Can somebody give me the whole explanation to this problem. Thanks beforehand!

Why number of digits of b is less than number of digits of a1 or a2,  and answer is not Impossible,  I output Impossible but WA. can we output leading zeroes??

I can't pass this test.

My English is poor, but I read it by a large time and I understood.
N=2,K=10  ==>it's a "2 digits 10-base"  10,11,...98,99... in Chinese "2位的十进制数字"

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;
}

problem description said dragon will not use breath two times, but test 21 there is at least three breath in the input file...

hope admin can fix it..

I can't pass this test.

Don't know, what else to test ...
Any ideas?

I think many people can not solve this problem because they do not have enough knowledge.
You should know "Problem LCA" and "Segment Tree" that to solve the problem.
My algo is O(n*sqrt(n)+m*log(n)), but i can solve for O(n+m*log(n)) (LCA can be solved for O(1) with preprocess or sqrt(n) without preprocess).

P.S. What is the optimal asymptotic behavior?

Edited by author 20.11.2016 23:27

Edited by author 20.11.2016 23:27

Edited by author 20.11.2016 23:28

import java.util.Scanner;

public class Sold_Out {
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);

        int n = scan.nextInt();
        int k = scan.nextInt();

        if(n<3){
            System.out.println(0);
            return;
        }

        System.out.println(Math.max(k - 1, n - k) - 2);
    }
}

Could anyone give me some sample input and output?
Thanks a lot.

isn't the maximul area of rectangle for the first input equal to 50 ? https://www.youtube.com/watch?v=UzftAa_sXYs

Calculates sum of divisors.
O(lim * log(lim)).
for(int i = 2; i <= lim; ++i){
    ++s[i];
    for(int j = i + i; j <= lim; j += i)
        s[j] += i;
}

In C++, you'd better use "long long" instead of "long".

4
BBAA
BABB
=
Bob

Почему в исходном примере выигрывает Боб? Если при первом сгибе получаем полоску АА, значит должна выиграть Алиса... Почему нет?

Can members of the director board be subordinates? For example, is 6 a subordinate of 3?

Для теста
5 5
что нужно вывести
0
2
5 5

или

0
1
5

A can't understand what's the problem and why I don't pass it. Help me, please.