Разбор Технокубок 2021 — Отборочный Раунд 2 (Codeforces Round 687 Div.1, Div.2)

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Разбор задач Технокубок 2021 - Отборочный Раунд 2

Разбор задач Codeforces Round 687 (Div. 1, основан на Отборочном раунде 2 Технокубка 2021)

Разбор задач Codeforces Round 687 (Div. 2, основан на Отборочном раунде 2 Технокубка 2021)

technocup, технокубок, технокубок 20201

+182

KAN
6 лет назад
76

Комментарии (54)

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aspirati0n

6 лет назад, скрыть # |

+26

Good!

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ThoughtNerd

6 лет назад, скрыть # ^ |

-6

Why?

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heath_cliff

6 лет назад, скрыть # |

For problem Div2 E, 99886275 is just so cool.

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Priyanshu14_

6 лет назад, скрыть # |

Thank You.

Nice Problemset — problems were to the point and good.

I really liked it.

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Feng

6 лет назад, скрыть # |

I used dictionary tree to passed the div2 D, I can guarantee its correctness, but I can't guarantee its time complexity,can anyone hack my solutions? https://codeforces.me/contest/1457/submission/99900717

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notAnAltAccount

6 лет назад, скрыть # |

In Div2D, why is n<=60 if the given condition is true?

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epsilon_573

6 лет назад, скрыть # ^ |

Because if n>60 , We can always find a contiguous triplet (a,b,c) such that a > b^c.

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Kofta

6 лет назад, скрыть # ^ |

Why we can always find that?

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C137

6 лет назад, скрыть # ^ |

← Rev. 3 →

In case $$$n \gt 60$$$ There will be three numbers $$$(a, b, c)$$$ all having the same msb (Most Significant Bit) set to 1. You can do one operation with numbers $$$b, c$$$ this will turn off the msb therefore $$$ a \gt (b \oplus c)$$$.

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Kofta

6 лет назад, скрыть # ^ |

Thanks

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EugeneJudo

6 лет назад, скрыть # ^ |

← Rev. 2 →

Because a_i <= 10^9 ~ 2^30. So of numbers which are k digits long in binary, we can have at most 2 to not get a triplet. And there are only 30 different sizes before we hit the cap of 10^9.

edit: I may have answered the wrong question

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Kofta

6 лет назад, скрыть # ^ |

Thanks

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JNU_dh_AK_WF

6 лет назад, скрыть # ^ |

← Rev. 2 →

Because these three numbers share the same highest bit $$$1$$$. And $$$1 \oplus 1=0$$$. Let's consider these three numbers as $$$a,b,c$$$ where $$$a \lt b \lt c$$$ , then the highest bit will become $$$0$$$ after we do $$$b \oplus c$$$, so $$$a \gt b \oplus c$$$ holds. Then we can always find that.

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Kofta

6 лет назад, скрыть # ^ |

Thanks

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minhcool

6 лет назад, скрыть # |

+52

Me when doing normal rounds: So I will try to do problems in order, if I can't do A in 30 minutes then I will consider B...

Me when doing russian rounds:

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Newton_01

6 лет назад, скрыть # |

For Problem B, n is <=10^5, test cases are <=10^4 and no. of colours are <=100. How is 10^11 (that's what I can interpret) solution works??

If some one has any explanation and would like to correct where I am wrong, it will be great :)

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naitikvarshney

6 лет назад, скрыть # ^ |

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

So when value of n is 10^5 then no. of test cases will definitely be 1. When test cases are 10^4, then value of n in all test cases are such that their sum will be atmost 10^5.

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Enigma20

6 лет назад, скрыть # ^ |

There is a statement, "Sum of $$$n$$$ over all test cases doesn't exeeds $$$10^5$$$". That basically means $$$t*n \lt =10^5$$$. So, overall operations doesn't exeed $$$100*10^5=10^7$$$ which is acceptable in the given time limit.

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huikang

6 лет назад, скрыть # |

I did not see that B had only 100 colors :/

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AdHocMan

6 лет назад, скрыть # |

I am still confused in Div 2D.

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ritesharma

6 лет назад, скрыть # |

plzz anyone can explain div2 D. why n<=60

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ritik_patel05

6 лет назад, скрыть # ^ |

For n <= 1e9, There can be atmost two numbers having same most significant bit. Therefore the bound will be atmost twice the number of bits i.e. 2 * log2(1e9) which is 60.

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ritesharma

6 лет назад, скрыть # ^ |

thanx bro!

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ritik_patel05

6 лет назад, скрыть # |

Thanks for detailed explanation of Div2E!

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devamg

6 лет назад, скрыть # |

For Div 2 Problem A, this is my code

Code


#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
 
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
 
	ll t;
	cin >> t;
 
	while(t--) {
		ll n, m, r, c;
		cin >> n >> m >> r >> c;
		ll v1 = abs((r + c) - 2);
		ll v2 = abs((r + c) - (1 + n));
		ll v3 = abs((r + c) - (1 + m));
		ll v4 = abs((r + c) - (m + n));
		cout << max(max(v1, v2), max(v3, v4)) << '\n';
		
	}
 
	return 0;
 
}

I have calculated the Manhattan Distance of the cell (r, c) from each of the corner cells, and printed the max distance. My code is giving wrong answer on test 2. Can anyone tell what's wrong in my code?

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munna_bhayia

6 лет назад, скрыть # ^ |

abs(a — b) + abs(c — d) is not equal to abs((a+c) — (b+d)).They both are different.

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ritesharma

6 лет назад, скрыть # ^ |

find abs of (row) and col seperate for all four corner

int ans=Math.max(row-1+col-1, Math.abs(n-row)+Math.abs(m-col));

       ans=Math.max(ans, row-1+Math.abs(m-col));
       ans=Math.max(ans, col-1+Math.abs(n-row));

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piddddgy

6 лет назад, скрыть # |

← Rev. 4 →

for D, I brute forced subarrays of size <= 31, then brute forced again on the "split point". for example, if i have a subarray i...j and a split point of k, we xor the range from i...k and k+1..j. then, we check if the sequence we made decreases at any point. why does this work? I used the claim that there does not exist a construction where we have to make > 29 moves.

https://codeforces.me/contest/1457/submission/99871104

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Xellos

6 лет назад, скрыть # |

+35

Nobody found the $$$O(N)$$$ for div1D?

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EmptySoulist

6 лет назад, скрыть # ^ |

how to solve div1D in $$$O(N)$$$ ?

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Xellos

6 лет назад, скрыть # ^ |

For each situation where you just took the $$$i$$$-th thing, find the set of positions of the clone. For each situation where the clone just took the $$$i$$$-th thing, find the set of your positions. In the first case, it's a contiguous range, which means it has to be up to 2 contiguous ranges in the second case. There's a lot of possible transitions from $$$i$$$ to $$$i+1$$$ but it's a constant number and each takes constant time.

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hoacrux

6 лет назад, скрыть # |

Can anybody explain the prefix suffix method used in division 2 in D-problem?

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teja349

6 лет назад, скрыть # |

Is it only me or Div1 E editorial is cryptic.

"Let call low bits that we don't need to care free bits." — can you maybe give an example.

What is A there?

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ruanxiaoyu

6 лет назад, скрыть # ^ |

← Rev. 3 →

For example, assume you have segment $$$(10, 20)$$$ and you set your number $$$011??$$$, your number will be in the segment whatever you set two lowest bits, then they are free.

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Conata

6 лет назад, скрыть # |

in div2C,why the answer of 2 2 1 10 11 1 is 11,if i make the 0 to 1,it just take 1 second

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NerfThis

6 лет назад, скрыть # ^ |

This is your test case:

n = 2, p = 2, k = 1; s = "10"; x = 11, y = 1.

x is the cost of adding a platform, y is the cost of removing the first cell. In your case, there is only 1 option: that is to add a platform at position 2, giving 11 as your answer.

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NerfThis

6 лет назад, скрыть # |

KAN For div2 D, the editorial states the following:

Indeed, there are no two integers with the same highest bit set. It is much easier to solve the problem in such constraints.

I think it should be stated as: there are at most two integers with the same highest bit set. Right? Because for input [1, 6, 7], the highest set bit of all numbers are different but 6 and 7 have the same highest set bit.

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Forrest_Gump

6 лет назад, скрыть # ^ |

Ya! Exactly. I too think it's a typo.

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KAN

6 лет назад, скрыть # ^ |

Thanks, corrected! I missed "more than".

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codinginveins

6 лет назад, скрыть # |

How is the space complexity for div2B O(N) ?

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NerfThis

6 лет назад, скрыть # ^ |

You are essentially keeping a separate index list for each possible color. The total houses is N, so the total entries in all index lists sum up to N.

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RajawatBanna

6 лет назад, скрыть # |

It is good tutorial.

Kindly do mention the code also. Sometime explanation is not sufficient

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jagerknight

6 лет назад, скрыть # |

Could someone please explain the dp approach to div2 C?

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X-O__O-X

6 лет назад, скрыть # ^ |

← Rev. 3 →

https://codeforces.me/blog/entry/85081#comment-727466

Spoiler

I will explain more the 4 and 5 hints. - Given that the first cell on which the ball will land is cell i, we know it will visit cell i, i+k, i+2k, ... , i+xk<n, right! so we want to count how many index j in {i, i+k, i+2k, ..., i+xk<n} such that s[i] = '0'. In other words, we need to how many cell in the group of cells the ball will visit, does not have a platform. we will of course need to put a platform in each one of them. we have no choice.

Let cnt[i] denote the min number of platforms we must add if the ball starts the journey at index i.
Let's consider the bruteforce first. It's a simple for loop.

we simply iterate over indecies i, i+k, i+2k, ... and if no platform we increment a counter. right!

for (int i = p; i < n; ++i) {
  for (int j = i; j < n; j += k ) {
      if (s[j] == '0') cnt[i]++; 
  }
}

In the previous code, we calculated cnt[i] for every index we are interested in.
This code is O(n^2)
Now, we need to optimize this code using DP
to do that, you need to realize the recursive nature of cnt[i].
cnt[i] clearly depends of cnt[i+k]

cnt[i] = (s[i] == '0' ? 1 : 0 ) + cnt[i+k]
// it means that number of platforms to add if ball starts at i is equal to number of balls to add if ball starts at i+k, plus 1 if cell i also needs a platform.

Now, we can write a recursive code.

int cnt(int i) {
    if (i>=n) return 0; 
    if (s[i] == '0') return cnt[i+k] + 1;  
    else return cnt[i+k]; 
}
int mem[n] = {};
for (int l=p; l<n; ++l) 
   mem[l] = cnt(l);

Now, It should very clear how dp or memoization should be used. we are simply calling the function cnt(l) for n-p times and in each call, it will need about (n-l) / k call. but the function has only 1 parameter with at most n values. so, there must be repetitive calls. use memoization. I hope it's clear now.

int mem[n]; 
int cnt(int i) {
    if (i>=n) return 0; 
    int &ret = mem[i]; 
    if (~ret) return ret; 
    if (s[i] == '0') return ret = cnt[i+k] + 1;  
    else return ret = cnt[i+k]; 
}


for (int l=p; l<n; ++l) 
   cnt(l);

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jagerknight

6 лет назад, скрыть # ^ |

Thank you!

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wasa855

6 лет назад, скрыть # |

+18

It seems that the standard program to the Div.1 D has something wrong
https://codeforces.ml/contest/1456/hacks

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KAN

6 лет назад, скрыть # ^ |

Fixed, some testers' solution got hacked too.

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Mridul323

6 лет назад, скрыть # |

What would be the difficulty level of XOR gun?

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WitchOfTruth

6 лет назад, скрыть # ^ |

I bet 1800

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charging

6 лет назад, скрыть # |

Nice trick in problem Div2D =)

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EmptySoulist

6 лет назад, скрыть # |

Sorry for my stupid, it's a little hard for me to understand Div1E solution, can someone explain it easily? thanks!

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SGrade

6 лет назад, скрыть # |

1415B - Repainting Street

Can somebody explain why test case 3 in test 2 (below) says right answer is 2 days? If we choose any color (e.g. 1), there are only 6 houses, which are not this color. So, with capacity k=6 one day should be enough to paint. Looks I am missing something, but don't get what.