Does anyon know, what is the longest test? I check a couple of numbers,with k= 99998 execution was the longest, but i'm not sure.

UPD: I check (write k=99998 instead a reading  number), and if WA#1 said truth, i reach the answere in 0,71. But this decision is still have TLE#41 :(

Maybe, there is a harder 'k' for this task?

Edited by author 07.08.2018 11:17

Edited by author 07.08.2018 11:17

i am doing (n%k)+(n/k) after putting restriction on k(k<=1000) and n(n>=1). but my answer is  said to be wrong.please help.

Edited by author 30.06.2013 01:11

"Some tram drivers are fans of fast driving, and they damage both rails and their trams. If a tram accelerates to a high speed, say 80 kilometers per hour, and then brakes sharply before a stop, it goes some distance skidding."
Is this "initial D" reference?!

Edited by author 19.08.2018 20:53

Had problem with 25th test because of integer overflow. My program used int64_t which was not enough with my approach. Casting computations to double hepled me.

The statement says

"The number of intermediate levels of employees between an arbitrary employee who has subordinates (an ordinary employee) and the employee who has no superiors (the main superior) is the same for all ordinary employees."

However, in the first paragraph, it says that "ordinary employee" means an employee who has NO subordinates. Therefore, I think the statement should say:

"The number of intermediate levels of employees between an arbitrary employee who has NO subordinates (an ordinary employee) and the employee who has no superiors (the main superior) is the same for all ordinary employees."

I looked at the Russian version of the statement with google translate and that seems to be the case.

Edited by author 06.08.2018 11:30

4*h *  n ( n + 1 )  > H
minimal  n - is answer!

but I doubt, that this is correct.

It's correct or not?

#include<stdio.h>
int main()
{
int n,m,sum;
scanf("%d \t %d", &n ,&m);
if(n<=4000 && m<=4000){
sum=n*(m+1);
printf("%d",sum);
}
return 0;
}

http://s020.radikal.ru/i702/1410/52/a428d102a994.jpg

Why?



Edited by author 11.10.2014 19:51

I Solved it with java BigInteger.
Is there any solutions without long arifmetic?

Ура! Решил с длинной арифметикой.
eps = 0.0000000000000000000000000000000000000000000000000000000000000000001

Задача геометрически простая, но реализация в целой арифметике неприятная (ответ всё равно будет вещественный, но алгоритм можно сделать абсолютно точным при сравнении точек). Лучше использовать длинную вещественную арифметику.

Edited by author 06.08.2018 12:36

The same algo gives tl 4 in java ... In C++ I use vector of vectors for the capacity matrix as well as in java and the input is with the Scanner class . How can I optimize ? Thanks in advance.

#!python3
s = """0 30
-25.9807621135 -15
25.9807621135 -15
3.711537444785714 10.714285714285715
-11.134612334357143 -2.142857142857144
7.423074889571431 -8.571428571428571
18.557687223928568 23.57142857142857
-29.69229955828571 4.285714285714285
11.134612334357145 -27.857142857142854"""

import matplotlib.pyplot as plt
from math import cos, sin, pi
fig = plt.figure()
ax = fig.add_subplot(111,aspect='equal')

r=[[float(a) for a in p.split()] for p in s.split("\n")]
def d(a,b): return (r[a][0]-r[b][0])**2 + (r[a][1]-r[b][1])**2
def z(a,b): return abs(a-b) < 1e-6

a = 0
for i in range(len(r)):
    for j in range(i+1,len(r)):
        for k in range(j+1,len(r)):
            if z(d(i,j),d(i,k)) and z(d(i,j),d(j,k)) and z(d(j,k),d(i,k)):
                a += 1
                x = [r[t][0] + 0.5*cos(2*pi*a/9) for t in (i,j,k,i)]
                y = [r[t][1] + 0.5*sin(2*pi*a/9) for t in (i,j,k,i)]
                plt.plot(x, y, '-')
for p in r: plt.plot(p[0],p[1],'bo')
plt.grid()
plt.show()
print(a)

var
s:string;
m:array[-50000..50000]of char;
i,min,max,u:longint;

begin
readln(s);
u:=0;min:=90001;max:=-90001;

for i:=1 to length(s) do
  begin
   if s[i]='<' then dec(u);
   if u<1 then u:=1;
   if s[i]='>' then inc(u);
   if (s[i]<>'>')and(s[i]<>'<') then
                                begin
                                 m[u]:=s[i];
                                 if min>u then min:=u;
                                 if max<u then max:=u;
                                 inc(u);
                                end;
  end;

for i:=min to max do write(m[i]);

end.

can't pass TEST#7, just can't,,,,

A test which help me to discover a bug in my code:

5 1 4 1
1 2
2 4
4 5
3 5

Answer:
3
1 2 3

import numpy
n,m=input().split()
a=[]
for i in range(int(m)):
    a.append(int(input()))
b=numpy.zeros(int(n))
for i in range(int(n)):
    b[i]=a.count(i+1)/int(m)*100
    print('%.2f' % b[i]+'%')

Answer from system:Restricted function

Getting wa @ 14, can somebody please give me test cases for test 14?

don't use scanf("%d")

Edited by author 19.07.2014 21:19

要用高精度。。。long long只对10个点。。。