#include <stdio.h>
int main()
{
   long int k,a,n;
   scanf("%ld",&n);
   while(n--)
   {


   scanf("%ld",&k);
   a=1;
   while(k-a>0)
    k=k-(a++);
   if(k==1)
    printf("1\n");
   else
    printf("0\n");
   }
}






Edited by author 27.12.2013 01:54

if((8*(input) - 7) -> integer)
 cout << 1;
else
 cout << 0;

#include <iostream>
#include <cmath>
using namespace std;

int main ()
{
    int x,e;
    double nk;
    cin >> x;
    int *b=new int[x];
    for (e=0; e<x; ++e)
    {
        cin >> b[e];
    }
    for (e=0 ; e<x; ++e )
    {
        nk = (pow((8*b[e]-7),0.5)-1)/2;
        long int k = static_cast< int >( nk );
        if ( nk==k )
        {
            cout << "1 ";
        }
        else
        {
            cout << "0 ";
        }
    }
    system ("pause > void");
}

Edited by author 05.12.2013 19:32

What's wrong with Test #1?
I've 0 0 1 0

#include <stdio.h>
#define c 65535
int main()
{
    long long k = 0;
    long long a[c];
    long n = 0,i,ot[c],j,t;

    a[0]=1;
    for (i=1; i<=c; i++)
    {
        a[i]=a[i-1]+i;
    }
    scanf("%ld",&n);
    for (i=0; i<n; i++)
    {
        scanf("%lld",&k);
        for (j=0; j<c; j++)
        {
            t=0;
            if (a[j]==k)
            {
                t=1;
                break;
            }
        }

        ot[i]=t;

    }
    for (i=0; i<n; i++)
    {
        printf("%ld ",ot[i]);
    }
    return 0;

}

Why it is so slowly??? I use standard formula with "sqrt(8*N-7)".

            int L = int.Parse(Console.ReadLine());
            string answer = "";
            for (int i = 0; i < L; i++)
            {
                if (Math.Sqrt(8 * long.Parse(Console.ReadLine()) - 7) % 1 == 0) answer += "1 ";
                else answer += "0 ";
            }
            Console.Write(answer);

I can't imagine shortest solution... What's the matter?
Many people has used C# successful in problem 1209.

Edited by author 28.11.2011 03:09

Делал на питоне:

import math
def solve():
    n = input()
    s = ''
    for i in xrange(n):
        k = input()
        d = math.sqrt((k << 3) - 7)
        result = int(d)
        if result == d and result & 1:
            s += '1 '
        else:
            s += '0 '
    print s

solve()

Выдает TL, как ускорить? Пробовал сразу ответ выводить - на моих тестах это дольше по времени.

k << 3 заменял на k * 8, result & 1 менял на result % 2 - все равно TL. Python 2.7

Для себя понял, что самое времязатратное - ввод/вывод. Поэтому AC получил так:

import math
import sys

stringArray = sys.stdin.read().split()
i = 0
s = ''
for string in stringArray:
    if i == 0:
        i += 1
        continue

    k = int(string)
    d = math.sqrt((k << 3) - 7)
    result = int(d)
    if result == d and result & 1:
        s += '1 '
    else:
        s += '0 '
print s

Но все равно интересует - есть ли способ ускорить первое решение?

for the given 5 sample how 4 values 0 0 1 0 are coming. please explain anybody

Edited by author 03.05.2013 12:20

My program(LANG c)is in the following:
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
int k[60000],value,i,n;
int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++)
       scanf("%d",&k[i]);
    for(i=1;i<=n;i++)
    {
         if(k[i]==0) printf("0 ");
         else
         {
              value=(int)sqrt((k[i]-1)*2);
              if((value*(value+1))==2*(k[i]-1))
                 if(i==n)  printf("1\n");
                 else printf("1 ");
              else
                 if(i==n) printf("0\n");
                 else printf("0 ");
              }
    }
    return 0;
}

BUT it can't compile,said that
"
d9b0f699-b537-45eb-b286-ce7765096c87d9b0f699-b537-45eb-b286-ce7765096c87(16) : error C2668: 'sqrt' : ambiguous call to overloaded functionS:\checker\compile\vc10\include\math.h(589): could be 'long double sqrt(long double)' S:\checker\compile\vc10\include\math.h(541): or 'float sqrt(float)' S:\checker\compile\vc10\include\math.h(127): or 'double sqrt(double)' while trying to match the argument list '(int)'
"

BUT in my computer(Windows 7),I can compile and run it,and the answer is right.
So I don't know where is wrong?And I don't know how to modify it.Help me,please!

Edited by author 22.02.2013 15:09

Hi! Please help russian shoolboy
Where the error is?
I realized that '1'-s are in the field of the sequence.
1 2 4 7 11 16
int main()
{
    long a,j,i,n;
    cin>>n;
    for(int x=0; x<n; x++)
    {
        cin>>a;
        i=1;
        j=1;
        while (a>i)
        {
           i=i+j;
           j++;
        }
        if (a==i)
            cout<<1;
        else
            cout<<0;
     }
    return 0;
}

What's wrong?

var n,p,k,i:longint;
    m:array [1..65535] of longint;
begin
 readln(n);
 p:=1;
 k:=0;
 for i:=1 to n do readln(m[i]);
 for i:=1 to n do
 begin
  while m[i]>p do
   begin
    p:=p+k;
    k:=k+1;
   end;
  if p=m[i] then write(1,' ')
  else write(0,' ');
  p:=1;
  k:=0;
 end;
 end.

Please help
Why am I getting floating-point invalid operation on test 3?
Would it be possible to see the test values?

program pr_1;
var n, i, x: longint;
a: real;
begin
read(n);
for i:=1 to n do
begin
  read(x);
  a:= (1 + sqrt(8*x-7))/2;
  if (Frac(a) = 0.0) then
    write('1 ')
  else
    write('0 ');
end;
end.

Could you  please explain where this formula comes     sqrt (8 * N-7)

Edited by author 05.03.2012 01:14

Edited by author 05.03.2012 01:17

Вот такой код. Хочу найти ошибку. Но найти не могу. Для всех перебранных случаев все вычисляется как надо. Не пойму как пройти тест.
#include <iostream>

using namespace std;

int foo(short pos);

int main()
{
    long i;
    short int y;
    cin>>i;
    for(int cnt_1 = 0; i > cnt_1; cnt_1++){
        cin>>y;
        cout<<foo(y)<<" ";
    }
}

int foo(short int pos)
{
    short int res(1);
    short int i(0);
    do
    {
        res = res + i;
        i++;
    }
    while(res<pos);

    if(res != pos)
        return 0;
    else
        return 1;
}

Какая тут ошибка?

#include <stdio.h>
#include <math.h>

main()
{
      int a,x[65535],i,t,delta;
      float x1;
      scanf("%d",&a);
      for( i = 0 ; i < a ; i++ )
           scanf("%d",&x[i]);
      for( i = 0 ; i < a ; i++ )
      {
            t = (x[i] - 1) * 2;
            delta = 1 + 4 * t;
            x1 = sqrt(delta);
            printf("%d ",fmod(x1,1.0) == 0);
      }
      return 0;
}

#include <iostream>
using namespace std;
int main ()
{
  unsigned int A[65536];
  unsigned short int N,Y;
  cin>>N;
     for (int i=1; i<=N;i++)
     cin>>A[i];
   for (int i=1;i<=N; i++)  {
       int S=0,j=1;
      bool k=true;
 while (k)
    {
      S+=j;
      j++;
     if (A[i]<=S)
      k=false;
     }
      Y=S-A[i];
        if (((j-1)-Y)==1)
            cout<<"1"<<" ";
          else
            cout<<"0"<<" "; }
            return 0;
 }

Edited by author 06.06.2012 18:42

Hi!
Why could be time limit on 3 test here:

    private void solve() {
        int N = nextInt();
        String res = "";
        for (int i = 0; i < N; i++) {
            long idx = nextLong();
            double f = Math.sqrt(8*idx - 7);
            if (f - (int)f == 0) {
                res += "1 ";
            } else {
                res += "0 ";
            }
        }
        out.print(res);
    }

May i don't know some fast way to find if integer is perfect square?
I know that on C++ is all right, but want to understand how to make it in java.

test ID: 4088875

Edited by author 23.01.2012 20:49