/*
* 解题思路:
* 这是道非常坑爹的题目、开始一直Runtime error、不知道为什么、后来发现是再输入的时候
* My dog ate my homework.
* Can you believe my dog died after eating my canary... AND MY HOMEWORK?
* This excuse is so good that it contain 0 keywords.
* 最后一行系统没有换行、不能单单通过判断 ‘\n’ 来停止输入( 这仅对前两行有效 ),还应增加EOF来停止最后一行输入、
* 否则系统一直认为你没有输入完全、进入死循环,导致Runtime error的错误!!
*/
#include <stdio.h>
#include <string.h>
#define A 30
#define B 100
int m,n;
char s[ A ][ B ],str[ A ][ B ],ss[B ];
char convert( char x )
{
if( x<=‘z‘ && x>=‘a‘ ) return x-‘a‘+‘A‘;
if( x<=‘Z‘ && x>=‘A‘ ) return x-‘A‘+‘a‘;
}
int search( )
{
int i,j;
int flag,len;
len = strlen( ss );
for( i=0;i<m;i++ )
{
j = flag = 0;
while( j<len )
{
if( s[ i ][ j ] == ss[ j ] || s[ i ][ j ] == convert( ss[ j ] ) ) j++;
else
{
flag = 1;
break;
}
}
if( !flag && s[ i ][ j ] == ‘\0‘) break;
}
if( flag ) return 0;
else return 1;
}
int main( )
{
int i,j;
int p,q,r,x,sum,total,maxSum;
int len[ A ],maxPos[ A ];
char c;
total = 1;
while( ~scanf("%d%d",&m,&n) )
{
for( i=0;i<m;i++ )
{
getchar();
scanf("%s",&s[ i ] );
}
getchar( );
maxSum = 0;
for( i=0;i<n;i++ )
{
p = 0;
while(( c = getchar( )) !=‘\n‘ && c != EOF )
str[ i ][ p++ ] = c;
len[ i ] = p;
}
maxSum = q = 0;
for( i=0;i<n;i++ )
{
p = sum = r = 0;
while( p<len[ i ] )
{
if( !((str[ i ][ p ]<=‘z‘ && str[ i ][ p ] >=‘a‘ ) || (str[ i ][ p ] <=‘Z‘ && str[ i ][ p ]>=‘A‘ ) ))
{
p++;
continue;
}
while( (str[ i ][ p ]<=‘z‘ && str[ i ][ p ] >=‘a‘ ) || (str[ i ][ p ] <=‘Z‘ && str[ i ][ p ]>=‘A‘ ) )
ss[ r++ ] = str[ i ][ p++ ];
x = search( );
if( x ) sum++;
p++;
r = 0;
memset(ss,‘\0‘,sizeof( ss ) );
}
if( sum > maxSum )
{
q = 0;
maxSum = sum;
maxPos[ q ] = i;
}
else if( sum == maxSum )
maxPos[ ++q ] = i;
}
printf("Excuse Set #%d",total++ );
for( i=0;i<=q;i++ )
for( puts(""),j=0;j<len[ maxPos[ i ] ];j++ )
printf("%c",str[ maxPos[ i ] ][ j ]);
puts("");
puts("");
}
return 0;
}