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По сути нужно выполнить то, что просят в условии. Находим в цикле максимальную высоту h, подсчитывая, сколько кубиков должно быть в i-м рядку и добавляя эти значения к результату. Выполняем цикл, пока результат не превышает n.

Авторское решение: 8924831

Отсортируем фонари в порядке возрастания. Дальше найдем максимально возможное расстояние между двумя соседними фонарями, пусть это будет maxdist. Также учтем границы улицы, т.е. найдем расстояния от крайних фонарей к концам улицы (a[0] - 0) и (l - a[n - 1]). Ответом будет max(maxdist / 2, max(a[0] - 0, l - a[n - 1]))

Асимптотика по времени O(nlogn), учитывая сортировку.

Авторское решение: 8924823

492C - Ваня и экзамены.

Сортируем пары (a_i, b_i) в порядке возрастания по количеству рефератов b_i, после этого проходимся от самого начала отсортированных пар и добавляем жадно по максимуму количество баллов, то есть добавляем значение min(avg * n - sum, r - a_i), пока суммарное количество баллов не будет не меньше, чем avg * n.

Асимптотика по времени O(nlogn), учитывая сортировку.

Авторское решение: 8924807

492D - Ваня и компьютерная игра.

Создадим вектор rez размером x + y, в котором будут находится последовательность ударов Вани и Вовы за первую секунду. Для этого заведем 2 переменные cntx = cnty = 0. Дальше пока cntx < x и cnty < y, мы проверяем 3 условия:

1) Если (cntx + 1) / x > (cnty + 1) / y, то добавляем в вектор “Vova”, cnty++.

2) Если (cntx + 1) / x < (cnty + 1) / y, то добавляем в вектор “Vanya”, cntx++.

3) Если (cntx + 1) / x = (cnty + 1) / y, то добавляем в вектор “Both” 2 раза, cntx++, cnty++.

Дальше можем отвечать на каждый запрос за О(1), ответом будет rez[(a_i - 1)mod(x + y)].

Асимптотика по времени O(x + y).

Авторское решение: 8924773

492E - Ваня и поле.

Поскольку gcd(dx, n) = gcd(dy, n) = 1, то Ваня сделает полный цикл за n движений. Сгруппируем все возможные пути в n групп, то есть те, что начинаются с точки (0, 0), (0, 1), …, (0, n - 1). Рассмотрим первый путь: он будет иметь вид (0, 0) - (dx, dy) - ((2 * dx) mod n, (2 * dy) mod n) - ... - (((n - 1) * dx) mod n, ((n - 1) * dy) mod n). Поскольку gcd(dx, n) = 1, то среди всех первых координат будут присутствовать все числа от 0 до n - 1. То есть можем занести в массив все соответствия между первой и второй координатой для пути с точки (0, 0), т.е. y[0] = 0, y[dx] = dy, ... , y[((n - 1) * dx) mod n] = ((n - 1) * dy) mod n. Теперь мы знаем, что все точки типа (i, y[i]), где 0 ≤ i ≤ n - 1, принадлежат к группе (0, 0). В таком случае точки типа (i, (y[i] + k)modn) принадлежат к группе (0, k). Тогда каждую точку (x_i, y_i) мы можем добавить к нужной группе k за О(1): (y[x_i] + k) mod n = y_i, k = (y_i - y[x_i] + n) mod n. Дальше просто ищем группу с максимальным количеством элементов, она и будет ответом.

Асимптотика по времени O(n).

Авторское решение: 8924746

Комментарии (45)

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TsunamiNoLetGo

10 лет назад, # |

I cant access the solution codes

→ Ответить

Wild_Hamster

10 лет назад, # ^ |

← Rev. 2 →

Yes, sorry, will be fixed soon. UPD: fixed

Borisov

В задаче С в формуле min(avg * n - sum, r - b_i) вместо b_i разве не a_i?

Поправьте если ошибаюсь.

Да, спасибо, исправлено.

0xFED

← Rev. 6 →

В пункте D нужно поменять местами "Vova" и "Vanya".

Исправлено.

pratyaksh

There'a typo above for 492D — Vanya and Computer Game. It should be cnty++ in case 1 and cntx++ in case 2.

Igorjan94

-16

Е: как-то и то же самое, но немного покороче 8925322

T0RRES

Круто, а я еще думал, что у меня код короткий :D

logic_max

← Rev. 4 →

Problem D can also be done using binary search.
Take an array of 0 to LCM(x,y). Player hitting at x would now hit at lx = LCM / x, similar for y. At every point P, number of hits = P/lx + P/ly. So, find the least point P such that number of hits is a % (x+y).
If both divide this point, then answer is both, otherwise the one who divides will be the answer.
Complexity -> O(log(x) + log(y)) per query.

intsashka

Хоть много человек и сдало Е, но ведь в условии неопределенность. Если я заканчиваю в той же точке где и начал, то я вижу снова те яблони, которые были в этой точке:)

jAckAL_1586

If (cntx + 1) / x = (cnty + 1) / y then why we add word “Both” 2 times and cntx++, cnty++. Can someone please explain.

Bayan

If i-th hit was in the same time with (i+1)-th hit, it mean that res[i]=both and it also mean that res[i+1]=both

Thanks bro :)

saurabhsuniljain

I was confused by this I thought only one monster gets the shot and it is consider only one shot during the contest

I also thought the same

Emruz_Hossain

Here vector rez contain who will give i th hit.By (cntx+1)/x or (cnty+1)/y we are calculating the time need to perform cntx+1 th hit for x or cnty+1 th hit for y. If the times are same thats means two hit will be performed simultaneously. Let x perform 5 hits and y perform 4 hits in same time. So 8th hit and 9th hit will be performed simultaneously. If the monster die in 8th hit who will hit last? The answer is Both,because they hits simultaneously. Same for the 9th hit. So for 8th and 9th hit we will add Both in our vector.That is why when (cntx + 1) / x = (cnty + 1) / y then we add word “Both” 2 times in the vector.

alone_wanderer

Deleted

pulkit_180894

Hi! Can anyone explain the solution for problem D for this test case? 7 5 20 26 27 28 29 30 31 32

Vova makes first two strikes at 1/20 and 2/20, knocks first monster. -"Vova" Makes the next strike at 3/20 attacking the second monster. Then at 4/20, both attack together and knock second monster. So, "Both" Vova then again makes two strikes and knocks the third. "Vova" Similarly, they both knock the fourth. "Both" Where am I going wrong? PS: This is test 3

← Rev. 5 →

Hi! Can anyone explain this test case for D? 7 5 20

Vova hits twice at 1/20 and 2/20 to knock out first monster. "Vova"

Then Vova strikes at 3/20 and Both strike at 4/20 to knock out the second monster. "Both"

Again, Vova strikes twice at 5/20 and 6/20 to knock out third.. and so on.

Why is the answer

Vova

Both

Vova ?

PS: This is test 3.

If Vova and Vanya hit at the same time, it will score as two hits. You could find it in notes of problem D:

''In the first sample Vanya makes the first hit at time 1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time 1''.

shanu.thakoor

I may be wrong, but in problem E:

shouldn't it be O(n+m)? I know that the bounds for n and m are of the same order so it shouldn't matter, just asking for technicality of the analysis.

Loved this contest, really nice problems! :)

Yes, but n is 10 times greater than m, so I decided not to include it to time complexity

testing_round

+12

For problem E, there is also an O(mlogm) solution.

Consider two points (x₁, y₁) and (x₂, y₂), if we start our move at point (x₁, y₁) and we can see point (x₂, y₂) along our path, then following equation will be true:

$\text{[math]}$

We can multiple each side by dx * dy and get this:

So we calculate value x_i * dy - y_i * dx for each apple tree and count frequency for each of these values by a data structure like map in C++. And answer is an apple tree at point (x_i, y_i) such that count[x_i * dy - y_i * dx] is maximum.

My submission: 8945734

good, we can do it without maps, since x1 * dy - y1 * dx(modn) < n, so it will be O(m). I tried to submit it and it got AC.

In your solution memory complexity is O(n), My main target was achieve an efficient solution that is independent of parameter n.

Type_62

I have the same solution like yours. In this case, we can archive an O(m log m) solution no matter what n is. :)

igor.pisarev

Nice solution :) Using unordered_map it should become clearly O(m)

compilation_error

For problem D I have different approach that I think is correct ,but it gets wrong answer on test 11 .

So,idea is following :
We know that Vanya attacks with frequency x hits in second, and Vova y hits in second , 1/x and 1/y for 1 hit respectively or ( lcm(x,y)/x ) / lcm(x,y) and ( lcm(x,y)/y )/lcm(x,y) .Lets x1=lcm(x,y)/x and y1=lcm(x,y)/y.It's easy to see that if(n%(x1+x2)==0 or (n+1)%(x1+x2)==0) than answer is "Both" , after that lets k=y1/x1(lets assume y1>=x1) than if(n%(k+1)==0) than answer is person who shots less productively. Otherways answer is person who shots more productively. this is my solution .
Can you help me? Is my approach wrong?

cookieme

x=3, y=5 ==> k=1 Hits by user:

1. Y
2. X
3. Y
4. Y
5. X
6. Y
7. Both
8. Both

Hits by your algorithm:

1. Y
2. X
3. Y
4. X
5. Y
6. X
7. Both?
8. Both?

late_riser

Problem C:

I got time limit exceeded during the system testing phase.

After contest, I compared my code with others and the only difference I found the array size.

I increased the array size and got accepted! But I don't understand why?

It is given N is 1<=N<=100000 and I took array of size 100005. Why this will give TLE? Can anyone explain please?

manetsus

Please, give your codes links for clear understanding.

As far I saw in your last submissions, the size of array is not the main matter. You did another change in your accepted code which caused TLE before. That is in line 14:

if(a.hw == b.hw) return a.score<=b.score; In the accepted code, you commented it. Thanks.

Finolis

I don't get why the output of Test#1 B is 2.5000.

Check my drawing :P

The answer should be 2.0 not 2.5. I saw some solutions sorting for (nlogn) then doing 14 — 9 = 5/2 = 2.500 but this makes no sense if you think at it as the drawing. Don't see why you need to sort, if you can just pass a sliding window to find the biggest gap and divide by 2.0 to find that d = 2.0.

Please enlighten me ^^

nic11

Latterns are points, not squares. So longest distance is 14 - 9 = 5, $\text{[math]}$

aboAdnan

7 лет назад, # ^ |

No, It's not that, because if it was like you said, the second sample's answer should be 2.5 also.

So, it's just that the answer provided in the editorial is wrong when either we consider the lanterns are squares or points in the middle of squares. So, that should be elaborated in the statement, then the editorial answer should be corrected.

Sorry for the annoyance if it's existed.

Problem D can be done in O(log(x+y) + n), preprocess costs O(log(x+y)), each query costs O(1). Here is my solution http://codeforces.me/contest/492/submission/9386556

mello

7 лет назад, # |

For problem C I think the note for the first sample input is wrong because it says he will get 1 point for writing 2 essays... Nevermind, it's 2 essays because that was the cost.

mailstomukulchauhan

4 года назад, # |

Please fix my code

its not printing double values

include<bits/stdc++.h>

using namespace std;

int main(){ int n,l; double * a=new double[n];

for(int i=0;i<n;i++){
    cin>>a[i];
}

sort(a,a+n);

double rez=max(a[0],l-a[n-1]);

for(int i=1;i<n;i++){
    rez=max(rez,a[i]-a[i-1]);
}
cout<<rez/2;

}