Almost everyone is sad about this lock down (including me), and if these contests give hope to people or put a smile on their face or excite them, there's no harm in saying a "thank you" as a good gesture. And I think that it's nearly impossible for such a large number of people to post it in an announcement. It's just a few, which doesn't flood these announcements. In my case, I just got excited seeing the announcement and that there isn't any comment yet, so I thought why not write a good thing for the organizers, there's nothing bad in it. It gives hope (or just puts a smile) to some, if not all.

I don't want to hurt anyone with my comments, and if I did, I am sorry about it.

I like to watch other people's rating graph when cf predictor says mine is going to be -50

Btw how did you generate such numbers .4%,5%,15%,146%. Any formulas for that or random. He is saying correct those who wasted time in solving that suffered. Googlers got Ac instantly. At least it should be not present on google

I think the best solution is just practice because reading is also an important skill

just make sure he is not on break lol

thank's , but letters not understand

please explain more

can you tell why it's only (n*H + (n-1)*C) and not (n*C + (n-1)*H)

You are not alone.

just saw your solution, can you explain why exactly:

lli i = (h-c)/(2*eff);
double boo = (h-c)/(2.0*eff);

In the statement, it says: "1≤c<h≤10^6". The temperature of the hot cup must be higher/larger than the temperature of the cold cup.

Try reading other comments. Comment

You do not need to calculate them as double, you can calculate as integers directly: either [t / (h + c + 1 — 2 * t)] or 1 + [t / (h + c + 1 — 2 * t)] where / is integer division. even if X is an integer, it would not hurt to calculate temperature difference at X + 1

I used double division followed by floor and passed fine. However you can calculate it exactly using integer division.

same problem subcase 66 (getting 2 instead of 1).

You have most probably missed out on cases where the the difference between h and t is the smallest possible difference and therefore 1 should be the output.
Tc:
1
7 1 6
Your code fails for this and other similar test cases the answer should be 1 and not 3.

1
10 4 9

here expected 1 found 3.

The Fraction class, fractions.Fraction worked for me (https://codeforces.net/contest/1359/submission/81820143) but only after the contest.

Yes, it sucked the soul out of my rating :(

I would have finished around rank 100 if I had solved C immediately without any errors...but instead I got 347 :(

Can relate. I wasn't able to overcome the precision error until after the time was over. :(

Yes, same problem. Discovered after the contest that the easiest way round this is to use the Python Fraction class (fractions.Fraction), see https://codeforces.net/contest/1359/submission/81820143. I wasn't aware of this class before, so I actually learnt something from this "Educational" round!

Not sure what one would do in other languages.

I have updated the comment with spoiler section(with explanation and code) and a bonus similar problem.

Note that there is no need to check the negative values, since if the biggest value of the subarray is negative, it is better to use a single element subarray which results in sum of 0.

there is the condition for input -30<=ai<=30.

this is useful to maintain O(N) if the diversity increases this algorithm reached O(N^2) as you said.

1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water and pour it into an infinitely deep barrel;
4. take one cup of the cold water and pour it into an infinitely deep barrel;
5. and so on...

Here for every cup of cold water you have already poured a cup of hot water already. So whenever the even number of cups poured, the temperature will be constant which is the avg(h, c).

After you pour the 1st cup of hot water. The temperature will be h. After you pour the 2nd cup of hot water. The temperature will be slightly low since you have poured one cup of cold water before.

Therefore, the temperature will be decreasing monotonically towards the avg(h, c) when ever you pour a cup of hot water. So you can use this property to binary search the appropriate value.

Link to my submission : 81771518 . Hope you find this useful

How to prove that if there is more than one number not the multiple of $$$k$$$ then the final modulo is different?

We can prove using mathematical induction. As BledDest showed that only those sequences may work which contain multiples of smallest number .

Notation : $$$x$$$ mod $$$(a_1,a_2,a_3)$$$ = $$$((x$$$ mod $$$a_1)$$$ mod $$$a_2)$$$ mod $$$a_3$$$

Base case : $$$size=2$$$ , consider $$$[a_i,v.a_i]$$$ where $$$v>1$$$ . Take any number $$$x$$$. Then $$$x = p.(v.a_i) + r$$$ , where $$$r<v.a_i$$$. Then $$$(x$$$ mod $$$v.a_i)$$$ mod $$$a_i$$$ = $$$r$$$ mod $$$a_i$$$ ,$$$(x$$$ mod $$$a_i )$$$ mod $$$(v.a_i)$$$ = $$$(r$$$ mod $$$a_i)$$$ mod $$$(v.a_i)$$$ = r mod $$$a_i$$$ .

Assumption : Suppose above is true for size $$$k$$$ i.e $$$[a_1,a_2 ... a_k]$$$. Note that all all elements are multiple of $$$a_1$$$.

Let us prove for size $$$k+1$$$ i.e $$$[a_1,a_2 ... a_{k+1}]$$$. consider $$$x$$$ mod $$$(a_i,...a_{k+1} ..a_j)$$$ (type 1) and $$$i$$$ not equal to $$$k+1$$$.It will make no difference since $$$a_{k+1}$$$ is not at first position.consider $$$x$$$ mod $$$(a_{k+1} ....a_j)$$$ (type 2) i.e $$$a_{k+1}$$$ is at first position. Then by assumption it will be same for all permutation where $$$a_{k+1}$$$ is at first position. Now we only need to prove that answer for any particular permutation of type 1 and type 2 are same. Let us choose $$$x$$$ mod $$$(a_{k+1},a_1,a_2,a_3.....a_k)$$$ and $$$x$$$ mod $$$(a_1,a_{k+1},a_2,a_3....a_k)$$$. Since $$$x$$$ mod $$$(a_{k+1},a_1)$$$ = $$$x$$$ mod $$$(a_1,a_{k+1})$$$ (from base case proof) hence both are equal . Thus answer is same for all permutation of length $$$k+1$$$.

I will be thankful if some one points out mistake in above proof.

pointless...its wrong..my submission is hacked

Because the largest number taken at each step (subtracted from the sum) maybe different when taking from the right.

Example:

11
3 0 1 -2 5 -5 -1 0 3 2 2

When taken only from left, the answer is 3 but when taken from right, the answer is 4 (3 2 2).

However, even this solution doesn't work for cases like,

8
4 -10 13 -13 2 2 -11 6

it is ok, if you use kadanes from left and right still you will end up in this case :P
7
30 -20 5 1 3 -20 30
ans would be 4 but by using two kadanes we get 0 :(