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! */

Can anybody give some test to judge my prog? I have WA on 5 test and cann't find my mistake.

test for wrong #14:
input
a
h
output
v
c

first test is not sample.

Edited by author 26.10.2010 19:06

I have wrong answer on test 2, but all tests that I checked have right answer. Please, give me some tests to check my code.

kek

Get me test, please! I use kmp.

Check (AB,CD)==0 (orthogonality).
Check (AB, CD, DA) ==0 (planarity).
Check AD>AC>AB,  AC>BC,  BD>BC (order).
Check whether the projections to XY,  YZ,  XZ craddle each other continuations.
It is sufficient to check only the projections to XY plane to get Accepted verdict.

I always get ML, and i cant understand what sould i do?

I tried to use Scanner and BufferedReader.

Please tell me method to solve this.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Task1787
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] kMinute = Console.ReadLine().Split(new char[] { ' ', '\t', '\n', '\r' }, StringSplitOptions.RemoveEmptyEntries);
            int k = int.Parse(kMinute[0]);
            int n = int.Parse(kMinute[1]);
            string[] nCam = Console.ReadLine().Split(new char[] { ' ', '\t', '\n', '\r' }, StringSplitOptions.RemoveEmptyEntries);
            int sum = 0;
            foreach (var item in nCam)
            {
                sum += int.Parse(item);
            }
            Console.WriteLine(sum - (k * n));
        }
    }
}

In this problem,many people use the BFS(breadth first search).However,you can't search it from the two numbers which are in the last line,you should search it from the plate you need.Because of this,I was wrong before at test#6.

If your can't understand it,you can test the sample input,the right answer is-
2
1
2
but not 2
4
2

And pay attention at the number of routes-1 to 2000.

Believe me!!!

Sorry to my English...I'm a Chinese...

Good luck to you.:)