1 1
1000000000

Edited by author 24.01.2011 11:02

Does any one know that case?

Edited by author 24.01.2011 10:45

RT

Greetings!
I've been reading FAQ about C# and noticed, that memory boundaries aren't specified via command-line arguments (as they are in Java). Isn't it a secret, how Timus team manage to set memory limits for c# then?

// The thing is that I will be first-time helping with organization of some contests in our region, so my interest has a practical nature. :-)

I do not understand when you have finished receiving data, and that if no effect then enter receive data never end, so never would show the results and would be an endless repetitive cycle.
now with that key, or as I have to request new data, since no effect enter then I have no idea how to stop the cycle does not specify how to stop it.
greetings.

Here's my solution :
- Input & reverse all the edges
- DFS from all vertexs
- If depth = N then output in reverse order

What's wrong w/ this one ????

I was WA#2 and now I get Ac,
because: Graph can be not connected

i have written algorithm that uses log2() which is in
c++(<cmath> library) and when i used it copmpilation error occured "error C3861: 'log2': identifier not found"
when i compile it in devc++ it works perfectly..
can everyone helm me? :(

Author "Progbeat"  -
http://acm.timus.ru/author.aspx?id=39102

Look at his recent accepted submissions.

for i := 2 to n do
    begin
      min:=maxlongint;
      for j := 1 to trunc(sqrt(i)) do
        if min>f[i-j*j] then
          min:=f[i-j*j];
      f[i]:=min+1;
    end;

rt
who can help me?
I need help!

What is the test number 3?

with code like this
if(lmastery-dmastery>maxdif)
    maxdif = lmastery - dmastery;

i got WA 10

but with
if(lmastery-dmastery>maxdif+1e-9)
    maxdif = lmastery - dmastery;
i got AC

Why I have worng anserw in test #2??

My code:

#include <cstdio>
#include <iostream>
#include <string>

using namespace std;

int main()
{
    int n;
    cin >> n;
    string nazwa[5000];
    int wartosc[1000], wartosc2[1000], x, koniec = n-1, pocz = 0, min = 1, max = 10000, summin = 0, summax = 0;
    for (int i=0; i<n; i++)
    {
            cin >> nazwa[i];
            cin >> x;
            if (x == 1)
               cin >> wartosc[i];
            if (x == 0)
                wartosc[i] = 0;
            wartosc2[i] = wartosc[i];
    }

    for (int i = koniec; i>=0; i--)
    {
        if (wartosc[i] == 0)
                       wartosc[i] = min;
        else
            min = wartosc[i];
        summin += wartosc[i];
    }

    for (int i = pocz; i < n; i++)
    {
        if (wartosc2[i] == 0)
                       wartosc2[i] = max;
        else
            max = wartosc2[i];
        summax += wartosc2[i];
    }

if ( summin < 10000 && 10000 < summax)
   cout << "YES";
else
    cout << "NO";
    getchar();
    getchar();
    return 0;
    system("PAUSE");
}


Edited by author 24.01.2011 01:13

There is theory of Ramsei.
if you want 4 you need 5 points.
if you want 5 you need 9 points.

Edited by author 03.02.2009 14:53

Use bitset !

here is my code
int main()
{

double l, h, w,N,angle;
double P=3.14;
double g=9.81,t,speed;
cin>>l>>h>>w;
t=sqrt(2*h/(g*100));
speed=w/60;
N=t*speed;
angle=N*360;
while(angle>360)
{
    angle=angle-1;
}
if(angle>90 && angle<270)
cout<<"bread";
else
cout<<"butter";


//system("pause");
}

is the problem as easy as it seems or there is some trick?
ACM problems r never easy except a very few which i have solved.please provide some test cases.