I didn't do anything for 'arbitary' sequence. But my solution got accepted.Can anybody help me clear this doubt?

Edited by author 22.01.2019 18:38

Edited by author 09.03.2021 22:26

10
6
1 2 even
1 1 even
3 4 odd
5 6 even
1 6 even
7 10 odd

i think the answer of this data should be 1,but my program answer is 4 but it return an accepted!

sort(a + 1, a + 1 + n);
  reverse(a + 1, a + 1 + n);

  for(int i = 1; i <= n; i++){
    if(sum1 > sum2){
      sum2 += a[i];
    }
    else{
      sum1 += a[i];
    }
  }

  cout << abs(sum1 - sum2);

If you write on c++ you can do this:

char dot;
unsigned int x,ip;

ip = 0;

for (int i = 24; i >= 8; i -= 8) {
  cin >> x >> dot;
  x <<= i;
  ip |= x;
}
cin >> x;
ip |= x;
------------------
Now you have ip, same you can read mask.

Edited by author 21.01.2019 14:50

Edited by author 21.01.2019 14:52

На Паскале не работает код если начать его так:

read(n); read(s); (или readln(s))

n - longint, s - string

Но если так:

readln(s);

А затем разрезать на n и s1, то всё ОК.
Разве так и должно быть?

#include <iostream>
using namespace std;
 long long  a,b,t,c,d,k,s,f;

int main ()
{
    c = 1; k = 0;
   cin >>t;
   for (t;t>=1;--t) {
     cin >>a;
      cin >>b;
   if ( b%a==0)
   {
       for( ; d != b; ++c )
       {
         d = a*c;
          if ( b % d == 0)
          {
              k= k+1;
              a = d;
              c = 1;
          }
       }
       cout << k<<"\n";
   }
   else
      cout << "0\n";
   c=1; k=0;
   }
    return 0;

}

Hi there!
I'm curious regarding submission metrics time and memory. Is this the avg time and avg memory of all the tests or maximum time and maximum memory used in the worst test?

Can anybody please reveal how does the system count time and memory for different languages? I would like to reproduce it locally before the submission. I use golang.
Thanks

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

namespace phrase
{
    class Program
    {
        static int First(string input, string[]array)
        {
            int col = Convert.ToInt32(input.Length);
            if (col>0&&col<=1000)
            {
                int countArr = 0,countStr = 0,c = Convert.ToInt32(array.Length),count = 0,indexStr=0;
                for (; countArr < c; countArr++)
                {
                    for (; (indexStr = input.IndexOf(array[countArr], countStr)) !=-1; countStr = indexStr + 1)
                    {
                        count++;
                    }
                    countStr = 0;
                }
                return count;
            }
            return 0;
        }
        static void Main(string[] args)
        {
            string[] one = new string[] { "a", "d", "g", "j", "m", "p", "s", "v", "y", ".", " " };
            string[] two= new string[] { "b", "e", "h", "k", "n", "q", "t", "w", "z", "," };
            string[] three = new string[] { "c", "f", "i", "l", "o", "r", "u", "x", "!" };
            string input = Console.ReadLine();
            int a = First(input,one);
            int b = First(input,two);
            int c = First(input,three);
            Console.WriteLine(a+(b*2)+(c*3));
        }
    }
}

var
  i,n,l,j:longword; found:boolean;
begin
  readln(n);
  repeat
  found:=False;
  for j:=3 to n div 2 do
  begin
    if n mod j=0 then
    begin
      found:=True;
      break;
    end;
  end;
  if found then n:=j;
  until not found;
  writeln(n-1);
  readln;
end.

Please help me to optimize my code. It's 0.405 sec

#include<stdio.h>
#include<stdlib.h>
#define x 100000

int main()
{
    long long int n , c=0,i;
    double ar[x] = {};

    while(scanf("%lld",&n)){
        if(n < 0) break;
        ar[c++] = n;
    }

    for(i=c-1; 0<=i; i--){
        printf("%.4lf\n",sqrt(ar[i]));
    }

    return 0;
}

var

  i, n: integer;
  c : real;

begin
  read(n);
  c := 0;
  if n > 0 then
  begin
    for i := 1 to n do
      c := n;
    write(c);
  end;
  if n <= 0 then
  begin
    for i := 1 downto n + 1  do
      c := (-((-n) * (1 - n) / 2) + 1);
    write(c);
  end;
end.

I think the answer is probablye Yes...

Be sure that(at least for the given interval, [4,2*10^9]) every even number can be shown as sum of two prime numbers.

1. if n is prime, {n}
2. if n is even, {n = p1 + p2}
3. if n is odd   {n = 3 + p1 + p2} ( n-3 is even, so that recall second case.)

For convenience, prime numbers under 10^5 can be initialized.

Edited by author 03.12.2012 03:00

Edited by author 03.12.2012 03:00

Weird.

Edited by author 12.01.2019 12:15

C++ costs 200 kb (Visual C++ and 400 kb other compilers).