Kolya and Tandem Repeat - CF#253 (Div. 2)B (443B) 哈希或水题

B. Kolya and Tandem Repeat

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

aaba2

output

6

input

aaabbbb2

output

6

input

abracadabra10

output

20

Note

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.

题意：在给定字符串后加任意长度为k的字符后，最长能形成多长的重复两次的字符串。

思路：用哈希是n方复杂度，其实n的三次方也可以过。

AC代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int M=100007,Hash[210],F[210];char s[210];bool solve(int a,int b,int c,int d){ return Hash[d]-Hash[c-1]*F[d-c+1]==Hash[b]-Hash[a-1]*F[b-a+1];}int main(){ int n,i,j,k,len,a,b,c,d,ans=0;  scanf("%s%d",s,&k);  len=strlen(s);  F[0]=1;  for(i=1;i<=len;i++)  { Hash[i]=Hash[i-1]*M+s[i-1];    F[i]=F[i-1]*M;  }  for(i=1;i<=len+k;i++)   for(j=i;j<=len+k;j++)   { a=i;b=j;c=j+1;d=2*j-i+1;     if(d>len+k)      break;     if(d<=len)      if(solve(a,b,c,d))       ans=max(ans,d-a+1);     else if(c<=len)      if(solve(a,a+len-c,c,len))       ans=max(ans,d-a+1);     else      ans=max(ans,d-a+1);   }  printf("%d\n",ans);}