昨天练习赛时,想的太复杂了...555.后来看到有个人的方法是只要匹配*号2边的内容就可以了,然后就是比较的串的长度必须是>=查询语句-1
#include
<
iostream
>
#include
<
string
>
using
namespace
std;
int
main()
{
int i, j, k, si, qi;
string s[101], q;
int n, m;
int oflag = false;
while (cin >> n)
{
if(!oflag)
oflag = true;
else
cout << endl;
for (i=0; i<n; ++i)
{
cin >> s[i];
}
cin >> m;
for (i=0; i<m; ++i)
{
cin >> q;
bool flag = false;
int pos = q.find('*');
for (j=0; j<n; ++j)
{
if(s[j].size() >= q.size()-1)
{
for (k=0; k<pos; ++k)
{
if(s[j][k]!=q[k])
break;
}
if(k>=pos)
{
for (qi=q.size()-1, si=s[j].size()-1; qi>pos; --qi,--si)
{
if(q[qi]!=s[j][si])
break;
}
if(qi<=pos)
{
if(!flag)
flag = true;
else
cout << ", ";
cout << s[j];
}
}
}
}
if(!flag)
cout << "FILE NOT FOUND";
cout << endl;
}
}
return 0;
}
#include
<
iostream
>
#include
<
string
>
using
namespace
std;
int
main()
{
int i, j, k, si, qi; string s[101], q; int n, m; int oflag = false; while (cin >> n) {
if(!oflag) oflag = true; else cout << endl; for (i=0; i<n; ++i) { cin >> s[i]; } cin >> m; for (i=0; i<m; ++i) { cin >> q; bool flag = false; int pos = q.find('*'); for (j=0; j<n; ++j) { if(s[j].size() >= q.size()-1) { for (k=0; k<pos; ++k) { if(s[j][k]!=q[k]) break; } if(k>=pos) { for (qi=q.size()-1, si=s[j].size()-1; qi>pos; --qi,--si) { if(q[qi]!=s[j][si]) break; } if(qi<=pos) { if(!flag) flag = true; else cout << ", "; cout << s[j]; } } } } if(!flag) cout << "FILE NOT FOUND"; cout << endl; } } return 0;}
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