who knows what the meaning of set of numbers is???
is it true that the user must specify how many numbers must be taken?

Edited by author 03.02.2015 23:09

Здравствуйте,
У меня возникают постоянные ошибки при сдаче программ на Py2.7, например задача: 1293

N = input()
B = input()
A = input()
print 2*N*A*B

Выдает RE.

Тот же код на Pascal сдается:
var
A, N, B: integer;
begin
read(N, A, B);
write(2*N*B*A);
end.

И это только единичный случай, так со всеми. Я даже не могу задачку a + b решить =\

in.nextToken();
        int n = (int) in.nval;
        byte ch1[] = new byte[n];
        byte ch2[] = new byte[n];
        while (n > 0) {
            in.nextToken();
            ch1[n - 1] = (byte) in.nval;
            in.nextToken();
            ch2[n - 1] = (byte) in.nval;
            n--;
        }
        int ost = 0;
        String answer = "";
        for (int i = 0; i < ch1.length; i++) {
            int a = ost + ch1[i] + ch2[i];
            if (a > 9) {
                answer = (a % 10) + answer;
                ost = a / 10;
            } else {
                answer = a + answer;
                ost = 0;
            }
        }
        out.print(answer);

Edited by author 02.02.2015 20:11

я ж решать хочу

A lot of the submissions are returning Fail(checker).

There seems to be some error with the judge, most of the solutions are returning a Fail(checker) status from the judge.

My program work correct with all of my test. But i have WA on third test. Can anyone give me some tests?

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

public class qw {
   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);
    PrintWriter out = new PrintWriter(System.out);

       int n = in.nextInt();
       int res = 0;

       if(n > 0) {
        for(int i = 2; i <= n; i++) {
           res += i;
        }
    } else if(n < 0) {
        for(int i = -2; i >= n; i--) {
           res += i;
        }
    } else res = 0;



       if(-10000 <= n & n <= 10000)out.println(res);
    out.flush();


   }
}

Edited by author 02.02.2015 00:14

include <iostream>
using namespace std;
int main()
{

    int N=0,a=0,b=0;
    cin>>N;
    a=(N/100000)+(N/10000-N/100000*10)+(N/1000-N/10000*10);
    b=(N/100-N/1000*10)+(N/10-N/100*10)+(N-N/10*10);
    if(a-1==b){
        cout << "YES\n";
    } else {
        cout << "NO\n";
    }
    return 0;
}

What's in test 2?
As I understand, it's not necessary that number of people in each team is the same.

my prog got AC but on 65,85,105... it answers 2,but I think the answer is 1.
yes,my second prog answers 1 and got AC too.
a 20-number period incorrect!

Edited by author 20.05.2013 20:57

#include<iostream>
#include<stack>
#include<math.h>
#include<iomanip>

float next_value = 0;
std::stack<float> my_stack;

int main()
{
    int a = 0;
    while (std::cin >> next_value)
    {
        if (a == 65536) break;
        if (0 <= next_value && next_value <= pow(10.0, 18))
        {
            my_stack.push(next_value);
        }
        else continue;
        a++;
    }

    while (!my_stack.empty())
    {
        std::cout << std::fixed << std::setprecision(4) << sqrt(my_stack.top()) << std::endl;
        my_stack.pop();
    }
    return 0;
}

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

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

        ArrayList<Long> list = new ArrayList<Long>();

        while (in.hasNextLine())
        {
                list.add(in.nextLong());
        }

        Collections.reverse(list);

        for (long x : list)
        {
            System.out.printf("%.4f", Math.sqrt(x));
            System.out.println();
        }
    }

}

Works on my computer with the given input.

Edited by author 31.01.2015 06:20

What is test case 6?

For example if I take N =4. then i write:
10 12
9 3
15 10
11 13

What will I get?

#include <stdio.h>

int main(void) {

    int car_num, min, i, arr[100], sum = 0;

    scanf("%d %d", &car_num, &min);

    for(i = 0; i < min; i++)
    {
        scanf("%d", &arr[i]);
        if(sum + arr[i] - car_num < 0)
            sum += 0;
        else
        sum = sum + arr[i] - car_num;
    }

    printf("%d", sum);

    return 0;


}

I've checked all the tests on the forum, but all of them actually wokred. What's wrong with the code?

type
Otrezok=record
  beginning,ending:real;
  colour:string;
end;

type
Point=record
  mean,predel:real;
end;

var
a:array[0..50000] of Otrezok;
b:array[0..50002] of Point;
i,N,j,s:integer;
k,j1,j2,max,f1,f2:real;
t,f3:string;

function colour(x:real;H:integer):string;
var j:integer;
begin

for j:=0 to H do
begin
if (x>=a[j].beginning) and (x<=a[j].ending) and (a[j].colour=' w') then colour:='w';
if (x>=a[j].beginning) and (x<=a[j].ending) and (a[j].colour=' b') then colour:='b';
end;
end;


begin

a[0].beginning:=0;
a[0].ending:=1000000000;
a[0].colour:=' w';
readln(N);
for i:=1 to N do
begin
readln(f1,f2,f3);
if f1<f2 then
begin
 a[i].beginning:=f1;
 a[i].ending:=f2;
 a[i].colour:=f3;
end;
end;
for j:=1 to N do
begin
b[j].predel:=0;
b[j+N].predel:=0;
b[j].mean:=a[j].beginning;
b[j+N].mean:=a[j].ending;
end;
b[0].predel:=0;
b[0].mean:=0;


  for i:=(2*N-1) downto 1 do
    for j:=1 to i do
    begin
        if b[j].mean>b[j+1].mean then
            begin
              k:=b[j+1].mean;
              b[j+1].mean:=b[j].mean;
              b[j].mean:=k;
            end;
       if (b[j].mean=b[j+1].mean) then b[j+1].mean:=0
       end;



for i:=1 to 2*N do
begin
if (b[i].mean<>0) and (colour(b[i].mean,N)<>colour(b[i].mean+0.1,N)) or (colour(b[i].mean,N)<>colour(b[i].mean-0.1,N)) then b[i].predel:=1;
if b[i].mean=0 then b[i].predel:=0;
end;
t:=colour(0,N);


s:=0;
for i:=0 to 2*N do
begin
if (b[i].predel=1) then
begin
s:=s+1;
k:=b[s].mean;
b[s].mean:=b[i].mean;
b[i].mean:=k;
end;
end;



for i:=s downto 0 do
b[i+1].mean:=b[i].mean;

b[s+2].mean:=1000000000;

max:=0;
b[0].mean:=0;


for i:=1 to s+2 do
begin
if t='w' then
begin
if (i mod 2)=0 then
begin
if (b[i].mean-b[i-1].mean)>max then
begin
max:=b[i].mean-b[i-1].mean;
j1:=b[i-1].mean;
j2:=b[i].mean;
end;
end;
end;
if t='b' then
begin
if (i mod 2)=1 then
begin
if b[i].mean-b[i-1].mean>max then
begin
max:=b[i].mean-b[i-1].mean;
j1:=b[i-1].mean;
j2:=b[i].mean;
end;
end;
end;
end;


write(Round(j1),' ',Round(j2));
end.

[deleted]

P.S. Don't EVER forget to initialize your variables.

Edited by author 30.01.2015 01:18

Edited by author 30.01.2015 01:19

This tests helped me with WA3, WA5 and WA10, but now I'm stuck at test 20. Any hints?

4 9
010000010
111010111
101111101
000101000
answer: 9

4 5
11000
01100
00110
00011
answer: 4

5 5
11100
11110
11111
01111
00111
answer: 3

4 9
100111100
100100100
100100100
111100111
answer: 6