals the end of the input. The character "#" is used only as a end-of-input marker and won't appear anywhere else in the input. A string may contain spaces, but is not in the beginning or end of the string, and there'll never be two or more consecutiv E spaces.

Output

For each input string, replace every occurrence of a reserved character in the table above from its percent-encoding, exactl Y as shown, and output the resulting string on a line by itself. Note This percent-encoding for a asterisk is%2a (with a lowercase "a") rather than (with a%2a "a").

Sample Input

Happy Joy joy!
http://icpc.baylor.edu/icpc/
Plain_vanilla
(* *)
?
The 7% Solution
#

Sample Output

happy%20joy%20joy%21
http://icpc.baylor.edu/icpc/
plain_vanilla
%28%2a%2a%29
?
The%207%25%20solution

Source mid-central USA 2007 You left, my world is left with rain ...

#include <stdio.h>
#include <string.h>
int main ()
{
    char str[80];
    while (gets (str) && strcmp (str, "#")!= 0)
    {
        int i, len = strlen (str);
        for (i = 0; i < len; i++)
        {
            if (str[i] = = ')
                printf ("%%20");
            else if (str[i] = = '! ')
                printf ("%%21");
            else if (str[i] = = ' $ ')
                printf ("%%24");
            else if (str[i] = = '% ')
                printf ("%%25");
            else if (str[i] = = ' (')
                printf ("%%28");
            else if (str[i] = = ")
                printf ("%%29 ");
            else if (str[i] = = ' * ')
                printf ("%%2a");
            else printf ("%c", Str[i]);
        printf ("\ n");
    }
    return 0;
}