A. Kolya and Tandem Repeat
Time Limit: 2000ms
Memory Limit: 262144KB
64-bit integer IO format: %I64d Java class name: Any
Kolya got string s for his birthday, the string consists of small English letters. He immediately added k more characters to the right of the string.
Then Borya came and said that the new string contained a tandem repeat of length l as a substring. How large could l be?
See notes for definition of a tandem repeat.
Input
The first line contains s (1 ≤ |s| ≤ 200). This string contains only small English letters. The second line contains number k (1 ≤ k ≤ 200) — the number of the added characters.
Output
Print a single number — the maximum length of the tandem repeat that could have occurred in the new string.
Sample Input
Input
aaba2
Output
6
Input
aaabbbb2
Output
6
Input
abracadabra10
Output
20
Hint
A tandem repeat of length 2n is string s, where for any position i (1 ≤ i ≤ n) the following condition fulfills: si = si + n.
In the first sample Kolya could obtain a string aabaab, in the second — aaabbbbbb, in the third — abracadabrabracadabra.
1 /*
2 2016年4月22日00:01:53
3
4 题意：给出一个字符串，给出k,可以向该字符串尾部添加k个字符串，求最长的连续重复两次的子串
5
6 看了题解，可以先把这k个字符填成‘*‘,再暴力枚举起点和长度，找出最大的长度
7 */
8
9 # include <iostream>
10 # include <cstdio>
11 # include <cstring>
12 # include <algorithm>
13 # include <queue>
14 # include <vector>
15 # define INF 0x3f3f3f3f
16 using namespace std;
17
18 int main(void)
19 {
20 char s[20005];
21 int n, len, i, j, k, flag, ans;
22 while (~scanf("%s %d", &s, &n)){
23 getchar();
24 len = strlen(s);
25 for (i = len; i < len+n; i++)
26 s[i] = ‘*‘;
27 len += n;
28 ans = -1;
29 for (i = 0; i < len; i++){
30 for (j = 1; 2*j+i<=len; j++){
31 flag = 1;
32 for (k = i; k < i+j; k++){
33 if (s[k]!=s[k+j] && s[k+j]!=‘*‘){
34 flag = 0;
35 break;
36 }
37 }
38 if (flag) ans = max(ans, 2*j);
39 }
40 }
41 printf("%d\n", ans);
42 }
43
44 return 0;
45 }