#include <iostream>
#include <conio.h>
using namespace std;

void find(int j, int i, int &k1,int &k2)
{
     if (j==i+1) return;
     else
     {
         int buf = k1;
         k1 = k1 + k2;
         k2 = buf;
         find(j-1,i,k1,k2);
    }
}

int main()
{
     long long Fi, Fj, f1,f2, f3;
     int i,j,n;
     cin>>i>>Fi>>j>>Fj>>n;
    if (j==n)  {cout<<Fj<<endl; return 0;}
    if (i==n)  {cout<<Fi<<endl; return 0;}
     if (i<j)
     {
               int buf = i;
               i = j;
               j = buf;
               buf = Fi;
               Fi = Fj;
               Fj = buf;
    }
     int k1 = 1, k2 = 1;
     f1 = Fj;
     if (i!=j+1)
     {
        find(i-1,j,k1,k2);
        f2 = (Fi - k2*Fj)/k1;
     }
    else
        f2 = Fi;

     int lim;
     if (j<0) lim = n - j -1;
     else
     if (j == 0 ) lim = n;
     else lim = n - j +1;
     for (int count = j +1; count < lim; count++)
    {
         f3 = f1 + f2;
         f1 = f2;
         f2 = f3;
    }
     cout<<f3<<endl;
     return 0;
}

I think I have correct solution.(DP)
but i got WA#11. pls give some tests.

Edited by author 29.01.2007 21:21

using System.IO;
using System;

class Program
{
    static void Main()
    {
        int size = Int32.Parse(Console.ReadLine());
        int[,] arr=new int[size,size];
        int count=1;
        for(int x=size;x>=0;x--)
        {
            int y=0;
            int xx=x;
            bool exc=false;
            while(!exc)
            {
                try
                {
                    arr[xx,y]=count;
                    count++;
                    xx++;
                    y++;
                }
                catch
                {
                    exc=true;
                    break;
                }
            }
        }

        for(int y=1;y<=size-1;y++)
        {
            int x=0;
            int yy=y;
            bool exc=false;
            while(!exc)
            {
                try
                {
                    arr[x,yy]=count;
                    count++;
                    x++;
                    yy++;
                }
                catch
                {
                    exc=true;
                    break;
                }
            }
        }

        for(int x=0;x<size;x++)
        {
            for(int y=0;y<size;y++)
            {
                Console.Write(arr[y,x]+" ");
            }
            Console.WriteLine();
        }

    }
}

Can Anyone give me some hints for test 3, I've been working on it for 3 days but still got test 3. I give the right answers for all the tests on forum

import java.io.BufferedReader;
import java.io.InputStreamReader;
public class AB {
    public static void main(String[] args) throws Exception
    {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int a = Integer.parseInt(reader.readLine());
        int b = Integer.parseInt(reader.readLine());
        System.out.println(a+b);

    }
}

Please, send me test 8

I got WA at #4 by using int type but got AC by using long long

I don't understand...
I think the max of date is 1e6(maybe n=1e4 and every ti=100, in this way the sum time is 1e6 and it's max)
Am I wrong? Who can tell me plz

Edited by author 15.03.2015 21:47

Здравствуйте, уважаемые администраторы! Я очень давно изучаю язык Action Script принадлежащий компании Adobe. У меня созрел такой вопрос: Вы добавите поддержку языков Action Script или отодвините всех программистов этой отрасли?

Can someone tell me the test case 2 please.

I am new in here and I need some help with 1017. Staircases.

Edited by author 14.03.2015 15:34

Edited by author 14.03.2015 15:34

What may it be? i've try any test i can think and any test in other discussins,but still WA #11.
I tested :
-------
WHAT ARE YOU DOING?
- I AM EATING.
- really?
-Yes,(new line)
REALLY!
- ok, i KNOW.
i TRUST YOU.
-------
and my program output :

-------
What are you doing?
- I am eating.
- Really?
-Yes,
really!
- Ok, i know.
I trust you.
-------
I don't know what's wrong, help, please.

For my first try I wrote this:

def sum_of_digits(n)
  sum = 0
  while (n > 0)
    sum += n % 10
    n /= 10
  end
  sum
end

num_digits = gets.chomp.to_i

exponent = num_digits/2
divisor = 10**exponent

limit = (10**num_digits) - 1
total = 0

0.upto(limit) do |cur|
  upper = cur / divisor
  lower = cur % divisor
  total += 1 if sum_of_digits(upper) == sum_of_digits(lower)
end

puts total

# eof

For 2, 4 and 6 digits it got the requisite answers of 10, 670, and 55252.  There was a bit of a pause while it crunched away at the 6-digit case, though.

For 8 digits it … well, I let it run and went to do something else.  Then I had the bright idea to time it (on mac OS X, using the time command):

4816030

real    1m42.596s
user    1m34.222s
sys    0m1.837s

The right answer, but the last time I checked, one minute anything was greater than two seconds.  What to do?

It’s pretty clear that calling sum_of_digits a bunch more times than necessary is the biggest problem. For try #2, I precomputed the sums and stuffed them into an array, and indexed the array in place
of calling sum_of_digits in the main loop.  Here’s the result:

time ruby lucky.rb < t8.txt
4816030

real    0m19.176s
user    0m18.017s
sys    0m0.313s

A better than 80% speedup.  Most times I would kill for that, but I still need to lose 95% of the remaining run time.  How to do it?

I checked … initializing the array of sums takes virtually no time, even if I write out the entire array contents to a file.

So the main loop is killing me, and unfortunately I need a dramatic change of algorithm but I
don’t see one. So …

1. Recode in C or C++
2. Hell if I know

Taking option 1, I got Accepted, with run times around 1.1 seconds or so for 8 digits.  It's a straightforward translation to C++, very brute force.

Try this test:
-1 -1
1

Answer:
0.00
0 0

Solved it with time of 0.5 sec and using "Trie" where I keep the name of each directory as a node.

Write me novopashinwm@mail.ru

Give me some input values with answers, please, so I could find out, what's wrong.

  if(s.equals("Blue")|| s.equals("blue"))
  if(s.equals("Red") || s.equals("red"))
  if(s.equals("Yellow") || s.equals("yellow"))

Can anybody explain the combinatorial approach of this problem ???
  Greetings

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

public class MathSqrt
{
    public static void main(String[] args) throws Exception
    {
        Scanner sc = new Scanner(System.in);
        double[] array = new double[4];

        for(int i = 0; i < array.length; i++)
        {
            array[i] = sc.nextDouble();
        }

        for(int i = array.length - 1; i >= 0 ; i--)
        {
            double b = Math.sqrt(array[i]);
            System.out.println(b);
        }

    }
}

Edited by author 18.02.2015 17:39

Edited by author 18.02.2015 17:39

idea...!