Judge Info
- Memory Limit: 32768KB
- Case Time Limit: 10000MS
- Time Limit: 10000MS
- Judger: Number Only Judger
Description
In the kingdom of frog, the most popular sport is jumping up and down. Frog likes to jump up(go higher) or jump down(go lower) but jump at the same height(not go higher,and not go lower). The frog king wants to have a sport path for jumping. The Sport path is made by same size blocks which may have different height,when the frog on the path, he can only jumps from one block to the next adjacent block. For frog, perfect path for jumping are those same-height-jumps can be avoided. The Sport path can be represented by an integer sequence which denotes the height of blocks.The length of the sport path is the numbers of integer in the sequence.
Task
Now, it is your turn. You will have a integer sequence represent a sport path, please output the longest perfect path we can find in the given sport path without cutting off.
Input
The first line of input contains?, the number of test cases. There are two lines for each test case. The first line contains an integer number?denoting how many blocks in the sport path. The second line contains? N ?integer numbers (the height of blocks in the range? [ ? 100,100] ).
Output
For each test case, print a line contains the solution.
Sample Input
2
8
1 1 3 5 5 7 9 8
5
1 2 3 4 5
Sample Output
4
5
解題思路：剛開始讀不懂題意，以為是消除相同數和其本身，經過朋友說下才明白，原來是找最長的子串，但是遇到重復數字就開始新的字符串序列。
剛開始用了兩個數組，之后用一個數組，而ACMser都不用開數組。
1
#include <stdio.h>
2
#include <
string
.h>
3
int
A[
103
];
4
5
int
main() {
6
int
t,i,n,pre,max,count,min;
7
scanf(
"
%d
"
,&
t);
8
while
(t--
) {
9
max =
1
;
10
scanf(
"
%d
"
,&
n);
11
for
(i=
0
;i<n;i++
) {
12
scanf(
"
%d
"
, &
A[i]);
13
}
14
pre =
0
;
15
count =
1
;
16
for
(i=
1
;i<n;++
i) {
17
18
if
(A[i]==
A[pre]){
19
if
(count>
max)
20
max =
count;
21
count =
1
;
22
continue
;
23
}
24
count++
;
25
pre=
i;
26
}
27
min =
count;
28
if
(min >
max)
29
max =
min;
30
printf(
"
%d\n
"
,max);
31
}
32
}
?
朋友的做法（不開數組）：
1
#include <stdio.h>
2
int
process(
int
n);
3
int
main()
4
{
5
int
testcase;
6
scanf(
"
%d
"
, &
testcase);
7
while
(testcase--
) {
8
int
n;
9
scanf(
"
%d
"
, &
n);
10
printf(
"
%d\n
"
, process(n));
11
}
12
}
13
int
process(
int
n)
14
{
15
int
i, pre, cur;;
16
scanf(
"
%d
"
, &
pre);
17
int
cnt =
1
, cur_max =
1
;
18
for
(i =
1
; i != n; i++
) {
19
scanf(
"
%d
"
, &
cur);
20
if
(cur ==
pre) {
21
if
(cur_max <
cnt)
22
cur_max =
cnt;
23
cnt =
1
;
24
}
25
else
{
26
cnt++
;
27
}
28
pre =
cur;
29
}
30
if
(cur_max <
cnt)
31
cur_max =
cnt;
32
return
cur_max;
33
}
?
?
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