hdu 1867 A + B for you again
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 670 Accepted Submission(s): 94
简单的kmp算法，由于Strstr要运算多次没有KMP灵活所以运用strstr有可能会超时。虽然本题不难，但是有一个地方需要注意。看了这组例子想必就该明白了
sdfg asdf asdfg
ghjk asdf asdfghjk
asdf ghj asdfghj
aghjks ghjk aghjksghjk
ghjk aghjks aghjksghjk
特别注意最后两组例子
代码：
代码
#include
<
stdio.h
>
#include
<
string
.h
>
int
next[
100005
];
void
getnext(
char
*
t)
{
int
i
=
1
,j
=
0
;next[
1
]
=
0
;
int
len
=
strlen(t
+
1
);
while
(i
<
len)
{
if
(j
==
0
||
t[i]
==
t[j])
{
++
i;
++
j;next[i]
=
j;
}
else
j
=
next[j];
}
}
int
kmp(
char
*
s,
char
*
t)
{
int
i,j,len1,len2;
getnext(t);
i
=
1
;j
=
1
;
len1
=
strlen(s
+
1
);len2
=
strlen(t
+
1
);
while
(i
<=
len1
&&
j
<=
len2)
{
if
(j
==
0
||
s[i]
==
t[j])
{
++
i;
++
j;
}
else
{
j
=
next[j];
}
}
if
(j
>
len2
&&
i
<=
len1)
return
0
;
return
j
-
1
;
}
char
str1[
100005
],str2[
100005
];
int
main()
{
int
mark1,mark2,j;
while
(scanf(
"
%s%s
"
,str1
+
1
,str2
+
1
)
!=
EOF)
{
mark1
=
kmp(str1,str2);
mark2
=
kmp(str2,str1);
if
(mark1
==
mark2)
{
if
(strcmp(str1
+
1
,str2
+
1
)
>=
0
)
{
printf(
"
%s
"
,str2
+
1
);
for
(j
=
mark1
+
1
;str1[j]
!=
'
\0
'
;j
++
)
printf(
"
%c
"
,str1[j]);
}
else
{
printf(
"
%s
"
,str1
+
1
);
for
(j
=
mark2
+
1
;str2[j]
!=
'
\0
'
;j
++
)
printf(
"
%c
"
,str2[j]);
}
}
else
if
(mark1
>
mark2)
{
printf(
"
%s
"
,str1
+
1
);
for
(j
=
mark1
+
1
;str2[j]
!=
'
\0
'
;j
++
)
printf(
"
%c
"
,str2[j]);
}
else
if
(mark2
>
mark1)
{
printf(
"
%s
"
,str2
+
1
);
for
(j
=
mark2
+
1
;str1[j]
!=
'
\0
'
;j
++
)
printf(
"
%c
"
,str1[j]);
}
printf(
"
\n
"
);
}
return
0
;
}