And why exactly you gave this guy so many negative reviews?

No announcement is in russian

Ofc codeforces's contests!!

Any sane person would prefer the former one.

no

UwU

me too !!

should be rating and not rank (rank may increase/decrease).

Best of Luck!!!

This was the lowest division and in my opinion, the problems were easier than other Div3 contests. I'm not saying this to discourage you but to give you advice. Maybe codeforces is a little too hard to start with. You can learn the basics from this website (https://www.pbinfo.ro/). It's a website from my country which takes you from zero to hero (or at least doing the first 2-3 problems in Div2 in less than 1 year, haha) You only have to translate the page to English when you enter the site. Good luck with your coding journey! I hope I gave some positive thoughts!

dont listen to alexadru7, just solve 800s instead, and if those are still too hard, do leetcode (but they shouldn't be that hard)

First take the at least number if l to r difference is 3 then take the number 1 2 3 if its greater than s then answer must be -1. other wise try to replace every number with large unused number as if sum of those number not exceed s. if sum==s you find the result.

implementation

you_ create new array have size = r-l+1. make array from 1 to r-l+1. add 1 to arr[pos]._ _You start with pos = size — 1. if arr[pos] == pos then pos--

Here $$$sum(x)$$$ denote sum of $$$1, 2, 3, ..., x$$$. Let $$$len=(r-l+1)$$$. A solution exists if and only if $$$s \gt =sum(len)$$$ && $$$s \lt =sum(n)-sum(n-len)$$$. Now to simulate the process, take numbers from $$$1$$$ to $$$len$$$ in an array (arr). And take numbers from $$$len+1$$$ to $$$n$$$ in a set (st). Let's denote $$$left$$$ as the sum remaining i.e. $$$left = s - sum(len)$$$. Now, you can start from last element of array do this while $$$left \gt 0$$$.

for i from len to 1:
    if(left==0)
        break
    y=arr[i]+left
    if(st.contain(y)):
        arr[i]=y
        break
    else:
        remove greatest element from st say g
        insert arr[i] in st
        left=left - (g-arr[i])
        replace arr[i] with g

Once you have got arr with sum s, you can add remaining elements at other positions. see my submission

Here's an example for n = 10, l = 3, r = 7, and s = 21.

We want to find the numbers that will be in the sum. First we try:

1 2 3 4 5 / 2 3 4 5 6 / ... 6 7 8 9 10

Then find the two sums that s is in between.

Over here, it is 2 3 4 5 6 and 3 4 5 6 7.

Now, since 3 4 5 6 7 is too large, we need to decrement some of the numbers one by one.

Since 25 — 21 = 4, we decrement 4 of the numbers, and to ensure no collisions occur we decrement 3, 4, 5, 6.

So the numbers included in the sum are 2 3 4 5 7.

Now, just put these numbers so that they occupy positions 3 ... 7.

LOL, ignore.

I did not understand from the statement k also depends on i.Can you please elaborate as to what you have taken as i and c and k

E was apparently available on stackoverflow https://stackoverflow.com/questions/46296588/distinct-n-numbers-so-that-sum-equals-to-n

I didn't read sum of $$$n$$$ over all testcase is $$$500$$$ and took a lot of time to solve in $$$O(nlogn)$$$. About your intended solution $$$O(n^3)$$$ I think we can solved it like $$$0-1$$$ Knapsack. Since sum $$$s \lt =n(n+1)/2$$$

no

Thats for trusted vs untrusted standings, not the rated part. Basically what it means is that if someone has peak rating $$$\geq 1900$$$, even if their current rating is $$$\lt 1600$$$, they won't appear in the official standings. Basically its meant to stop people from smurfing to specialist to appear at the top of the official standings of a Div3.

I also think so. but it's not dp. you_ create new array have size = r-l+1. make array from 1 to r-l+1. add 1 to arr[pos]. _You start with pos = size — 1. if arr[pos] == pos then pos--.

you can prove that in the optimal solution,you will spend minimal days reaching a position then stop and only earn tugriks on that position until you pass c,so iterate over which position you stop at

I don't think so. It can be solved by using Greedy.

Assume Polycarp last at position i, then for position before i, upgrade to next position as soon as possible will be the best way.

My approach was array day[i] carries the number of days to reach position i and the rem[i] the number of money i start with at position i

then i get the minimum of days[i] + ceil(( c — rem[i]) /arr[i])

i really don't know if it consider dp or not

I think so, but the problem is that you should consider amount of taken items, it is fixed

I get AC with greedy

Maybe, but if so its tougher than other solutions in my opinion. I thought about this at first, but the best state I could come up with was $$$dp[\text{current number(1 to n)}][\text{count of taken}][\text{sum of taken}] = 1 \text{ if it is possible}$$$, which is $$$O(n ^ 4)$$$. I can't see any trivial way to reduce the state.

I refuse to believe that this is your first account. Also wtf is this ??

Every square is a rectangle!

Yeah TL was super tight, my first submission with 64 bit integers ran in 1980ms, after replacing it with 32 bit ints it ran in around 1s.

Mission successful !! :)

Invalid inputs are passing as hacks. I think the validator doesn't have a check for 1 different element.

I submitted

1
3
1 1 1

and got successful hack.

Yup, hacked someone with this case as well. Seems like validator doesn't actually check that property. Should be fixed fairly soon.

1
3
1 1 1

[deleted]

we are not going to lose rating lmao. I don't know if you are delusioned or you are joking it's most probably the latter.

you have made a mistake in your first_y and second_y points in the code. I corrected it (commented on the line with mistake). 112571015

your solution fails on this test

1
4 4
??00????

after the first iteration, your string will be0?00???0, which is incorrect, since these new zeros must be next to the already existing zeros. to fix this, you can first process the pairs about which it is already obvious which character should be set (pairs 1?, ?1, 0?, ?0), and then the remaining ones (which consist only of ?) 112623883

Today people like me (Newbie) tried to hack GEOTHERMAL(I.G.M) solutons :):)

overflow b_i <= 10^9 & n <= 10 ^ 5 worst case : Sum <= 10^14