As you can see the task is to output the sequence consists of 1000000
symbols. I output exactly 1000000 symbols one by one (...Write
(sym)...Write(sym)...Write(sym)...), buf after 0.04 seconds I
receive "Output Limit". What does it mean and how should I output the
sequence?
Please, help me!

Hi, Maybe I misunderstood the problem. With the given numbers, you could actually make two piles 27,8 in one and 13,14,5,5 in the other, and the difference would be just 2. Where did I misunderstand?

My code sorts all the weights in descending order and then arranges them into two piles using algorythm:

1. Start with the heaviest stone;
2. If weight(pile1)<Weight(pile2) add current stone to pile1;
   Else add current stone to pile2;
3. If there are more stones current stone = next stone; Goto 1;

Edited by author 30.01.2011 13:57

Edited by author 30.01.2011 13:57

Edited by author 30.01.2011 13:58

Edited by author 30.01.2011 14:41

I had WA28. After two hints I get AC.
1) check if A=B
2) check if sphere has intersections with segment using long long integers, not doubles

Edited by author 06.02.2011 08:52

[hidden code:)]

Edited by author 03.02.2011 14:33

Here is my solution. It works fine in eclipse, but it crashes in here.
Could you tell me what is wrong?

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Sum {

    public static void main(String[] args) {

    int x,y;
    long z;

    try {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        x = Integer.parseInt(bf.readLine());
        y = Integer.parseInt(bf.readLine());
        bf.close();
        z = x+y;
        System.out.println(z);
    } catch (IOException e) {
        System.out.println(e);
    }

    }
}

x=4*a+1;
s=d+2*a;

Does anybody knows what is the test 6?

HELP!!!






#include <iostream>

using namespace std;

int main()
{
    int const N=1024;
    int n, a[N], b[N], P=1;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        b[i]=a[i];
    }


    for(;;)
    {
        int s=0;
        for(int i=0;i<n;i++)
            if(a[i]==i+1)
                s++;

        if(s==n)
            break;

        else
        {
            P++;
            for(int j=0;j<n;j++)
                a[j]=b[a[j]-1];
        }
    }

    cout<<P<<endl;



    return 0;
}

HELP!!!

using System;
using System.Collections.Generic;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            double i, j,a;
            double display=0;
            Console.WriteLine("enter the number:");
            double n = double.Parse(Console.ReadLine());

            try
            {

                for (i = 1; i <= n; i++)
                {
                    for (j = 1; j <= n; j++)
                    {
                        if ((i * j) == n)
                        {
                            a = double.Parse(string.Concat((object)i, (object)j));

                            if (i == 1)
                            {
                                if (n == 1)
                                    display = 1;
                                else
                                    display = a;
                            }
                            else
                            {
                                if (a < display)
                                {
                                    display = a;
                                }
                            }
                        }
                    }
                }

                Console.WriteLine(display);
            }
            catch (NotFiniteNumberException)
            {
                Console.WriteLine(-1);
            }
        }
    }
}

Edited by author 01.02.2011 15:27

I used dp and this is my program:
var f:array[0..500,0..500]of extended;
    i,j,n:longint;
function min(a,b:longint):longint;
 begin
  if a<b then exit(a);
  exit(b);
 end;
begin
 readln(n);
 fillchar(f,sizeof(f),0);
 f[0,0]:=1;
 for i:=1 to n do
  for j:=1 to i do
   f[i,j]:=f[i,j-1]+f[i-j,min(j-1,i-j)];
 writeln(f[n,n]:0:0);
end.

But I want to know a faster algorithm,
could you please tell me?

#include<iostream>
using namespace std;
int main(void)
{
    int ary1[1000],ary2[1000],n=5,input,result=1,m=0;
    cin>>n;
    if(n<1&&n>1000)
        exit(0);
    for(int i=0;i<n;i++)
    {
        cin>>input;
        ary2[i]=ary1[i]=input;
    }
    for(int x=0;x<n;x++)
    {
        if(ary1[x]!=(x+1))
            break;
        else
            m++;
    }
    if(m==n)
    {
        cout<<result;
        exit(0);
    }
    else
        m=0;
    while(1)
    {
        for(int j=0;j<n;j++)
        {
            ary2[j]=ary1[ary2[j]-1];

        }
        result++;
        for(int k=0;k<n;k++)
        {
            if(ary2[k]!=(k+1))
                break;
            else
                m++;
        }
        if(m==n)
        {
            cout<<result;
            exit(0);
        }
        else
            m=0;
    }
    return 0;
}

I thought the algo is to find the loop of continued fraction.
But it seems that if the loop is too small, 10^9 would make this algo fail.

Give some hint plz.

Hello,

The International Institute of Information Technology, Hyderabad, India
cordially invites you to be a part of Felicity, the annual cultural and
technical festival, to be held from 18th - 20th February, 2011.

1) Codecraft 2011

CodeCraft 2011, the online programming competition is all set to start on
6th February,2011. Codecraft requires eight hours of brainstorming over
some of the toughest algorithms challenges from various fields of Computer
Science.

Where : http://felicity.iiit.ac.in/codecraft/
When  : 6th February 2011, 2 pm - 10 pm IST(GMT +5:30).

2) Time Limit Exceeded

This is an unusual programming event which emphasizes on solving problems
under special constraints. The event is based on C/C++ with the scoring
depending on factors like code length, deviation from the best available
solution, or even the number of semi-colons and whitespaces used.

Where : http://felicity.iiit.ac.in/tle/
When  : 11th February 2011 2pm to 13th February 2011 2pm (48 hours [2 days])

IST (GMT +5:30).

You need to REGISTER online to participate in these event. There are
prizes in each of the mentioned events, details of which will be put on
the site in due time. Kindly visit the respective sites for more details.

For queries , mailto : codecraft.11@gmail.com , felicity11.tle@gmail.com.

Can send SMS from a to b through z, (a -> z -> b), or can only
use one direct edge... if possible the cost would be the sum, or the maximum on the path???

Hello what should I print if the answer is integer, say 1:

1/1

or

1
?

Thank you.