cf443B Kolya and Tandem Repeat

 

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

aaba
2

output

6

input

aaabbbb
2

output

6

input

abracadabra
10

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.

 

其实就是个很水的模拟……只要看懂题意应该就没什么问题

不过要注意k>s.length的情况

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,len,k,ans;
int main()
{
    string s;
    char ch[1001];
    cin>>s;
    scanf("%d",&k);
    len=s.length();
    for (int i=len;i>=1;i--)
      ch[i]=s[i-1];
    for (int s=1;s<=len;s++)
    for (int i=1;i<=len-s+1;i++)
      {
        if (s-1+2*i>len+k) break;
        bool mark=0;
        for (int j=s+i;j<=min(len,s+i*2-1);j++)
          if (ch[j-i]!=ch[j]) {mark=1;break; }
        if (!mark) ans=max(i,ans);
      }
    if (k>len) ans=max(ans,(k+len)>>1);
    printf("%d",ans*2);
}