We will hold Tokio Marine & Nichido Fire Insurance Programming Contest 2021（AtCoder Regular Contest 122).

- Contest URL: https://atcoder.jp/contests/arc122
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20210612T2100&p1=248
- Duration: 120 minutes
- Number of Tasks: 6
- Writer: maroonrk
- Tester: maspy, yamunaku
- Rated range: — 2799

The point values will be 400-500-600-700-800-1200.

We are looking forward to your participation!

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it's been 45 min and I have done NOTHING.

F*ckin' precision errors >:O

its been 1:30 hours ..and I have done only B :(

I used to treat ARCs as Div. 2 contests.

I was wrong.

rng_58 says they are more of a Div 1.5 https://codeforces.com/blog/entry/75163?#comment-592623

Can we solve Problem B — Insurance , when the scenarios don't happen with equal probability?

I can't think of any possible solution for that case.

I think the solution will be pretty much the same.

It would be exactly the same, but when you are computing the expected value you multiply by the probability, not by $$$1/n$$$.

In this code, in function

`f`

you can just replace`E += p * v;`

with`E += p[i] * v;`

.C had a cancer implementation.

My (unproven) solution is quite easy to implement. Just apply Euclidean algorithm with pairs $$$(n, b)$$$ such that $$$b$$$ is close to $$$n / \varphi$$$, until you find a solution with less than $$$130$$$ operations.

what is n/φ and how did you arrive at this conclusion?

$$$\varphi = (1 + \sqrt 5) / 2$$$. The last operations (of type $$$3$$$ and $$$4$$$) are executed only once (alternately) for each side, so $$$a, b$$$ grow "fast". The only bottleneck is that there may be many initial operations of type $$$1$$$ and $$$2$$$ (because of the error propagation of $$$b - n / \varphi$$$), but the total operations are about $$$200$$$ on average with a very high deviation.

that was my first solution, then I couldn't find a way to get it down below 150 moves.

For which input did your code reach $$$150$$$ moves?

1e18 gave exactly 150 moves (for my initial code)

`b = 618033988749894686`

works in`106`

moves.`b = 618033988749894689`

works in`127`

moves.`b = 618033988749894693`

works in`119`

moves.The density of $$$b$$$ that work seems quite high.

UPD: actually it's not necessary to choose $$$b \approx n / \varphi$$$, just using a random $$$b$$$ seems to work.

Problem D: Try and fail to write the solution by yourself. Remember there's a XOR MST problem in codeforces, google it and copy a solution from there

https://codeforces.com/contest/888/submission/32165432

https://atcoder.jp/contests/arc122/submissions/23373118 ;)

I don't get the relationship between today's problem and MST. Is "the graph with edges $$$\leq x$$$ is connected" equivalent to "Bob can get a score $$$\leq x$$$"? (Why?)

It isn't. We take only the edges that merge components of not both even size. Think about how things work in a trie, if you have both subtrees of even size then you'll solve the subtrees independently otherwise there'll be one number going to the other subtree and that'll dominate everything.

How To Lose Rating

`lower_bound()`

and`upper_bound()`

to find the closest elements in $$$b$$$ doesn't work)How to prove that the function in B is convex? I can't get it?

It's a sum of convex functions.

There should've been a red alert going in your head when you tried the first step and realised you don't have a given array, but 2 given arrays.

My thinking flow

`upper_bound()`

and it worksSame by sorting the two given arrays and get WA. Luckily I came up with a counterexample soon and implement Trie after that.

Can you share that counter-example? I got WA using that too lol

For example if $$$a=[3],b=[4,7]$$$, by only checking consecutive pairs you get the answer $$$7$$$, but the real answer is $$$4$$$.

I haven't come up with counter-example for

`upper_bound`

ones yet :PThat takes an array as argument. Which array and why not the other way around? Hmm.

It's why I want to at least intuitively understand the algorithms I copy. If I get wrong results, might as well throw it away and start over otherwise.

how to solve C?

https://atcoder.jp/contests/arc122/editorial/2071

Its a very nonintuitive way (atleast for me) , if you have solved it , can you tell me your intuition or how u solved it.

I didn't solve it :P

While exploring the possible solutions, I've decided to see what happens in some easy scenarios. One of those scenarios was to start from $$$x=1$$$, $$$y=0$$$ and then do operations $$$x=x+y$$$ and $$$y=x+y$$$ alternatively. I've noticed that in this case $$$x$$$ and $$$y$$$ after some steps are always Fibonacci numbers, so we need just to represent $$$N$$$ as a sum of Fibonacci numbers.

Okkk thanks .

I used Fibonacci triangle rows in A

https://oeis.org/A144154

https://atcoder.jp/contests/arc122/submissions/23371678

With so many solves probably not the intended solution, but I was happy to have found the relation

can you please explain the theory?is it pascal triangle?

I drew some small examples of how many + and — we can have at each position. For example with 4 signs we have:

for 5 signs:

Assuming the 1st and last columns always contain fibonacci pairs, how to generate the ones in between? Well at each step each of the — factors will turn into + on the next turn since we cannot have two consecutive -, and also we can notice from drawing smaller examples that the number of + factors also break into fibonacci pairs — 8+ breaks into 5+ 3-, 5+ breaks into 3+ and 2- and so on.

Then we can see in the above example 8/5, the second column is generated as

8+ breaking into 5+ and 3-, and 5- turning into 5+

therefore second column contains 2*5 + and 1*3 -

similar transition happens for the next column, 2*5+ breaks into 2*3+ and 2*2-

and 1*3- turns into 1*3+, combined that is 3*3+ and 2*2-

and we can generalize to

1*8 2*5 3*3 5*2 8*1

1*5 1*3 2*2 3*1 5*1

which are also fibonacci numbers!

So in my solution I generate these two rows, compute each index's coefficient and add it to my result.

This "linearity" type solution is used in the video https://codeforces.com/blog/entry/91672?#comment-803151

I wish I had more clarity of mind while implementing problems like C :D

Can someone please explain me the the implementation of dp in problem A?

let dp[i][0] be sum of values of equation end with +a_i, dp[i][1] = end with -a_i, and cnt[i][0] is the number of equations end with +a_i, cnt[i][1] = end with -a_i

then

dp[i][0] = (dp[i-1][0] + dp[i-1][1]) + (cnt[i-1][0]+cnt[i-1][1]) * a_i

dp[i][1] = dp[i-1][0] — cnt[i-1][0] * a_i

cnt[i][0] = cnt[i-1][0] + cnt[i-1][1]

cnt[i][1] = cnt[i-1][0]

the answer is dp[N][0] + dp[N][1]

Code : https://atcoder.jp/contests/arc122/submissions/23365039

thanks

Hello, i have read your code, your code is amazing, but i have some questions. what's the meaning of "#include<atcoder/all>" in line 2. Why did I remove it and change the "mint" into "long long" and get the wrong answer

see https://atcoder.github.io/ac-library/production/document_en/

oh thanks

B was easier for me than A as B was simple math.But I approached B very late sadly

Managed to solve C only after the contest :(

Here's my Randomised solution:

https://atcoder.jp/contests/arc122/submissions/23376359

When you realize, for E, that arbitrarily choosing the last element satisfying the condition works, and you spent last half hour thinking how to select last element if multiple element satisfies condition.

I implemented a different solution for B than the one in the official editorial. Hopefully it will prove interesting for some:

We should find minimum after x and y for:

`(n * x + PS[n] - PS[y] - 2 * x * (n - y)) / n`

, where PS[i] is the sum of the first i elements of the sorted input array. Basically, y is an index such that all elements`1 <= i <= y`

will provide`min(A[i], 2x) = A[i]`

and all elements`y < i <= n`

will provide`min(A[i], 2x) = 2x`

. This also suggests a constraint for a fixed y:`V[y] / 2 <= x <= V[y + 1] / 2`

.Minimizing the formula above is the same as minimizing:

`n * x - PS[y] - 2 * x * (n - y) = x * (2 * y - n) - PS[y]`

. In order to minimize this, we can iterate`1 <= y <= n`

and find a suitable x. In fact, if`2 * y - n`

is negative, x should be as large as possible, so`x = V[y + 1] / 2`

. Otherwise (`2 * y - n`

is positive), than x should be as small as possible, so`x = V[y] / 2`

.can you give a link to your submitted solution?

Screencast with commentary

How much can we rate A in terms of CF ratings ?

I suppose 1600/1700.

In B, I first wrote a binary search, but the ans was increasing with decreasing precision. I think, the side it was going was not correct when precision was small and numbers were large. But, ternary search was consistent on this. Can someone clarify more on this?

I did it using binary search on the

`x`

Solution. The`loss vs x function`

is initially non-increasing and then non-decreasing, so we'll have to find x corresponding to the trough in the loss function [Start with`low = 0`

and`high = max(Ai)`

, then check the losses for`mid`

,`mid-1`

,`mid+1`

to half the search space]Later checked the editorial, turns out this is just equivalent to finding the median.

I was checking on mid and mid + e(the error term), I think that caused the issue

For problem B my solution tries to perform binary search to find when the derivative flips from negative to positive.

$$$f(x) = \sum\limits_{i=1}^N (x + A_{i} - min(A_{i}, 2x)) = \sum\limits_{i=1}^N (x + A_{i} - \frac{Ai + 2x - abs(A_i - 2x)}{2})$$$

$$$f'(x) = \sum\limits_{i=1}^N \frac{2x - A_i}{abs(2x - Ai)}$$$

Points where we get NaNs because of division by zero are skipped.

Thanks.

This just proves that median observation in editorial since that function is sum of signum functions derivate will change sign at median only.

Nice observation! That way to write $$$min(a,b)$$$ is very clever.

ARC E -

If one didnt realises second observation in E. Then $$$LCM_{j \neq i}(A_j)$$$ can be computed in $$$O(n \log max(A_i))$$$ using prefix and suffix LCMs with bigints.

Python Submission

Time complexity $$$O(n^3 \log max(A_i))$$$ remains same.

More specifically how did I come up with this particular solution was -

Since there exists one last no from index $$$i$$$ such that $$$LCM_{j \neq i}(A_j) \neq LCM_{j}(A_j)$$$. This python solution just brute forces to find one such index which can be last. Then recursively solves for the rest of $$$n-1$$$ numbers.

In D I thought it was ok to remove all pairs of equal elements initially, so that all the elements are distinct but it was failing on one test case and on removing this it passes all test cases, can someone point out a case where it fails(the test cases are not available on the dropbox yet)

WA AC

$$${1,19,19,20}$$$

Answer is 18 (1 xor 19 = 18, 19 xor 20 = 7) but removing 19 makes it (1 xor 20=21).

Thanks!

C:

spend hour to write solution which uses up to 250 operations, delete it, write proper solution in 15 minutes

Can anybody explain the curve in Question B

can anyone pls point out the mistake it is showing WA for 5 testcases for A many Formulae Problem

## include<bits/stdc++.h>

using namespace std;

## define lln long long int

## define ss second

## define ff first

## define vi vector

## define all(v) v.begin(),v.end()

## define vpi vector<pair<int,int>>

## define fr(i,a,n) for(int i=a;i<n;i++)

## define pb push_back

## define ub upper_bound

## define lb lower_bound

## define pi pair<int,int>

int gcd(int a,int b){ if(a<b)swap(a,b); if(b==0)return a; return gcd(b,a%b);

}

int main(){

int n; cin>>n; lln a[n]; lln m=1e9+7; fr(i,0,n)cin>>a[i];

lln add=a[0]%m,sub=0,ca=1,cs=0;

fr(i,1,n){

}

cout<<(add%m+sub%m)%m<<endl; }