s = input() + 'X' * int(input())
for v in range(len(s) - len(s) % 2, 0, -2):
for i in range(len(s) - v + 1):
if all(s[j + v // 2] in ('X', s[j]) for j in range(i, i + v // 2)):
print(v), exit()#include <array>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <functional>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <sstream>
typedef long long ll;
using namespace std;
void solve() {
string s;
int k;
cin >> s >> k;
int n = s.size();
if(k >= n){
cout << (n+k)/2*2 << endl;
return;
}
int ret = 2 * k;
for(int i = n-k-1; i >= 0; i--){
int max_h = (n + k - i) / 2;
for(int h = 1; h <= max_h; h++){
bool found = true;
for(int j = 0; j < h; j++){
if(i+j+h >= n) break;
if(s[i+j] != s[i+j+h]){
found = false;
break;
}
}
if(found){
ret = max(ret, 2 * h);
}
}
}
cout << ret << endl;
return;
}
int main(){
// ios_base::sync_with_stdio(false);
// cin.tie(NULL);
int t = 1;
// cin >> t;
while(t--)
solve();
return 0;
}822B - Crossword solving | 1623A - Robot Cleaner |
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