In fact need to do what is asked in the statement. We need to find in a cycle the maximum height h, counting, how many blocks must be in i-th row and adding these values to the result. Iterate until the result is not greater than n.

Jury's solution: 8924831

Sort lanterns in non-decreasing order. Then we need to find maximal distance between two neighbour lanterns, let it be maxdist. Also we need to consider street bounds and count distances from outside lanterns to street bounds, it will be (a[0] - 0) and (l - a[n - 1]). The answer will be max(maxdist / 2, max(a[0] - 0, l - a[n - 1]))

Time complexity O(nlogn).

Jury's solution: 8924823

492C - Vanya and Exams.

Sort (a_i, b_i) in non-decreasing order for number of essays b_i, after that go from the beginning of this sorted pairs and add greedily the maximal number of points we can, i.e. add value min(avg * n - sum, r - a_i), while total amount of points will not be greater, than avg * n.

Time complexity O(nlogn).

Jury's solution: 8924807

492D - Vanya and Computer Game.

Let's create vector rez with size x + y, in which there will be a sequence of Vanya's and Vova's strikes for the first second. To do this, we can take 2 variables cntx = cnty = 0. Then while cntx < x and cnty < y, we will check 3 conditions:

1) If (cntx + 1) / x > (cnty + 1) / y, then add into the vector word “Vova”, cnty++.

2) If (cntx + 1) / x < (cnty + 1) / y, then add into the vector word “Vanya”, cntx++.

3) If (cntx + 1) / x = (cnty + 1) / y, then add into the vector word “Both” 2 times, cntx++, cnty++.

Then we are able to respond on each query for О(1), the answer will be rez[(a_i - 1)mod(x + y)].

Time complexity O(x + y).

Jury's solution: 8924773

492E - Vanya and Field.

As long as gcd(dx, n) = gcd(dy, n) = 1, Vanya will do full cycle for n moves. Let's group all possible pathes into n groups, where 1 - th, 2 - nd, ... , n - th path will be started from points (0, 0), (0, 1), …, (0, n - 1). Let's look on first path: (0, 0) - (dx, dy) - ((2 * dx) mod n, (2 * dy) mod n) - ... - (((n - 1) * dx) mod n, ((n - 1) * dy) mod n). As long as gcd(dx, n) = 1, among the first coordinates of points of the path there will be all the numbers from 0 to n - 1. So we can write in the array all relations between the first and second coordinate in points for the path, that starts in the point (0, 0), i.e. y[0] = 0, y[dx] = dy, ... , y[((n - 1) * dx) mod n] = ((n - 1) * dy) mod n. Now we know, that all points with type (i, y[i]), where 0 ≤ i ≤ n - 1, belong to the group with start point (0, 0). In that case, points with type (i, (y[i] + k)modn) belong to the group with start point (0, k). Then we can add every point (x_i, y_i) to required group k for О(1): (y[x_i] + k) mod n = y_i, k = (y_i - y[x_i] + n) mod n. Then we need just to find group with the maximal amount of elements, it will be the answer.

Time complexity O(n).

Jury's solution: 8924746

P.S. Sorry for my bad English, I hope, I will correct this editorial as much, as possible.

Comments (35)

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TsunamiNoLetGo

9 years ago, # |

I cant access the solution codes

→ Reply

Wild_Hamster

9 years ago, # ^ |

← Rev. 2 →

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

pratyaksh

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

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.

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.

pulkit_180894

← 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

6 years ago, # ^ |

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.com/contest/492/submission/9386556

mello

6 years ago, # |

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 years ago, # |

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;

}