Есть ли возможность сдавать задачи на языке Фортран

Each crossroads in the input is the binary OR of its branches, so a 5 (binary 101) at some crossroads in the input means there are branches 4 (100) and 1 (001) there.

As for the output, for each crossroads where there are no branches (each 0, let's call this crossroads P), repeat this:
Find the Manhattan (vertical+horizontal) distance to the closest branch from C. In the description, this closest distance is 2. Next, find all branches that lie on this exact distance from C. In the description, these are branches with number 16, 8, and 4. The output at location P should reflect all types of branches at distance 2 by ORing these numbers together, and in the description we get 28 (binary 10000 | 1000 | 100).

import java.util.Scanner;
import java.io.*;

public class HelloWorld {

    public static void main(String args[]) {

        int a = 0;
        int total = 0;
        Scanner input = new Scanner(System.in);
        a = input.nextInt();


            total = convertFunc(a);
            System.out.println(total);
    }

        public static int convertFunc(int n){

        int sum = 0;
        if(n > 0 && n <= 10000 )
        {
            for(int i = 1 ; i <= n ; i++ )
            {
                sum += i;
            }
        }
        else
        {
            for(int i = 1 ; i >= n ; i--)
            {
                sum += i;
            }
        }
        return sum;
    }
}

import java.text.DecimalFormat;
import java.text.DecimalFormatSymbols;
import java.util.ArrayList;
import java.util.Scanner;


public class Task1001
{
    public static void main(String[] args)
    {
        Scanner scanner = new Scanner(System.in);
        ArrayList<Long> list = new ArrayList<Long>();
        for (int i = 0; i < 4 ; i++)
        {
            long a = scanner.nextLong();
            if (a >= 0 && a <= 10E18)
            list.add(a);
        }
        DecimalFormatSymbols symbol = new DecimalFormatSymbols();
        symbol.setDecimalSeparator('.');
        DecimalFormat f = new DecimalFormat("#,##0.0000", symbol);

        for (int i = list.size(); i > 0 ; i--)
        {
            double c = (Math.sqrt(list.get(i - 1)));
            System.out.println((f.format(c)));
        }
    }
}



I need help please!!!

Edited by author 04.04.2015 17:02

Problem was solved.

Edited by author 04.06.2015 18:02

Please give me some tests

var a, b : array[1..1000] of integer;
    n, i, k : integer;
begin
  readln(n);

  for i := 1 to n do begin
    read(a[i]);
  end;
   i := 2;
   k := 1;
   while i <= n + 2 do begin
     if a[i] = a[i - 1] then begin
     k := k + 1;
     i := i + 1;
     end
     else
     begin
       write(k,' ',a[i - 1],' ');
       i := i + 1;
       k := 1;
     end;
   end;


end.

you need to write answer with two digits after point, so don't use cout!
use printf("%.2f\n",  );
and don't forget #include <cstdio>

i've been finding this mistake for 2 days!!

Does someone have any idea what's test 3?

i understand problem !!!

Edited by author 03.06.2015 12:14

если жарить 3 бифштекса на сковороде, где умещается всего 2, то потребуется не 3, а 4 минуты, поскольку последний бифштекс на 2 части не разделить и не положить на сковороду одновременно обоими сторонами

java

public class MainClass
{
    public static void main(String[] args)
    {
        System.out.println("A=\"'\";B='A=\"\";B=;?$(B,0,4)+A+$(B,4,4)+A+B+A+$(B,8,37)';?$(B,0,4)+A+$(B,4,4)+A+B+A+$(B,8,37)");
    }
}

This is my code
   #include <iostream>
   #include <stdio.h>
   #include <math.h>
   using namespace std;
long n,x[201],y[201];

   int main()
  {double xa,xb,xc,ya,yb,yc,l,a,b,c,x,h;
   cin>>xa>>ya>>xb>>yb>>xc>>yc>>l;
  b=sqrt((xc-xa)*(xc-xa)+(yc-ya)*(yc-ya));
  a=sqrt((xc-xb)*(xc-xb)+(yc-yb)*(yc-yb));
  c=sqrt((xa-xb)*(xa-xb)+(ya-yb)*(ya-yb));
  if(c!=0){
  x=(c*c+b*b-a*a)/(2.0*c);
  h=sqrt(b*b-x*x);
  if(b!=0&&a!=0){
  if((b*b+c*c-a*a)/(2.0*b*c)<=0) h=b;
  if((a*a+c*c-b*b)/(2.0*a*c)<=0) h=a;}

  if(h<=l) h=0; else h=h-l;}
  else h=a;
  printf("%0.2f\n",h);

  if(a>=b) {if(a<=l) h=0; else h=a-l;}
  else {if(b<=l) h=0; else h=b-l;}
  printf("%0.2f\n",h);


      return 0;
   }

Please give me some test

Edited by author 11.06.2015 18:48

Edited by author 11.06.2015 18:48

pls give me some test data, dont know what s wrong with code.
greetings.

My code:
a = readlines
a = a.join
a = a.split(/\s+/)
a.reverse!
a.map!{|elem|elem.to_i}
a.map!{|elem|(elem**0.5).round(4)}
puts a

I've tried code from FAQ page:
tokens = []
while string = gets do
   string.split().each do |token|
      tokens << token
   end
end
tokens.reverse.each do |token|
   puts (token.to_i ** 0.5).round(4)
end

Both variants return right results in my terminal, but in both cases i've got "wrong answer" here.
Why?

I passed all the tests, described above in forum, but Wrong answer3.

Give me, please, the test#3?
nikolaievgen@gmail.com
thanks!

Hi everyone, I was trying to solve this problem, and I use an algorithm that works on "codeblocks", but doesn't work here. I would like that someone help me. This is the code, what is wrong with it? It's Inefficient? Thanks.

#include<iostream>
#include<vector>
#include<string>

using namespace std;

string convertir(string s){
    string conv="";

    for(int i=0;i<s.size();i++){
        if(s[i]=='i'||s[i]=='j') conv+="1";
        else if(s[i]=='a'||s[i]=='b'||s[i]=='c') conv+="2";
        else if(s[i]=='d'||s[i]=='e'||s[i]=='f') conv+="3";
        else if(s[i]=='g'||s[i]=='h') conv+="4";
        else if(s[i]=='k'||s[i]=='l') conv+="5";
        else if(s[i]=='m'||s[i]=='n') conv+="6";
        else if(s[i]=='p'||s[i]=='r'||s[i]=='s') conv+="7";
        else if(s[i]=='t'||s[i]=='u'||s[i]=='v') conv+="8";
        else if(s[i]=='w'||s[i]=='x'||s[i]=='y') conv+="9";
        else conv+="0";
    }

    return conv;
}

int main(){
    string number,aux;
    vector<string> dictionary;
    vector<string> converted;
    int n;

    while(cin>>number){
        if(number=="-1") break;

        cin>>n;

        dictionary.clear();
        converted.clear();

        for(int i=0;i<n;i++){
            cin>>aux;
            dictionary.push_back(aux);
            converted.push_back(convertir(aux));
        }

        bool max[100];
        for(int i = 1; i<100; i++){
            max[i]=false;
        }
        max[0]=true;
        vector<string> sequence(number.size()+1,"");

        for(int i=1;i<=number.size();i++){
            for(int j=0;j<n;j++){
                if(converted[j].size()<=i && number.substr(i-converted[j].size(),converted[j].size())==converted[j]){
                        max[i]=true && max[i-converted[j].size()];
                        sequence[i]=sequence[i-converted[j].size()]+dictionary[j]+" ";
                }
            }
        }

        if(max[number.size()]==false) cout<<"No solution."<<endl;
        else{
            aux=sequence[number.size()];
            aux.erase(aux.size()-1);
            cout<<aux<<endl;
        }
    }


    return 0;
}

I can't understand this problem mean, someone can explain it for me? Why first test and second test have this result?