### This blog is for personal use. You can check it out if you want.

#### As many people are checking out this blog, I thought of adding problems for some of the tricks.

# Tricks

- Sum-Xor property: $$$ a+b = a \oplus b + 2(a\And b) $$$. Extended Version with two equations: $$$a+b=a\vert b + a\And b$$$
**AND**$$$a \oplus b = a\vert b - a\And b$$$ Problem 1 Problem 2 - Upto $$$10^{12}$$$ there can be atmost $$$300$$$ non-prime numbers between any two consecutive prime numbers.
- Any even number greater than $$$2$$$ can be split into two prime numbers. Problem1 Problem 2
- Sometimes it is better to write a brute force / linear search solution because its overall complexity can be less. Problem 1
- When $$$A \leq B$$$ then $$$\lfloor \frac{B-1}{A} \rfloor \leq N \leq \lceil \frac{B-1}{A} \rceil$$$ where $$$N$$$ is the number of multiples of A between any two multiples of B. Problem 1
- Coordinate Compression Technique when value of numbers doesn't matter. It can be done with the help of mapping shortest number to $$$1$$$, next greater to $$$2$$$ and so on. Problem 1
- Event method: When there is a problem in which two kinds of events are there (say $$$start$$$ and $$$end$$$ events), then you can give $$$-ve$$$ values to $$$start$$$ events and $$$+ve$$$ values to $$$end$$$ events, put them in a vector of pairs, sort them and then use as required. Problem 1 Problem 2
- When applying binary search on $$$doubles$$$ / $$$floats$$$ just run a loop upto 100 times instead of comparing $$$l$$$ and $$$r$$$. It will make things easier.
- For binary search you can also do binary lifting sort of thing, see this for more details. (I don't know how to add that code without messing up the list, that's why the link). Problem 1 Problem 2
- Sometimes, it is useful to visualize array into a number of blocks to move towards a solution.
- $$$gcd(F_n,F_m)=F_{gcd(n,m)}$$$, where $$$F_x$$$ is the $$$x_{th}$$$ fibonacci numbers and first two terms are $$$0,1$$$. Problem 1
- In ternary search, always keep in mind that when $$$r=l+2$$$ then $$$mid1=l$$$ and $$$mid2=r$$$ or we can say that $$$l$$$ & $$$r$$$ will remain same $$$i.e.$$$ $$$r=l+2$$$ and the condition $$$l+1$$$ will never get checked, so you have to check it explicitly. see this problem

Also, if you know any tricks / methods that you want to share and are not in the blog then you can write in the comment section, I will add them to the blog.

I will update this blog if I come across any more general methods / tricks.

If you know the proof of #3, please tell me.

I don't know the proof. It is not proved yet. But it is also not proved wrong (I think).

I used google and got this: https://arxiv.org/pdf/1811.02415.pdf and https://vixra.org/pdf/1702.0150v1.pdf

By the way, it's a fine exercise. Find mistakes in the second one (the first one is significantly longer).

After 11 paragraphs of the second "proof", we just reached this impressive conclusion:

"Therefore, we only need to prove that 2k, for every integer k > 1,is the sum of two primes."

Yup, that was the funniest part IMO.

Those papers are from arxiv.. they are not peer reviewed papers..the proofs in those articles have either been proven to be false or else no has bothered to verify it being patently flawed.

The proof called the goldbach conjecture. Many of mathematics are working hard to prove it. As a competitive programmer. The prove is simple green text "Accept".

There is no proof , only Goldbach Conjecture

I have discovered a truly marvelous proof of this, which this webpage is too narrow to contain.

For those who don't get the reference;

just like fermat.

Why will I tell you if I can win one million dollars? :v Anyway you can use it in this problem.

The proof is the green text that says "Accepted".

Depends on how strong are the system tests.

FWIW it's definitely been verified for numbers below $$$10^{18}$$$, so it's completely fine to use that property for any long longs (which is basically always the case for us).

On this topic, there are some other similar conjectures. ARC 080 D makes use of some more "Goldbach-like" hypotheses.

We know that any prime number except 2 and 3 can be represented as 6*n+1 or 6*n-1.

So, Just add two prime numbers:

1. (6*a+1) + (6*b+1) = (6*a+6*b) +2 //even number

2. (6*a+1) + (6*b-1) = 6*a+6*b //even number

3. (6*a-1) + (6*b-1) = (6*a+6*b) -2 //even number

Let me know if I am wrong.

You don't prove that every number can not be represented this way. This is Goldbach's conjecture which is an open problem.

https://math.stackexchange.com/questions/895992/6n1-and-6n-1-prime-format#:~:text=So%20a%20prime%20number%20cannot,than%206n%C2%B11.

https://primes.utm.edu/notes/faq/six.html

What about these?

We can represent a number in form of

6*n,6*n+1,6*n+2,6*n+3,6*n+4 and 6*n+5. So, considering a number prime it is one out of 6*n+1 or 6*n+5 because we can see other forms are divisible by 2 or 3.What has this got to do with anything? Are you sire you understood the problem "statement".

Ok, My bad. It does not imply "every".

Thanks

Trick 7 is my personal favourite

Suggest a problem plox

You can try this problem.

Here is one

I don't know, in Russia it is very popular and we call it "scanning line" or "scanline".

Hm yea I think it's also called sweep line algorithm? At least that's what is chapter 30 of the cses book.

Yes, that's the English name.

Can you share some easy problems that make use of this algorithm?

https://codeforces.com/gym/100099

It seems like the problems are in russian

You can use this service to get the text from the picture and then you can translate it with this or Google translate.

https://translate.yandex.ru/

nice

This trick is quite well known probably but still maybe it is useful for beginners:

Ways of initializing a global array :

memset(a,0,sizeof(a)) ; // initialize with 0

memset(a,-1,sizeof(a)); // initializing with -1

memset(a,0xc0,sizeof(a)); // initializing with negative infinity

memset(a,0x3f,sizeof(a)); // initializing with positive infinity

I didn't know about the negative infinity and positive infinity, I guess I will put it in the blog.

but be careful, if you have something like

`const int INF = [something];`

then

`INF = 0x3f3f3f3f`

, not`INF = 1e9`

I prefer

`std::fill()`

, because you can just fill the array using any arbitrary values that you want, despite being slowerFor $$$7$$$, you can also do this: for start values, store $$$2*start$$$, and for end values, store $$$2*end+1$$$. If it's even, it will be a start value, else an end value. And start values will always come before end values.

Nice.

I think 6 is coordinate compression

extended versions of

trick 1can be obtained from this a+b = a|b + a&b and a xor b = a|b — a&b ,for eg trick 1 can be obtained by subtracting eqn 2 from eqn 1.(btw sorry for not formatting it properly.)It would be great if everybody could share a few of the peculiar tricks they know

Can you provide links of some problems related to every tricks. It will be very helpful.

For trick 1, you can use it on this problem.

A different trick is that for queries, you can sometimes split it into 2 types of queries (when query is greater than sqrt(n) and less than sqrt(n)) so instead of using O(n) time you use O(sqrt(n)). You can use it in these problems 1 2.

how do we solve this problem? I saw few submissions of this problem, there was dp involved. Can you explain?

Be careful using

unordered_map, it can give TLE or your solution may get hacked , see the blogpost by neal about this link also unordered_map doesn't let you use complex data type likepair<int,int>,int. to make it work see the blog by arpa about it linkWith custom hash, you can configure it so that it allows

Time to add the blog to favorite and read it after eternity.

Any problems related to 7?

check this problems

https://codeforces.com/contest/1424/problem/G

https://codeforces.com/contest/1420/problem/D

problem G is for rule 7 , thanks a lot bro for giving the sample questions

I think there are two problems in the CSES problem set in the searching and sorting section.

Can we make this blog like every coder shares their one trick which is not mentioned in the post. It will be a great learning experience for all of us

Yes, If it is nice I will add it.

LOL stop treating cp as a game of remembering tricks...

LOL Read the heading... This blog is for personal use. Everyone has a way of doing things.

LOL

That’s what practice is. See https://codeforces.com/blog/entry/82893?#comment-700332

HAHA i think you misinterpreted what that comment was about

No, I don’t think I did. But I’ll also stop feeding the troll behind the fake account. (:

Hm I'm curious, how do you deal with adhoc problems then?

For trick #8, I usually loop up to 200 times just to be safe. :P

I heard this somewhere that, if we wanted to find your name in the sorted list of names of people around the globe, it would take no more than 64 operations/iterations for binary search to figure it out. I think that 100 times is more than safe unless the epislon is really small like -1e18 or smth.

That statistic is probably right because $$$2^{64} > 7 \texttt{ billion}$$$. I think that binary search would take at most $$$33 - 34$$$ iterations because $$$log_2{7000000000} \sim 33$$$

Can You suggest a problem for the same?

In CF EDU binary search section, there is a lot of problems about binary search on real values.

can you explain 5 one, i mean what is the operator being used?

Do you mean floor and ceil operator? or anything else?

oh ok got it

Hey,when i open my profile it is showing that this page is temporarily blocked by the administrator,do you know anything related to it.

Yeah, there were some problems in rating changes in the latest contest. That's why.

oh ok thanks

Oh, man!! Point-8 is the best I was getting TLE for some weird reason but this trick got me accepted.

With doubles, you should do

`while(abs(l - r) > EPS)`

instead of`while (l < r)`

.How small should you set EPS?

I usually set it as 1e-8 sometimes 1e-9.

This approach also works but I think you should go with 100 iterations it's just safe.

Hey bro @nubir345 can u please give the question according to the trick just below each trick , would be very easy for others , the questions given below in comments .

When I started with CP, I struggled with Binary search a lot. Nowadays I use these rules-of-thumb for BS.

Say we are doing BS on some variable

`x.`

And There is some function`bool check(int x)`

which is`monotonic.`

And

`check`

returns $$$[False, False, ..., False, True, ..., True]$$$ for different values of x and 0 <= x <= 1e9CodeCheck how Errichto Link makes use of an extra variable in BS to store the answer. I think that makes it further easy and avoid corner case mistakes.

I'm sure there have been tutorials etc about this trick but since people are talking about binary search, I want to add this. I like to use iterative binary search that looks like this:

This assumes 0 doesn't have "some property". In the end,

`ans`

will be the largest integer that doesn't have "some property".Using this, I have been able to avoid guessing about one-off errors for 6 years already. It is short to write, intuitive and generalizes well to floats and bigints. I'm not sure exactly what your "trick 8" accomplishes, but I suspect iterative binary search also makes that unnecessary.

Isn't this Binary Lifting?

I've only ever heard the term "binary lifting" used on trees (problems like "find the $$$k$$$-th ancestor" or LCA). But yes, both of those are essentially binary search.

I don't agree with this. You can do binary jumps instead of binary search but I wouldn't call them the same thing.

how does this works in case of floats?

`for (double k = MAX_N; k > EPS; k /= 2)`

hi, could you explain how we can use this binary search solve a problem like, "What is the smallest number whose square is strictly larger that a given number N?" Thanks.

Find the largest one that isn't and add 1.

Since you have been using this for 6 years, have you never run into a binary search problem(not insanely hard) where doing some sort of easy trick(like the +1 thing here) is not applicable? I'm asking because I really like this approach but am kinda scared that if the question is complex enough, i will simply not be able to find the trick so that i can apply this method(obviously in those scenarios going back to the original implementation is always an option).

Obviously all kinds of things can happen in insanely hard problems. But no, this can (almost?) always be used in place of binary search. The way I see it, there are 4 use cases for binary search:

and all of these can be handled by this implementation.

How to get the inversion of just one number?

By having memorization, it can be linear $$$O(n)$$$ to get all inversion in $$$[1 \dots n]$$$ (with high constant for modulo ofcourse)

But if just need to find inversion of one number without using power function, what will be the complexity of the code ?

I testest for one thousand random integer number (approximately $$$10^9$$$) and I think it is about $$$O(2 * log(n))$$$ am I correct ?

Open problem. But it's really fast!

What is the difference between your method and the one people usually use (using binary exponentiation)?

Faster and easier (maybe :) ).

I found this on cp-algorithms. Apparently, it can only be used if

`x`

is less than`p`

.inv(x) = inv(x % p)

Thanks for clarifying, I've just known about that. By the way, there is a slight difference between your code and cp-algorithms' code, your code computes the

`p - p / x`

first, before multiplying the result with`get_inv(p % x)`

; while cp-algorithms' code computes`p / x * get_inv(p % x) % p`

first, then subtracting the result from`p`

. How can both of them produce the same result?`(p - a) * b % p = (p * b - a * b) % p = (-a * b) % p = p - a * b % p`

If you add memorization to that function, in theory to say the complexity by this way will be Linear (with high constant for taking modulo), while using binary exponentiation is $$$O(n log n)$$$ (with lighter constant)

For trick 1, you can add this great problem.

Given a strictly increasing array $$$a_1 < a_2 < a_3 < .... < a_n$$$

Do this transformation $$$a_i := a_i - i$$$

Then the array becomes non- decreasing $$$a_1 <= a_2 <= a_3 <= .... <= a_n$$$

Fairly simple trick but quite useful sometimes :)

Can you give link to any problem that use this trick?

This one

perhaps i didn't understood this ,but is this applicable to every strictly increasing array? can you show some examples.

$$$1, 2, 3, 5, 7$$$ becomes $$$0, 0, 0, 1, 2$$$

what about 2 5 9 23 ?

Strictly increasing is a subset of non-decreasing.

ok , i thought that it will make the array non decreasing surely, it can make an array non decreasing but we cant be sure.(non decreasing in the sense that some element are equal)

[Deleted]

cause if a<b then a<=b-1 and so a-i<=b-(i+1)

Guys I'm not getting the point 8. Can anyone help me with that?

Sometimes precision errors lead to weird behaviour in the

`while (left <= right)`

loop in binary search.Can anyone please help me with Problem 2 in 1st point? Thank you!

please keep updating this blog it is very helpful

for $$$2$$$ and $$$3$$$, every number up to $$${10}^{12}$$$ accepts a goldbach decomposition with one of the numbers less than or equal to $$${10}^6$$$ (I'm not sure if the upper limit is $$${10}^5$$$).

Problems for 9)

1354D - Multiset

1404C - Fixed Point Removal

The second problem in #3 is not related to the Goldbach conjecture.

They use that every even positive integer is a

differenceof two primes.If you have any implication between these two conjectures, it would be a big breakthrough in number theory. See e.g. this paper.

Can anyone help me with problem 1 of point 1 — Xor sum on Atcoder. I failed to solve it but could not find any English editorial for the problem. I found a recursive formula online

However, the explanation is in Chinese and GG translate only makes things messier.

For each u,v<=n,a xor b=u,a+b=v,consider that if a,b are both even number,then a xor b=u/2,a/2+b/2=v/2,and u/2,v/2<=n/2,so there is f(n/2).

If a is even and b is odd(or b is even and a is odd),then a/2 xor (b-1)/2=(u-1)/2,a/2+(b-1)/2=(v-1)/2,so there is f((n-1)/2).

For the same reason,if a,b are both odd,then (a-1)/2 xor (b-1)/2=u/2,(a-1)/2+(b-1)/2=(v-2)/2,u/2<=(v-2)/2(because the reason of xor is always no more than sum)<=(n-2)/2,so there is f((n-2)/2).

That's why f(n)=f(n/2)+f((n-1)/2)+f((n-2)/2).

(It's fun that i'm also chinese and i hate messy GG translate too.However,i have to use GG translate to write this explanation lol)

And we can also prove these 3 cases are mutually exclusive(i mean that,each u,v is exactly in one case):

For each u,v(easy to see they are both even or both odd),if u,v are both odd,then it's in case2(an odd and an even in a,b).And for the case u,v are both even,if (v-u)/2 is even,means that a+b dont produce a carry on the last bit,then it's in case1(a,b are both even).For the same reason,if (v-u)/2 is odd,means that a+b produce a carry on the last bit,it's in case3(a,b are both odd).

It really helped thanks a lot

Thanks a lot.

(Problem specific trick) If x%a < (2*x)%a then x < a <= 2*x Otherwise x < 2*x < a . Problem Link : https://codeforces.com/contest/1104/problem/D

https://codeforces.com/contest/1350/problem/C

How did you use method 7 in the problem 1? I used the method of difference array and prefix sum.

Solution using Method 7(Scanline).

Wow, this blog is quite useful for me. Upvoted :)

$$$F[gcd(x,y)]=gcd(F[x],F[y])$$$

$$$F$$$ is the Fibonacci sequence that $$$F(0)=0$$$, $$$F(1)=1$$$, $$$F(n)=F(n-1)+F(n-2)$$$ for $$$n \ge 2 $$$

Any problem associated with this trick?

$$$CF226C$$$

I think it is the one.

Nice blog!. Other related problems for Trick 7.

ABC-183D (Water Heater) and CSES (Restaurant Customer)

You can also add one more property.

gcd ( x , y ) = gcd ( x-y , y );

This results in, gcd ( x , y , z , w ,.....) = gcd ( x-y , y , z , w,....)

Also,

for any a, b, if b > a, a%b = a. Additionally, if a ≥ b > [a/2] , a%b = a - b.

For Trick # 2 you can add this Problem

Using point 2 can we state that for checking if n is prime we just need to check if n is not divisible by any number from 2 to 300 rather than sqrt(n) if n<=1e12?

Ofc not: any product of primes greater then 300 not prime but still not divisible by any number from 2 to 300.

oh right, thanks for pointing out my mistake!

It just means that when you have to find greatest prime less than 10^12 than you just have to check for 300 numbers (10^12 — 300 to 10^12) if they are prime or not. Max Number of operations would be 300*10^6.

Here I got one trick , If n>1, then there is always at least one odd prime number x satisfying n < x < 2n

You can check this problem 1178D - Prime Graph

More specifically that there is always a prime number in the interval [n,3n/2] for n>1.

This has been very helpful, thank you so much!

Another recent problem that can be solved using trick 1: 1556D - Take a Guess