WA5 can sb give me some hints?
wa5 again and again

61 submits lost AC verdict (many because of tricky 12 test).
Try to find a bug in your source code. Good luck!

P.S. Thanks to Alias (Alexander Prudaev) for idea of new tests.

I have o(n * n) algorithm, not many calculations inside cycle, but still TLE12. Imho, everybody writes such a solutions, but in c++ in can be accepted in 0.015, but i have 1.031 (or more) in java. That's not fairly. I ask autors to increase TL up to 2 seconds, or maybe change your time checking system to prevent this...

My program passed the tests, published in this thread, but it couldn't passed test 5.
What is in this test?

Edited by author 05.03.2008 20:18

What precision should we use in output?

Could I have WA#17 because of it?

Is double (long double) enough for us?

Even after anti-greedy tests were added, it is possibly to solve the problem with greedy algo. Just randomize directions in which you make "greedy walk" of the incidence matrix. The best thing is that it is impossible to kill such an algo with any anti-greesy tests! =)

RT.

TXs.

#include <complex>

using namespace std;

complex<double> X;

I've found a simple formula for this integral, but it uses p1, q1, p2, q2 such that P(x)= a4 ( x^2 + p1 · x + q1) · ( x^2 + p2 · x + q2).
These values are always real, but I don't know how to find them.

Maybe there is a solution which doesn't use them?

In the problem description we know that all output should be positive integers.But I submit a solution which contains '0' but it got accepted.I think the Check Program has something wrong.

Anyone with a an idea for a test that can fix this?

public static void main(String[] args){
        Scanner in = new Scanner(System.in);
    int k,ans=0,start,stop;
    double r;
    boolean p;

    start=in.nextInt();
    stop=in.nextInt();
    r=2147483647;
    if (start==1) System.out.println(1); else {

        for (int i=stop; i>=start; i-- ){
            p=true;
            for (int j=2; j<=Math.round(Math.sqrt(i)); j++) {
                if ((i % j)==0) {
                    p=false;
                    break;
                }
            }
            if (p) {
                ans=i;
                r=1/i;
                break;
            }
        }

        for (int i=stop; i>=start; i--){
            if (Math.abs(Math.sqrt(i)-Math.round(Math.sqrt(i)))<(1e-8)) {
                if (r>(1+Math.sqrt(i))/i)  ans=i;
                break;
            }
        }
        if (ans!=0) {
            System.out.println(ans);
        } else {
            for (int i=start; i<=stop; i++){
                k=1;
                for (int j=2; j<=Math.round(Math.sqrt(i)); j++) {
                    if ((i % j)==0) {
                        k+=(j+(i / j));
                        if (j==(i / j))  k-=j;
                    }
                    if (r>(k/i)) {
                        ans=i;
                        r=k/i;
                    }
                }

            }
            System.out.println(ans);
        }
    }
    }


Where is my mistake&

Edited by author 25.03.2008 18:19

Hi, i'm having problems with the input of this task.

How does my program know when the input is over?

It would be very helpful if you could give me a code sample (only the input, not the actual task :) ). Something like:

#what libraries the input needs

while(way to know the input is over)
{
way_to_read (scanf, cin ...something else?)
}

Thanks, I believe if i get this part, the rest will work first time :)

Edited by author 13.03.2008 12:29

like
1 1
.
when there is one room,output what?

1. Precision of 1e-10 seconds is enough to compare boil-time with wake-time.

2. If some guy wakes up at the same time the water is ready (I mean exact equality up to precision), the water is disposed to cups first. So he will fill up an empty tank in this case.

3. Output times after rounding them UP. It is not quite clear from problem statement, but it's the closest guess if we assume that drinking can start only at integer number of seconds. At least this approach gives AC.

4. Output rounded times modulo 24:00:00 (i.e. if cup is ready the next day, subtract 24 from HH).

Can you tell me the 2nd test

I am new here, so i am not understanding the output format of this problem ( how many digit after the decimal point).
could anyone please tell me about this...........

Edited by author 17.08.2008 16:38

#include <iostream>
using namespace std;
int main()
{
    int n,k,m;
    cin>>n;
    while(n)
    {
        cin>>k>>m;
        if(!k && !m)
            cout<<"1\n";
        else if(k<m)
            cout<<"0\n";
        else
            cout<<(float)k/(k+m)<<endl;
        n--;
    }
    return 0;
}

Edited by author 17.08.2008 15:09