Problem:
Score : 900 points

Problem Statement

Takahashi has decided to give a string to his mother.

The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T.
That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).

Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value.
Find the highest possible value achievable.

Constraints

1 \leq |S| \leq 300

0 \leq K \leq |S|

S consists of lowercase English letters.

K is an integer.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the highest possible value achievable.

Sample Input 1

abcabcabc
1

Sample Output 1

7

Changing the first character to c results in cbcabcabc.
Let this tring be T, then one longest common subsequence of T and T' is cbabcbc, whose length is 7.

Sample Input 2

atcodergrandcontest
3

Sample Output 2

15

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define fr first
#define sc second
#define ll  long long
#define mp make_pair
#define mod 1000000007

using namespace std;

string s;
int dp[330][330][330];
int aft[330][30];
int bef[330][30];
int f(int st,int en,int k){
    if(st == en){
        return 1;
    }
    int &ret = dp[st][en][k];
    if(ret != -1)return ret;
    int r =0;
    r = max(r,f(st+1,en,k));
    r = max(r,f(st,en-1,k));
    if(st +1 != en &&k){
        r = max(r,f(st+1,en-1,k-1)+2);
    }
    for(int i=0 ;i < 26 ; i++){
        if(aft[st][i] <= bef[en][i]){
            r = max(r,f(aft[st][i],bef[en][i],k)+2);
        }

    }
    return ret = r;
}

int v[100100];
int main()  {
    int k;
    ios::sync_with_stdio(0);
    cin>>s>>k;
    int n = s.size();
    memset(dp,-1,sizeof dp);
    for(int j=0 ;j <26 ; j++){
        v[j] = -1;
    }

    for(int i=0 ;i <n ;i ++){
        for(int j=0 ;j <26  ; j++){
            bef[i][j] = v[j];
        }
        v[s[i]-'a'] = i;
    }

    for(int j=0 ;j <26 ; j++){
        v[j] = n+2;
    }

    for(int i=n-1 ;i >=0 ;i --){
        for(int j=0 ;j <26  ; j++){
            aft[i][j] = v[j];
        }
        v[s[i]-'a'] = i;
    }
    int mx =0 ;
    for(int i=0 ; i<n ;i ++){
        for(int j=i ;j <n ;j ++){
            if(k == 0 && s[i] != s[j])continue;
            mx = max(mx,f(i,j,k-(s[i] != s[j])));
        }
    }
    cout<<mx<<endl;
    return 0;
}