This is one of the easiest problem in timus and I can't believe i'm getting WA1 for this problem. Im solving this problem using C++ and I'm using getline method from sstream to get the input. I'm storing the names in vectors and printing them with empty lines between the houses. I know Test 1 is usually the given data, but im getting WA1 even though I'm getting the same answer in my computer as given in the problem description. Does anyone know how to overcome this WA1 problem?

Edited by author 03.01.2012 15:53

My solution jumped from TL 19 to AC in under 0.3s when i replaced output via printf with writing each digit of a number via putchar. That just blew my mind.

Edited by author 09.12.2013 11:44

Could you tell me what is test N9?

O(k*(sqrt(n)/log(n) + log(n))).
hint: precalc primes table(so, we can factorize number using O(sqrt(N)/log(N)) operations).
gl.

Can somebody help me?
I don't know what to do. I tried a lot of tests and all answers were correct.
My code:
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<stdlib.h>
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
int main()
{
    long a, b, x, y, z, p, q;
    char ch;
    scanf("%d%d", &a, &b);
    scanf("%s", &ch);
    scanf("%d%d%d", &x, &y, &z);
    a=min(a, 100);
    if(ch=='A') scanf("%d%d", &p, &q);
    if(ch=='P') scanf("%d", &p);
    if(ch=='P') if(x>0) ch='D';
    else
    {
        if((abs(p)>20)&&(abs(p)<160)){
        if(((p>20)&&(p<=(90)))||((p<=-90)&&(p>-160)))
        printf("LEFT %d", a);
        else printf("RIGHT %d", a);
     }
     else if(abs(p)<=20) printf("FRONT %d", a);
     else printf("BACKWARD %d", a);
    }
    if(ch=='D') if((x>0)&&((x*20)>=b))
    if(abs(z)<5) printf("FIRE %d", min(20, b));
    else printf("BACKWARD %d", a);
    else ch='G';
    if(ch=='G') {
        if(x==0) printf("STOP");
        else {
            if(abs(z)<5) printf("FIRE %d", min(20, b));
            if(z>=5) printf("LEFT %d", a);
            if(z<=-5) printf("RIGHT %d", a);
        }
    }
    if(ch=='A') if((p*q)>(3*(x*y)))
    if((x>0)&&(abs(z)<10)) printf("FIRE %d", min(20, b));
    else printf("FRONT %d", a);
    else if((x>0)&&(abs(z)<5)) printf("FIRE %d", min(20, b));
    else printf("BACKWARD %d", a);
}

My code:

why WA#6

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

public class misol1607
{
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int a = in.nextInt();
        int b = in.nextInt();
        int c = in.nextInt();
        int d = in.nextInt();
        int r=c;
        while (a<c)
        {
            if ((a+b)>=c)
            {
                r=c;
                break;
            }
            else
            {
                a=a+b;
            }

            if (a>=(c-d))
            {
                r=a;
                break;
            }
            else
            {
                c=c-d;
            }
        }

        out.println(r);

        out.flush();
   }
}

double h[131072],a,s;
int main() {
    int i=0,j;
   freopen("input.txt", "rt", stdin);
    while(cin>>a){
    h[i]=a;
    s=sqrt(a);
    h[i]=s;
    i++;
    }
    for(int j=(i-1);j>=0;j--){
        cout<<setiosflags(ios::fixed)<<setprecision(4)<<h[j]<<endl;
    }
    return 0;
}

Can anyone explain me test #1;
London keeps 1st from day 12 to 21 (10 days) and 23 to 25 (3 days). Day 22th London = StPetersburg (=15b). Why output is London 14?

uses
  SysUtils;
var
  a : array[0..10000] of longint;
  n, m, i, x, y : longint;
begin
  readln(n);
  if n = 1 then
    writeln(1)
  else if n mod 4 = 0 then
    writeln(0)
  else if (n mod 4 = 3) or (n = 25) or (n = 5) then
    writeln(2)
  else
    writeln(1);
  readln;
end.

15
<>><<>
<<><<<
>><

answer is :
28.

my program is accepted.

I have solved it using max bipartite matching. Is there any better idea?

I'm wrong with test 7, can someone give me some test?
Thanks.

Edited by author 06.12.2013 09:29

Edited by author 06.12.2013 09:29

Огорчает одинаковый лимит времени для компиляторов и интерпретаторов. На Джаве приходится уже не оптимизировать, а извращаться. Может, есть смысл ввести разумный коэффициент ?

#include <iostream>
using namespace std;

int main ()
{

    int n[1755],z=2,a=0,y,t=0;
    for (z=2;z<=15000;z++)
    {
        for (int i=1; i<=z; ++i)
        {
            y=z/i;
            if (y*i==z)
            {
                a=a+1;
            }
        }
        if (a==2)
        {
            n[t]=z;
            t++;
        }
        a=0;
    }

    int x;
    cin >> x;
    int *b=new int[x];
    for (int e=0; e<x; ++e)
    {
        cin >> b[e];
    }
    cout << "\n" << "\n";
    for (int e=0; e<x; ++e)
    {
        cout << n[b[e]-1] << " ";
    }
    system ("pause > void");
}

#include <iostream>
#include <cmath>
using namespace std;

int main ()
{
    int x,e;
    double nk;
    cin >> x;
    int *b=new int[x];
    for (e=0; e<x; ++e)
    {
        cin >> b[e];
    }
    for (e=0 ; e<x; ++e )
    {
        nk = (pow((8*b[e]-7),0.5)-1)/2;
        long int k = static_cast< int >( nk );
        if ( nk==k )
        {
            cout << "1 ";
        }
        else
        {
            cout << "0 ";
        }
    }
    system ("pause > void");
}

Edited by author 05.12.2013 19:32

Please write answer for N = 169, 841, 674041.

please)