We will hold AtCoder Beginner Contest 206（Sponsored by Panasonic）.

- Contest URL: https://atcoder.jp/contests/abc206
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20210619T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: physics0523
- Tester: kyopro_friends
- Rated range: ~ 1999

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

How to do F ?

We can assign Grundy values to each game. Let's denote $$$g(x, y)$$$ as the Grundy value corresponding to the game where all intervals lying entirely in $$$[x, y)$$$ are considered. We find:

Alice wins if $$$g(1, 100) > 0$$$ otherwise Bob wins.

Code

Hey, just a small request...

I tried to solve F (https://atcoder.jp/contests/abc206/tasks/abc206_f) by DP.

This is what I tried.

My understanding for intersecting is, Two intervals A, B are said to intersect if there is atleast one real number x such that x belongs to both A and B. (i.e. intervals which lie completely within each other do intersect).

With this definition, I tried the following logic. First for each position from 1 to 100 note the end positions of the interval starting from that position. Similarly find the minimum end position of all the intervals starting at or after the current position.

Then do a reverse dp (states 0, 1 : 0 -> considering all the intervals starting from >= current position, whether the 1st player can win. 1 -> considering all the intervals starting from current position (only this current position), whether the 1st player can win).

The transition is for all the intervals starting from current position, dp[i][1] = dp[i][1] | !dp[end_position][0]. Then For j from i until the minimum end position -1, dp[i][0] = dp[i][0] | dp[j][1]

https://atcoder.jp/contests/abc206/submissions/23636854

I could not think of a case where it fails. Could you please suggest a case where it does not work ?

How to use Mobius inversion to find the number of coprime pairs in [L,R] ?

The idea is from this problem, instead of taking input make a array with element [L,R]

You can also use the idea stated here : https://math.stackexchange.com/questions/218890/how-many-numbers-in-a-given-range-are-coprime-to-n

I implemented this and it works well within time-limits

Btw, what is the upper bound on the number of distinct prime factors of any integer N (something that we can generally assume while solving problems)?

for finding coprime pairs between [L,R] you can iterate over i=[L,R] range and get all factors of i use principle of inclusion and exclusion to get number of coprime with i which lies in range [1,R] and subtract coprime which lies in range [1,L].

Finding coprime in range using inclusion and exclusion principle link

Atcoder Solution Link

But how would $$$(R - L + 1) * 7 * 2^7$$$ fit in the constraints. As integer $$$i$$$ at worst case could be a product of at max $$$7$$$ primes.??

it is 10^6*max(2^7,sqrt(10^6)/2) = 5*10^8 which passes in 1.5 sec for me. Definitely it is tight solution.

I guess on average, there are not that many numbers with 7 distinct primes, for instance, the smallest number that is a product of 7 distinct primes is 2*3*5*7*11*13*17 = 510510, so the amortized analysis of the time complexity would be less than that.

i saw submissions from the leaderboard do this after computing the number of coprimes [L, R]

I know it has something to do with removing

`g, where g = gcd(x,y) and either x/g = 1 or y/g = 1`

but can't wrap my head around it.example submissions: AnandOza SSRS

I walked through the example test case, which might be helpful to whoever is still confused

Okay so I finally understood the method. Basically after computing the co-rime pairs, we loop

`for i in range [max(2, l), r]`

, and for every i, we substract`2 * n_coprime - 1`

from answer.Why

`2 * n_coprime - 1`

? Let's look at some example.we have a convincing pattern, the top and left borders are the

`g, where g = gcd(x,y) and either x/g = 1 or y/g = 1`

that we need to remove.let's see another example

it looks like the pattern persists, and this is the reason of doing

`ans -= 2 * n_coprime(i) - 1`

, as`2 * n - 1`

computes the number of elements in the top and left border of a matrix.detailed explanation of example test case [3, 8]Hope this helps, highly suggesting reading this as well if you want to understand what mobius functions does!

It was available on gfg :

https://www.geeksforgeeks.org/find-the-number-of-pairs-such-that-their-gcd-is-equals-to-1/ Just make an array with every element from L to R.

Also check out my comment :https://codeforces.com/blog/entry/91945?#comment-806252

[PROBLEM -D]can anyone help me to find out where am i doing wrong. or can you give some test cases where my solution is failing. MY SUBMISSIONWhat's the Idea behind D ?

The idea was to find the sum of depths of all connected components.

Create edges between all $$$s_{i}$$$ not equal to $$$s_{n - i - 1}$$$. The answer is the sum of $$$size - 1$$$ for each connected component.

why this is working, any proof?? can you check my approach, and give some testcases where my solution will fail.

https://codeforces.com/blog/entry/91945?#comment-806252 see this.

Yes , I understand after reading the code what's done to arrive at the answe , I want to know the why ?

you can also use disjoint set union to find the answer my submission: https://atcoder.jp/contests/abc206/submissions/23594043

D : Observation + Graph

Make all a graph on all pairs of numbers which are not equal in a string palindromically. i.e. all pairs s.t. : s[i]!=s[n-1-i]

Now find size of connected component and add sz — 1 to answer because we pick a number, and make atleast sz — 1 operations to make all other numbers in that component equal to that number.

CodeE using Mobius :

Idea : Tot pairs — (pairs with atleast one element = gcd) — coprime pairs(apply mobius here)

Corner case : When L = 1 we subtract it twice so we need to add 2*R — 2 once to ans.

Code :

CodeFor E what I did

Ans=Total Pairs- pairsHavingGcd1- 2*(pairs(x,y)having gcd=x)

n=r-l+1

Total Pairs = n*(n-1)

pairs having gcd1=https://www.geeksforgeeks.org/find-the-number-of-pairs-such-that-their-gcd-is-equals-to-1/

pairs(x,y)having gcd=x, iterate for every number(i) from max(2,l) to r-1 keep on adding r/i — 1

max(2,l) because if we use 1 then it will be subtracted twice as we have already counted it while calculating gcd==1

since (x,y) and (y,x) needs to be counted therefore multiply by 2

In the last 3 ABCs, the problem F are very educational which is great.

Educational screencast. Slow, detailed explanations. Three different solutions for B, two different solutions for D. AC on F during the last minute. HD soon

physics0523 when you write a contest, do you try to include physics problems whenever possible? Just curious because I'd love to solve some physics (kinematics based geometry) problems.

https://atcoder.jp/contests/abc206/submissions/23628260

pls tell what is wrong in this D sol?

`if(find(arr[i] != find(arr[n-i+1])))`

Is this intentional?

Just to check whether they have same parent or not?

Is there something wrong in it? Can u provide any test case?

Seems like a syntax error. Read carefully. Maybe you wanted

`if( find(arr[i]) != find(arr[n-i+1]) )`

instead of`if(find(arr[i] != find(arr[n-i+1])))`

Sums up my rating graph... Thank u so much tho...

Hey Could someone help me on why i am getting wrong answer on F ? I tried thinking for case on where it would fail but cannot seem to fine.

My understanding for intersecting is, Two intervals A, B are said to intersect if there is atleast one real number x such that x belongs to A and B.

With this definition, I tried the following logic. First for each position from 1 to 100 note the end position of the interval starting from that position. Similarly find the minimum end position of all the intervals starting at or after the current position.

Then do a reverse dp (states 0, 1 : 0 -> considering all the intervals starting from >= current position, whether the 1st player can win. 1 -> considering all the intervals starting from current position (only this current position), whether the 1st player can win).

The transition is for all the intervals starting from current position, dp[i][1] = dp[i][1] | !dp[end_position][0]. Then For j from i until the minimum end position -1, dp[i][0] = dp[i][0] | dp[j][1]

https://atcoder.jp/contests/abc206/submissions/23636854

Could you please suggest a case where it does not work ?

Correct Answer: Alice

Your Code: Bob

So thank you for giving a testcase where it gives wrong answer.