WpKRwvgKenn
output shold be
"WpKRwvgKenneKgvwRKpW"
not
"WpKRwvgKennneKgvwRKpW"

WA4: s1 is already palindrome.
ex: aba ababa
WA7: s1 consist of only one character.
ex: a aa

abbb
answer: abbba
abbbb
answer: abbbba

Edited by author 25.11.2020 01:48

Edited by author 25.11.2020 01:50

An easy and simple o(n) solution exists. pretty basic

has anybody got WA8 too? I tried all tests in forum but it didn't help :(

once I've solved this exercise, but know a vontto make my algorithm better. I've forced whith WA#12. Could you tell me this test?

It may be caused by conflicts...
(Sorry for my poor English

I've been wa on this case for a long time:
abaabaaba
The right answer is abaabaabaaba

Try this test:
a

Your output should be:
aa

Good luck!

It's just bruteforce!

I had WA25 when used char[20002];
After changed to char[30002] got AC.

Help me WA4?
#include <iostream>
#include <string>
#include <string.h>
#include <algorithm>
using namespace std;
string S, ss, s;
int k, l, r, mid;
int main(){
    cin >> S;
    int n = S.size();
    if(n == 1){
        cout << S;
        return 0;
    }
    for (int i = n - 1; i >= 0; i --){
        k = n - i;
        if(i - k >= 0){
            ss = S.substr(i, k);
            s = S.substr(i - k, k);
            reverse(s.begin(), s.end());
            //cout << ss << ' ' << s << endl;
            if(ss == s){
                l = i - k;
                mid = i;
                r = i + k;
                continue;
            }
        }
        k --;
        if(i - k >= 0){
            ss = S.substr(i + 1, k);
            s = S.substr(i - k, k);
            reverse(s.begin(), s.end());
            //cout << ss << ' ' << s << endl;
            if(ss == s){
                l = i - k;
                mid = i;
                r = i + k;
            }
        }
    }
    //cout << l << ' ' << r << endl;
    if(!l && !r || (l == 0 && r == n - 1)){
        s = S.substr(0, n - 1);
        cout << S;
        reverse(s.begin(), s.end());
        cout << s;
    }
    else{
        s = S.substr(0, l);
        reverse(s.begin(), s.end());
        cout << S << s;
    }
    return 0;
}

Hello,
I've not found a topic about this test. Does anyone know what is it?
I'm just searching from the half of the string to find where can be the center of a palindom.
Thank you

Can anyone please suggest the test case which might give me WA35...

I have used string , also tried with character array...  and KMP algorithm

.....
please do reply

ADMINSTRATOR
HELP PLEASE !!!!!!!!!!!!
IT'S MY COD
var
    a:array [1..20010] of char;
    i,j,k,n,w:integer;
    r:char;
    s:string;
    f:boolean;
    g:text;
    t:integer;
procedure solve;
label 1;
var e:integer;
begin
     f:=false;
     t:=0;
     for k:=1 to n+i do
     if (a[k]<>a[n+i-k+1]) then goto 1;
     f:=true;
     1:
end;
    {IMPORTANT PART}
Begin
      assign(g,'input.txt');
      reset(g);
      while not eof(g) do
      begin
          readln(g,s);
          n:=0;
          for i:=1 to length(s) do
              begin
                  n:=n+1;
                  a[n]:=s[i];
              end;
              i:=0;
       repeat
                solve;
                if not(f) then
                begin
                     i:=i+1;
                     for j:=n+i downto n+1 do
                       begin
                            a[j]:=a[j-1];
                       end;
                  a[n+1]:=a[i];
                  solve;
                end;
        until f;
    end;
        // close(g);
          for j:=1 to n+i do
               write(a[j]);
               readln;readln;
end.

REAL MADRID LUCK
THE BEST CLUB

Tested with already palindrome (as said before), any ideas?
For input: aba it will write: ababa and for: abba -> abbabba

I've tried this problem with hashing but I get WA on test 12. Anyone else has tried this with Hashing?? or even better get AC

Hi,

Does anyone know what the test case is? I've tried my solution against the following, and they seem to be correct. A little lost right now.

"No", "OnLine", "AbabaAab", "abaabaaba", "2101221012", "122212221", "A", "AA", "aba", "saaasaaas", "babab"

Solutions
=========
NoN
OnLineniLnO
AbabaAababA
abaabaabaaba
210122101221012
1222122212221
AA
AAA
ababa
saaasaaasaaas
bababab


Thanks!

Edited by author 03.03.2012 08:35

I have WA#1 but algo was right.
I use this construction:

import qualified Data.ByteString.Char8 as C8

...

input <- C8.getLine

C8.putStr $ result input


----------

But I have always WA#1. At my PC (Linux) I have right answers.

then i tried:


input' <- C8.getLine

let
      input = if (C8.index input' (-1 + C8.length input')) == '\r' then C8.init input' else input'

--------

So, i removed last character '\r' from line if it exists. And after this my code was accepted!

Could you tell me what is test N9?