poj3650

题目大意：

Character Encoding
” ” (space) %20
“!” (exclamation point) %21
“$” (dollar sign) %24
“%” (percent sign) %25
“(” (left parenthesis) %28
“)” (right parenthesis) %29
“*” (asterisk) %2a
输入一个字符串，遇到上述左边字符，替换成右边字符

解题思路：

无

代码如下：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main()
{
  char ch[100];
  int i;

  while(gets(ch)&&(ch[0]!='#'))
  {
    for(i=0;i<strlen(ch);i++)
    {
      if(ch[i]==' ')
      {
        printf("%%20");
      }
      else if(ch[i]=='!')
      {
        printf("%%21");
      }
      else if(ch[i]=='$')
      {
        printf("%%24");
      }
      else if(ch[i]=='%')
      {
        printf("%%25");
      }
      else if(ch[i]=='(')
      {
        printf("%%28");
      }
      else if(ch[i]==')')
      {
        printf("%%29");
      }
      else if(ch[i]=='*')
      {
        printf("%%2a");
      }
      else 
      {
        printf("%c",ch[i]);
      }
    }
    printf("\n");
  }
  return 0;
}