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! */

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.