a= input().split()

l = int(a[0])
h = int(a[1])
ω = int(a[2])

g = 9.81
ω = ω * 2 * 3.14159 / 60
t = (h / 100) / (g * 0.5)
x = ω * t * l

if x % l <= l/2:
    print("butter")
else:
    print("bread")

Edited by author 21.10.2021 22:47

w = N/t
h = g*t*t/2
this is also if l/2 >= h you must print "BUTTER" and you stop your program else continue

1 1 1
butter

1 1 100
butter

1 1 900
bread

1 1 1000
bread

100 10 1000
butter

{bu-ga-ga};

Edited by author 29.01.2011 00:33

I can't understand what i do not correctly. Help me, please.

At first I wrote a tricky solution using trigonometry and binary search.  And I was unable to pass through test case No.  12.
Then I wrote simple two-step solution and got AC.

Can anybody help me with this test case?

I was trying many times to get past first test and fixed a few mistakes in the algorithm to no avail.
And the reason to fail was that the output was in brackets.

I think when the toast rotates an angle of w after t seconds, it will land if only the distance it traveled ( 0.5 * 981 * t * t ) plus its projection on the orthogonal ( 0.5 * l * |sin(w)| ) equals to h.

But I got WA when implementing this approach, then change the latter expression into l/2 and got AC. Could anyone explain this?



Edited by author 18.12.2011 02:22

Where is the error in my code?
Please, help...

#include <stdio.h>
#include <conio.h>
#include <math.h>

int main() {
    float l, h, w, n, t;

    scanf("%f%f%f", &l, &h, &w);

    h /= 100;
    l /= 100;
    w /= 60;

        h -= l/2;
        t = sqrt(2*h/9.81);
        n = t*w;
        n -= (int) n;
        if(n <= 0.25 || n >= 0.75) printf("butter");
        else printf("bread");


    return 0;
}

Got AC 0.015 with this magic code

#define forn(i,n) for (int i = 0; i < (int)(n); i++)
typedef long double ld;
ld g = 9.81l;
int main() {
    ld l, h, w;
    cin >> l >> h >> w;
    l /= 100, h /= 100, w *= acosl(0) / 15;
    ld t = 0;
    forn (i, 1000) {
        ld d = fabsl(sinl(w * t) * l / 2);
        //g x^2 /2 + (x-t)wl/2 + d = h
        //x^2 g/2 + x wl/2 - wlt/2+d-h = 0
        t = (-w * l / 2 + sqrtl(w * w * l * l / 4 + 2 * g * (w * l * t / 2 - d + h))) / g;
    }
    cout << ((cosl(w * t) > 0) ? "butter": "bread");
}

the next moment found the normal solution.

Edited by author 19.10.2011 12:17

  l/=100;
    h/=100;
    w/=60;
    if (l/2 >= h)
       cout<<"butter"<<endl;
    else
    {
      h -= l/2;
       t = sqrt(2*h/g);
        ob = t * w * 360;

      if (ob >= 90 && ob <= 270 )
          cout<<"bread"<<endl;
       else
          cout<<"butter"<<endl;

    }

Edited by author 29.04.2011 01:12

Edited by author 27.04.2011 00:25

var w,g,h,l,t:real;
begin
g:=981;
read(l,h,w);
if l/2>h then begin write('butter'); halt end;
t:=sqrt(2*(h-l/2)/g);
w:=frac(w*t/60);
if ((w>=0) and (w<=1/4)) or ((w>=3/4) and (w<=1)) then write('butter') else write('bread');
end.

I thing that I don't need length of sandvich.
t=sqrt(2*h/g)
n=(t/60)*w
May be it's wrong? Please answer.

do anybody know why its wrong
#include <cstdlib>
#include <iostream>
#include <math.h>
using namespace std;

int main(int argc, char *argv[])
{
    float l, h, omega;
    double long wynik, obroty_wykonane;

    cin>> l >> h >> omega;
          obroty_wykonane=((omega/60)*sqrt(((h-(0.5*l))/450.5)));
          wynik=obroty_wykonane-floor(obroty_wykonane);
    if (wynik>0.25 && wynik<0.75) {cout << "bread"<< endl;}
    else {cout << "butter"<< endl;}
    return EXIT_SUCCESS;
}

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");
}

Edited by author 28.08.2011 02:03

В обоих случаях из примера бутерброд за время падения не совершает ни 1 полного оборота, соответственно он падает в обоих случаях маслом вниз.

Скорость вращения - 1/4 оборота в секунду.

Время падения 1 раз = 0,9.
Количество оборотов 1 раз = 0,2258.

Время падения 2 раз = 1,01.
Количество оборотов 2 раз = 0,2524.

В обоих случаях бутерброд НЕ совершает ни 1 полного оборота и должен упасть как и был, маслом вниз.

Почему же ответы для 1 примера - "маслом вниз", а для 2 примера - "хлебом вниз"?