One of the solutions would involve sorting and stacks. I've tried to use reb-black tree, but it seems to have problems with dublicate-keys.

Edited by author 14.08.2015 22:05

HINT1: solution always exist
HINT2: try to solve this problem with end.
HINT3: this problem can solve <= O(n logn)

I tried implementing mergesort which probably had recursion depth runtime error; bisect insertion to save some memory, but too slow for python list (perhaps a good alternative if I have access to pointer?); heapq is much faster but instable. What are other options?

import java.io.InputStreamReader;
import java.text.DecimalFormat;
import java.util.Scanner;

public class Reverse {

    private static void printer(double value){
        String pattern = "###0.0000";
        DecimalFormat myFormatter = new DecimalFormat(pattern);
        String output = myFormatter.format(value);
        System.out.println(output);
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(new InputStreamReader(System.in));
        long a,b,c,d;

        a = sc.nextLong();
        b = sc.nextLong();
        c = sc.nextLong();
        d = sc.nextLong();

        printer(Math.sqrt(d));
        printer(Math.sqrt(c));
        printer(Math.sqrt(b));
        printer(Math.sqrt(a));
    }
}

It's very easy problem, but many people don't understand how to solve it ... :)

7 lines :

#include <iostream>
using namespace std;
int main(){
unsigned long long N;
cin >> N;
cout << N*N << endl << N;
return 0; }

My code fails with WA#1 although it passes every test cases that I've found in comments. Any hints?

Why isn't the output for 1 just:

1

?

It is equivalent to:

0
X
*
1
+

... but it is shorter. The problem asks for minimal length doesn't it?

Problem solved. Was using a greedy solution which omitted some of the possibilities.

Edited by author 26.03.2018 11:18

Всем привет. Пишу на с++. И возникла проблема с вводом широты и долготы без перехода на новую строку.
Выходит:

<X1>
^<X2>
'...

что не очень красиво. Хотелось бы так

<X1>^<X2>'<X3>" <NL/SL>

но не знаю как реализовать.

Спасибо.

how can i see my submission list?

Edited by author 23.03.2018 17:43

Edited by author 23.03.2018 17:43

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Iterator;
import java.util.Stack;


public class problem1654 {

    public static void main(String[] args) throws IOException {
    BufferedReader inp = new BufferedReader(new InputStreamReader(System.in));
    String str=inp.readLine();
    Stack<Character> st=new Stack<Character>();
    st.push(str.charAt(0));
    String s="";
    for(int i=1;i<str.length();i++){
    if(!st.empty()&&st.peek()==str.charAt(i))
        st.pop();
    else{
    st.push(str.charAt(i));

    }
    }



while(!st.empty()){
    s=s+st.pop();
}
int n=s.length();
for(int i=0;i<n;i++){
    System.out.print(s.charAt(n-i-1));


    }

    }}

Here is my code and here is my output. Everything should work! What's wrong ? What requirements have I not fulfilled?
http://lpaste.net/1392484352197132288
7325189087
5
it
your
reality
real
our
reality our
4294967296
5
it
your
reality
real
our
No solution.
-1

The code is written in pure C. The compiler is gcc

for all-integer comparison you have to use int64_t

do not derive the inequality by hand, use Maxima:

rot(c,s) := matrix([c,-s],[s,c]);
/* D: sqrt(a^2+b^2); */
ma: rot(e/D, sqrt(D^2-e^2)/D);
mb: rot(b/D, a/D);
dd: (ma.mb.[a,-b])[1][1],ratsimp;

load(ineq)$
assume(D>0)$ assume(a>0)$ assume(b>0)$ assume(d>0)$ assume(e>0)$
ne: d >= dd;
ne: (ne * D^2)-2*a*b*e,ratsimp;

import java.util.Scanner;
public class Sum{
  public static void main(String args[]){
    Scanner sc= new Scanner(System.in);
    int N, sum=0;
    System.out.print("Enter the value of N:");
    N= sc.nextInt();
    if(N<10000){
      if(N<0){
        for(int x=N; x<=1; x++){
          sum+=x;
        }
      }
        else{
          for(int x=1; x<=N; x++){
            sum+=x;
          }
        }
    }
    System.out.println(sum);
  }
}

Edited by author 20.03.2018 00:35

Define d[i][j] to be the maximum number of floors that can be checked with i eggs after j steps.

To answer the query one should do the binary search to find number of steps X such that
 d[input_eggs][X] >= input_floors with X being smallest possible.

d[i][i] = i, d[2][5] = 15 ,d[2][6] = 21 ( 1 -> 6 -> 11 -> 15 -> 18 -> 20 -> 21 )

d[i][j] = d[i - 1][j] + d[i - 1][j - 1] + 1;

Explanation of the formula:

Suppose we know d[i][j - 1] and d[i - 1][j - 1]

Throw an egg from (d[i - 1][j - 1] + 1) floor. Did it break? If yes, we can definitely check all floors from the first to d[i - 1][j - 1]th with remaining i - 1 eggs within remaining j - 1 steps. If it didn't then you just check how many floors you can check at most with i eggs within j - 1 steps starting from ((d[i - 1][j - 1] + 1) + 1) th floor

d[i][j] can be calculated in N^2
Query is answered in log(N) time

My Java 1.8 solution runs in 0.062s.

Edited by author 19.03.2018 20:45

 while (n>0)
      {
            b=a;
            b=b/pow(10,n-1) - rezult/pow(10,n-1);
            for (int i=1;i<10;i++)
            if ((b-i>=0)&(b-i-1<0))  rezult += (i)*pow(10,n-1);
            n--;
      }
/*
if delete :

b=b/pow(10,n-1) - rezult/pow(10,n-1);
            for (int i=1;i<10;i++)
            if ((b-i>=0)&(b-i-1<0))  rezult += (i)*pow(10,n-1);

- work! */