B. Petya and Exam
time limit per test
memory limit per test
input
output
It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy to Petya, but he thinks he lacks time to finish them all, so he asks you to help with one..
?" and "*").
It is known that character "*" occurs no more than once
n
the special pattern characters differ from their usual meaning.
?" with one good lowercase
English letter, and the character "*" (if there is one) with any, including empty, string of bad
The good letters are given to Petya. All the others are bad.
Input
1 to 26
s of lowercase English letters, characters "?"
and "*" (1 ≤ |s| ≤ 105).
It is guaranteed that character "*" occurs in s
n (1 ≤ n ≤ 105) —
the number of query strings.
n
105.
Output
n lines: in the i-th
of them print "YES" if the pattern matches the i-th
query string, and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrary.
Examples
input
output
input
abc
a?a?a*
4
abacaba
abaca
apapa
aaaaax
output
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>
using namespace std;
typedef long long int LL;
const int maxn=1e5;
char a[maxn+5];
char b[maxn+5];
int c[30];
char s[30];
int n;
int main()
{
memset(c,0,sizeof(c));
scanf("%s",s);
for(int i=0;s[i];i++)
{
c[s[i]-'a']=1;
}
scanf("%s",a);
scanf("%d\n",&n);
int len=strlen(a);
for(int i=0;i<n;i++)
{
scanf("%s",b);
int len2=strlen(b);
int res=len2-len;
if(res<-1)
{
printf("NO\n");
continue;
}
int j=0,k=0;
bool ans=true;
while(j<len||k<len2)
{
if(a[j]!='?'&&a[j]!='*')
{
if(a[j]!=b[k])
{
ans=false;
break;
}
else
{
j++,k++;
continue;
}
}
else if(a[j]=='?')
{
if(c[b[k]-'a']==1)
{
j++,k++;
continue;
}
else
{
ans=false;
break;
}
}
else
{
if(res==-1)
{
j++;
continue;
}
else
{
for(int p=k;p<=k+res;p++)
{
if(c[b[p]-'a']==1)
{
ans=false;
}
}
if(ans==false) break;
else
{
j++;
k+=(res+1);
continue;
}
}
}
}
if(ans==true)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
