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普通的自动机能过……RT……
p是pattern,now是输入字符串,1300MS通过……
bool go(char now[],char p[])
{
if (p[0]=='*' && p[1]==0) return 1;
if (now[0]==0 || p[0]==0)
{
if (now[0]==0 && p[0]==0) return 1;
return 0;
}
if (p[0]=='?'|| p[0]==now[0])
return go(now+1,p+1);
if (p[0]=='*')
{
for (char *i=now;;i++)
{
if (go(i,p+1))
return 1;
if (*i==0) return 0;
}
return 0;
}
return 0;
}
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