acm.jlu.edu.cn-2005-The Dir Command

The Dir Command
We all know the dos command "dir" . It can list the files in the computer . When we use the 
file name with "?" , "*" as the parameter , it can list the files you need . 

                
 Input Specification       
         There are several lines (at most 100) each contain a file name and terminated by a line 
containing '*' . Then a line contains a integer N and followed N lines . Each line contain
a command "dir" with parameter.             
 Output Specification       
         For each "dir" print the result , each file name in a line in lower case . If nothing found
print "no such files" in a single line . Print a blank line after each result . Note that
the command "dir" will treat the capital letter and small letter as the same .             
 Sample Input       
     
autoexec.bat
boot.ini
bootfont.bin
config.sys
io.sys
msdos.sys
pagefile.sys
*
4
dir autoexec.bat
dir boot*.*
dir ??.sys
dir *.*
      
     
 Sample Output       
     
autoexec.bat

boot.ini
bootfont.bin

io.sys

autoexec.bat
boot.ini
bootfont.bin
config.sys
io.sys
msdos.sys
pagefile.sys

 



开始打算只用C的　后来发现对输入的了解不够　没办法了


 
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

void main()
{
	char files[100][20];
	int idx=0;
	int idxf=0;
	char ch;
	scanf("%c",&ch);
	while(ch!='*')
	{
		if(ch>='A'&&ch<='Z')
			ch='a'+ch-'A';
		files[idxf][idx++]=ch;
		if(ch=='/n')
		{
			files[idxf][idx]='/0';
			idx=0;
			idxf++;
		}
		ch=getchar();
	}
	int n;
	scanf("%d/n",&n);
	for(int i=0;i<n;i++)
	{
		char com[20];//输入的命令保存在com
		cin>>com;//滤掉dir
		cin>>com;
		for(int j=0;j<strlen(com);j++)
			if(com[j]>='A'&&com[j]<='Z')
				com[j]='a'+com[j]-'A';
		bool found=false;
		for(int j=0;j<idxf;j++)
		{
			int p=0,q=0;//p指向当前一个文件名中字字符，q指向命令中的字符
L1:			if(files[j][p]==com[q])//字符相等，pass
			{
				p++;q++;
				goto L1;
			}
			if(files[j][p]=='/n'&&strlen(com)==q)//两个字符串同时到头了，ok
			{
				printf("%s",files[j]);
				found=true;
				continue;
			}
			if(com[q]=='*')//'*'的匹配
			{
				if(com[q+1]=='/0')//如果*是命令的最后一个字符,ok
				{
					cout<<files[j];
					found=true;
					continue;
				}
				while(p<strlen(files[j])&&files[j][p]!=com[q+1])//匹配多个字符，直到有相同字符
					p++;
				q++;
				goto L1;
			}
			if(files[j][p]!='/n'&&com[q]=='?')//匹配?，要保证文件名没到结束
			{
				p++;q++;
				goto L1;
			}
		}
		if(!found)
			printf("no such files/n");
		cout<<endl;
	}
}