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There are a couple of facts that we can try to reason about.Submission count: As you have mentioned, they are submitting a huge number of questions every day, it is possible they have them solved and they are just submitting for showing off. Or, it’s not a single person that is solving it.The name of the user: There is a `bot` in their name and with a number suffix. If you google the term `luogu_bot3 codeforces`, you will see another account. Which means it’s an automated process, i.e. a bot, and many of them.Coding style: If you look at each submission, you can see the difference. There are usually 2 kinds of people. One who uses a template for their code and others write the program from scratch. Even if they write it from scratch every time, there are bound to be similarities in them. But if you see the solutions, it’s completely different. eg:The starting template:#include<cstdio> using namespace std; char c[200005]; vs#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=2e5+5; The spacing, holy war of tabs v spacesint main() {  cin >> n;  for(int i=1;i<=n;i++)  scanf... vsint main(){  scanf("%d%d",&n,&k);  scanf("%s",c); This will be more evident if you actually copy-paste the code into sublime that can highlight the difference between the actual snippets from codeforces (not from the above snippet)And, then the final evidence:AtCoder への提出回数ランキングを見てみたら bot vs. golfer vs. 異常精進勢という構図になってて面白い1位 vjudge1 (18682回)2位 kotatsugame (17292回)3-8位 luogu_bot{1..5}9-11位 vjudge{2..5}12位 leafmoon (7038回)13位 c_r_5 (6900回)14位 kmjp (6608回)15位 ei13333 (6470回)— うさぎ (@a3VtYQo) June 18, 2019which translates to:If you look at the ranking of submissions to AtCoder, it is interesting to see that bot vs. golfer vs. abnormal progress  1st vjudge1 (18682 times)  2nd place kotatsugame (17292 times)  3-8th place luogu_bot {1..5}  9th-11th vjudge {2..5}  12.Leafmoon (7038 times)  13th place c_r_5 (6900 times)  14th place kmjp (6608 times)  15th ei13333 (6470 times) It is mostly a contest that is run on AtCoder and submissions are taken on codeforces through a bot and one of the users kmjp - Codeforces is quite popular.

Avoiding or at least initially attempting to avoid a brute force approach in mathematics is a useful habit to develop.Case in point.How can we generate an alternative proof that avoids a direct manipulation of four consecutive integers?Well. Sometimes, when a number of terms involved in a given problem is odd, it is possible to exercise a symmetry of an arrangement by recentering the said terms on the one that sits in the dead middle of the formation (Fig. 1):In this problem, however, we are given a number of factors which is even and which is awkward to arrange symmetrically with respect to a central term.For example, if we take [math]n[/math] to be the second from left factor then the corresponding product will take the following shape*:[math](n - 1)\cdot n\cdot(n + 1)\cdot(n + 2) \tag*{}[/math]and if we take [math]n[/math] to be the second from right factor then the corresponding product will take another awkward shape:[math](n - 2)\cdot(n - 1)\cdot n\cdot(n + 1) \tag*{}[/math]Moving [math]n[/math] all the way to the right, we get:[math](n - 3)\cdot(n - 2)\cdot(n - 1)\cdot n \tag*{}[/math]While trying to generate a non brute-forcish proof, we inquire: is it possible to arrange the objects at hand in a way that closely resembles the arrangement of objects shown in Fig. 1?More interestingly - does such an arrangement even exist?Turns out that the answer to the above question is a yes and a tactical maneuver that allows us to unmask the hidden symmetry in this particular case is the introduction of an arithmetic mean [math]m[/math] of our four consecutive integers:[math]m = \dfrac{n+n+1+n+2+n+3}{4} = \dfrac{4n+6}{4} \tag{1}[/math]from where it follows that:[math]m = n + \dfrac{3}{2} \tag{2}[/math]But then from (2) it follows that:[math]n = m - \dfrac{3}{2} \tag{3}[/math]and, consequently:[math]n + 1 = m - \dfrac{1}{2} \tag{4}[/math]and:[math]n + 2 = m + \dfrac{1}{2} \tag{5}[/math]and:[math]n + 3 = m + \dfrac{3}{2} \tag{6}[/math]Now notice an interesting phenomenon - while it was a challenge to center, intuitively speaking, our original four consecutive integers in such a way so as to have some nice symmetry, in terms of the arithmetic mean [math]m[/math], as is evident in (3, 4, 5, 6), we do have a nice symmetry (Fig. 2):Lastly, we construct the requested formation [math]f[/math] as follows:[math]f = n\cdot(n + 1)\cdot(n + 2)\cdot(n + 3) + 1 \tag*{}[/math]To unwind [math]f[/math] in a nice way, we first pair up the middle two factors that are equidistant from [math]m[/math]:[math](n + 1)\cdot(n + 2) = \left(m-\dfrac{1}{2}\right)\cdot\left(m+\dfrac{1}{2}\right) = m^2 - \dfrac{1}{4} \tag{7}[/math]We then pair up the peripheral two factors that are also equidistant from [math]m[/math]:[math]n\cdot(n + 3) = \left(m-\dfrac{3}{2}\right)\cdot\left(m+\dfrac{3}{2}\right) = m^2 - \dfrac{9}{4} \tag{8}[/math]and then we multiply out the results:[math]\left(m^2-\dfrac{1}{4}\right)\cdot\left(m^2-\dfrac{9}{4}\right) = \tag*{}[/math][math]\left(m^2\right)^2 - 2\cdot m^2\cdot\dfrac{5}{4} + \left(\dfrac{3}{4}\right)^2 \tag*{}[/math]observing that the last term ruins the perfect square party - that is why we need to add a [math]1[/math] to the mix:[math]f = \left(m^2\right)^2 - 2\cdot m^2\cdot\dfrac{5}{4} + \left(\dfrac{3}{4}\right)^2 + 1 = \tag*{}[/math][math]\left(m^2\right)^2 - 2\cdot m^2\cdot\dfrac{5}{4} + \left(\dfrac{5}{4}\right)^2 = \tag*{}[/math][math]\left(m^2 - \dfrac{5}{4}\right)^2 \tag{9}[/math]Technically, the number on the RHS of (9) is rational. But the set of positive integers is closed under both multiplication and addition. That is, a product of any two positive integers is another positive integer and a sum of any two positive integers is back a positive integer. Hence, the number [math]f[/math] is a positive integer. Hence, the number on the RHS of (9) must be 1) a positive integer and 2) a perfect square of one. [math]\blacksquare[/math]A seemingly natural inquiry into whether the above method of arithmetic mean , let us pretentiously call it that, is scalable or not is a waste of time and a futile effort. As best as I can recall from memory we have the following interesting result on the books: for any fixed natural number [math]k\neq 2,4[/math] the polynomial[math]x\cdot(x+1)\cdot(x+2)\cdot\ldots \cdot(x+k-1) + c \tag*{}[/math]where [math]x[/math] is a positive integer and [math]c[/math] is a rational number is never a square of a polynomial.Consider the above an exotic, albeit possibly elegant, find at the dead end.*If we take the suggested recentering of[math](n - 1)\cdot n\cdot(n + 1)\cdot(n + 2) \tag*{}[/math]as a starting point then it is possible to manipulate it into submission but then that would be qualified as a brute force approach which we have set out to avoid in the first place.For two real numbers [math]a[/math] and [math]b[/math] it is a matter of verification by hand that the following identity holds:[math]\left(a^2 - 1\right)\cdot\left(b^2 - 1\right) = (ab - 1)^2 - (a - b)^2 \tag{10}[/math]Using (10) as a template, we unwind the above product as follows:[math](n - 1)\cdot n\cdot(n + 1)\cdot(n + 2) = \tag*{}[/math][math](n - 1)\cdot (n + 1)\cdot n\cdot(n + 2) = \tag*{}[/math][math]\left(n^2 - 1\right)\cdot\left(n^2 + 2n\right) = \tag*{}[/math][math]\left(n^2 - 1\right)\cdot\left((n + 1)^2 - 1\right) = \tag*{}[/math][math]\left(n(n + 1) - 1\right)^2 - (n - n - 1)^2 = \tag*{}[/math][math]\left(n(n + 1) - 1\right)^2 - 1 \tag*{}[/math]as claimed.

The procedure of publishing a research paper can be split into three major steps - carrying out research, writing a manuscript and submission of manuscript.Once you have carried out enough research work and have sufficient materials on some topic of interest, you should start writing your manuscript. Although every journal has its own guidelines and format for submitted manuscripts, a good manuscript generally consists of the following sections - Title, Abstract, Introduction, Methods, Results/Discussion and Conclusion.Readers' first level of interaction with a paper takes place through the title. It is recommended that the title be short and concise and should avoid technological jargons or abbreviations. The abstract is the summary of the entire paper usually within 250 words. The editor may do some first level of screening based on the abstract. The abstract should highlight the novelty of your study and important findings. The introduction section should contain literature review with proper references to give a detailed background of the research problem you have addressed. The methods section should describe the experimental and/or simulation techniques used to carry out research. The results should be discussed in an orderly fashion in the results/discussion section. You should have a fraction of results which would validate your experimental and simulation techniques by comparing with the existing literature. You may discuss the limitations of our study if you have any in this section. The Conclusion section summarizes the major findings of the study and what they mean altogether. Some journals do not allow any separate Conclusion section.Once you have your manuscript ready, you must pick a journal. It is very essential that your work matches with the scope of the journal you have picked. Also, make sure that you have prepared your manuscript as per the journal guidelines and template. Finally, follow the submission procedure described in the journal site to submit the manuscript. If your manuscript gets positive reviews, it will get published in the journal after you complete the necessary formalities.Dr. U. K. Basak,ProfessorTechno India University.Ex Senior Research Fellow, Saha Institute of Nuclear Physics