b = input()
s = ''
d = []
for i in b:
    s += i
    if s == s[::-1]:
        d.append(s)
    if s != s[::-1] and len(s) % 2 != 0:
        s = s[1:]
    if s == s[::-1]:
        d.append(s)
    if s != s[::-1] and len(s) % 2 != 0:
        s = s[1:]
m = max(d, key=len)
if len(m) <= 1000:
    print(m)

see what your program outputs for an integer answer
ps it should output .00

посмотрите, что выводит ваша программа при целом ответе
ps она должна выводить .00

1)
4 4
1 2 9 15
1 4 0 8
2 3 20 30
3 1 31 0
1 4 9 30

->
4
1 3 4 2

2)
2 3
1 2 0 5
1 1 110 80
1 1 90 0
1 2 100 6

->
3
2 3 1

3)
3 6
1 2 50 55
2 1 55 40
1 2 0 1
1 3 41 80
3 2 80 12
2 1 15 0
1 2 49 7

->
6
1 2 4 5 6 3

Finally I've solved this problem using very hard optimised Main & Lorentz's algo, and 0.405s only.

But I noticed many people solved this problem very fast and using much less memory.
What algo do you use? I doubt it's possible to speed-up Main & Lorentz or Crochemore algos so much.

Is it some alternative (like suffix tree) solution, or just some strong idea can be applied to this particular problem? Is it possible to solve it in O(n)?

1 5
/\/\X
.\X\\
ans: 4

what is wrong with my code:

#include <iostream>
#include <algorithm>
#include <climits>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <iomanip>
#include <unordered_map>
#define ll long long
#define fri(a, b) for (ll i = a; i < b; i++)
#define frj(a, b) for (ll j = a; j < b; j++)
#define frk(a, b) for (int k = a; k < b; k++)
#define frh(a, b) for (int h = a; h < b; h++)
#define frz(a, b) for (int z = a; z < b; z++)
#define rfri(a, b) for (int i = a; i >= b; i--)
#define rfrj(a, b) for (int j = a; j >= b; j--)
#define yes cout << "YES" << "\n";
#define no cout << "NO" << "\n";
#define fast                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
const int mod=100000007;
using namespace std;
ll dp[50][1005];

ll func(ll n,ll s){
  if(dp[n][s] != -1) return dp[n][s];
  if(s == 0) return 1;
  if(n == 0){
    if(s == 0) return 1;
    else return 0;
  }
  ll ans = 0;
  fri(0,10){
    if((s - i) >= 0) ans = (ans + func(n-1,s-i)) ;
  }
  return dp[n][s] = ans;
}

int main() {
    fast
    ll T = 1;
    // cin >> T;

    frz(0,T){
      ll n,s;
      cin >> n >> s;
      memset(dp,-1,sizeof(dp));
      if(s%2) cout << 0 << "\n";
      else {
        ll ans = func(n,s/2);
        cout << (ans*ans) << "\n";
      }
    }
}

For example:
16
9001 9001
8999 8999
9001 8999
8999 9001
-9001 -9001
-9001 -8999
-8999 -9001
-8999 -8999
9001 -9001
8999 -8999
9001 -8999
8999 -9001
-9001 9001
-8999 8999
-9001 8999
-8999 9001
5
9000 9000
9000 -9000
-9000 -9000
-9000 9000
0 0

My AC program gives answer:
1 2 3 4
9 10 11 12
5 6 7 8
13 14 15 16
2

But right answer:
1 2 3 4
9 10 11 12
5 6 7 8
13 14 15 16
2 8 10 14

 First I wrote the program in Java, but I got TLE on test #9, below is my program that gets TLE:

import java.util.Scanner;

public class Main {

    public static int read(int idx, int[] tree){
        int sum = 0;
        while(idx > 0){
            sum += tree[idx];
            idx -= idx & (-idx);
        }
        return sum;
    }

    public static void update(int idx, int maxIndex, int[] tree){
        while(idx <= maxIndex){
            tree[idx]++;
            idx += idx & (-idx);
        }
    }

    public static void main(String[] args) {
        int N;
        int maxIndex = 32005;

        int[] level;
        int[] binaryIndexTree;

        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        level = new int[N+1];
        binaryIndexTree = new int[maxIndex];

        for(int i=0; i<N; i++){
            int x = sc.nextInt();
            int y = sc.nextInt();
            x++;
            level[ read(x, binaryIndexTree) ]++;
            update(x, maxIndex, binaryIndexTree);
        }
        for(int i=0; i<N; i++){
            System.out.println(level[i]);
        }

    }
}

Then I write in C++ and get AC?
@admins: Please extend time limit for Java.

my solution takes 0.031 seconds.

Can Anyone be kind enough to explain this ques to me? I'll be grateful to you..

DO your own solve then check this do not cheat yourself
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl "\n"
#define FastAF ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

template <typename T> // cin >> vector<T>
istream &operator>>(istream &istream, vector<T> &v){for (auto &it : v) cin >> it;return istream;}
template <typename T> // cout << vector<T>
ostream &operator<<(ostream &ostream, const vector<T> &c){for (auto &it : c) cout << it << " ";return ostream;}

const int mx=2e5;

int ar[mx];
void d(){
    ar[0]=2;
    int k=1;
    for(int i=3;i<mx;i++){
        bool f=true;
        for(int j=2;j*j<=i;j++){
            if(i%j==0){
                f=false;
                break;
            }
        }
        if(f){
            ar[k++]=i;
        }
    }
}

int main(){
    FastAF
    d();
    int n;
    cin>>n;
    while(n--){
        int a;cin>>a;
        cout<<ar[--a]<<endl;
    }
return 0;
}

Algo: sqrt with precomputation

Edited by author 11.05.2025 11:58

Edited by author 11.05.2025 12:02

6
0 0 0 0 12 40
ANS: 52

8 7

0 0
16
1 1
2 1
2 0
3 0
3 1
2 2
3 2
4 1
4 3
0 3
0 4
1 4
1 5
-1 5
-1 -1
1 -1
12.00000000000000000000

0 0
5
-1 -1
1 0
0 -1
2 -2
2 4
5.50000000000000000000

0 0
17
-1 -1
5 -1
1 0
1 1
2 2
2 1
3 1
3 3
4 4
4 3
5 3
5 5
0 5
0 6
6 6
6 7
-1 7
24.00000000000000000000

0 0
4
0 -1
1 0
0 1
-1 0
2.00000000000000000000

0 0
4
-1 2
-1 -1
1 0
3 -1
3.6666666666

0 0
3
-1 -1
1 -1
0 1
2.00000000000000000000

0 0
11
0 -1
3 0
2 2
1 1
2 3
0 4
0 3
-1 5
-2 3
0 2
-1 0
10.50000000000000000000

I would appreciate hints(or solution) a lot!

Please e-mail me at addflash@dmc.chat

Why is the sample 50.211, shouldn't the sample be 50.198 because the path (45, 0) -> (50, 1) -> (5, 1) -> (0, 0) is length 50.198?

Is there some constructive approach? I solved it with some tricky bruteforce with optimizations

sum 2e6

find out how to solve the problem for n <= 68. for bigger n just do n = (n - 35) % 34 + 35 and solve the problem for this n.

i think the memory limit should be higher