include

9 месяцев назад, # |

I think I will do three questions.

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tl_xujiayi

9 месяцев назад, # |

I hope it can be simpler this time.

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thelastdinner

9 месяцев назад, # |

it's better than a well-known problem in ex.

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Kevin_pbw

9 месяцев назад, # |

But only I can't connect to AtCoder?

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no_oneplb

9 месяцев назад, # |

+41

What the hell is problem statement of C?

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rsiloniya

9 месяцев назад, # ^ |

Same problem, if the cells are given completely random then, total possibilities will be 9! and disappointments would be a small fraction. If someone understood or solved it correctly, please write in this thread.

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jason1234567890

9 месяцев назад, # |

B is so hard.

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44_FeiDing

9 месяцев назад, # ^ |

-14

just enumerate

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9 месяцев назад, # |

+13

The atcoder crashed...

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44_FeiDing

9 месяцев назад, # ^ |

My Atcoder is fine.

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dsanurag520

9 месяцев назад, # |

dude can you explain qC

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9 месяцев назад, # ^ |

I don't know what does C mean too.

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kabu

9 месяцев назад, # ^ |

I don't know what does C mean too.

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junjun5653-2

9 месяцев назад, # |

-8

I don't have any ideas about C-False Hope.

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9 месяцев назад, # ^ |

-8

This question is very vague.

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frostwood

9 месяцев назад, # ^ |

-8

agree.I can't get the writter's idea

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9 месяцев назад, # ^ |

You can consider all permutations of the order Takahashi sees the grids, check if it satisfies the condition easily. I agree the statement is not so clear, and that reading is the hardest part in solving the problem xd.

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sukh_x

9 месяцев назад, # ^ |

How we're calculating the probability tough!

→ Ответить

9 месяцев назад, # ^ |

wdym? isn't the probability equal to "total number of permutations that satisfy condition"/(1*2...*9) ?

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sukh_x

9 месяцев назад, # ^ |

oh...alrgt will try it out

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44_FeiDing

9 месяцев назад, # |

The difficulty of problems A-F will not be changed

But I think it's harder than usual

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VMaksimoski008

9 месяцев назад, # |

C has the worst problem statement I've ever seen.

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Hosen_ba

9 месяцев назад, # |

← Rev. 2 →

-33

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9 месяцев назад, # ^ |

-11

Agree.The description of the title of this question is terrible.I don't know the origin of 2/3 at all!

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plagiarism2012

9 месяцев назад, # |

Literal False hope

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9 месяцев назад, # |

+17

After reading C:

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Sreedhar3210

9 месяцев назад, # |

The Problem statement and explanation for C is too bad

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uks_28

9 месяцев назад, # |

Task C was really a disaster.

Never seen such a bad description of a problem ,Even the Explanation of test cases are not clear.

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baiganio

9 месяцев назад, # ^ |

-14

I think the explanation is quite clear but the problem is too hard for C.

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WeaponizedAutist

9 месяцев назад, # |

-8

Am I the only one who thought E >>> (F and G)

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9 месяцев назад, # ^ |

how F please

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WeaponizedAutist

9 месяцев назад, # ^ |

← Rev. 2 →

you can do bitmask dp (bitmask denotes consumed medicines), notice that it's always optimal to defeat any enemy as soon as you can defeat it

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9 месяцев назад, # ^ |

Thank you very much! I didn't see the medicine constraint during the contest :(

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sahaun

9 месяцев назад, # ^ |

Probably because $$$G$$$ is a kinda-standard problem?

But I think E is easier than F. $$$1\leq P_i\leq 8$$$ gives it away.

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g-s01

9 месяцев назад, # ^ |

How did you do E?

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9 месяцев назад, # ^ |

← Rev. 3 →

Consider lcm(1 ... 8) = 840, so all the starting time that have the same value after mod 840 has the same min time to transport from station 1 to n. Just preprocess the min time of station transportation starting at time x(mod 840) and answer the questions after this.

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g-s01

9 месяцев назад, # ^ |

Could you provide with a proof for this?

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9 месяцев назад, # ^ |

← Rev. 2 →

sort of obvious rly... I am really sorry, but I dont know how to put the explanation into mathematical terms :(

but maybe you can figure out by examining all the time mod p[i] after visiting station i and you will see.

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WORTH

9 месяцев назад, # ^ |

← Rev. 2 →

Let's say you have $$$2$$$ buses having $$$p$$$ $$$=$$$ $$$[5, 6]$$$ and $$$t$$$ $$$=$$$ $$$[3, 4]$$$

If you reach the first bus stand at $$$0$$$, you can take the first bus at $$$0$$$, reach the bus stop after $$$3$$$ units of time, start bus two at $$$6$$$, and reach the final bus stop at $$$10$$$.

You can find the same for different values when you reach the first bus stop. It will be different for starting times (in this case) $$$6, 11, 16, ...$$$

Now try to find the bus travel time if you reach the first bus stop at time equal to $$$lcm(5, 6) = 30$$$. You'll realize that the position of buses is exactly the same as it was at the time $$$= 0$$$. So the value is same and this pattern will continue.

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sahaun

9 месяцев назад, # ^ |

Hint 1

Hint 2

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aditya_108

9 месяцев назад, # |

Can anybody explain C ?

→ Ответить

WeaponizedAutist

9 месяцев назад, # ^ |

just find all valid orders by brute forcing over all permutations of length 9

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Imran435

9 месяцев назад, # ^ |

Doing just as you suggested but getting WA on sample testcase #2. Any idea what is wrong with my code?

#include <bits/stdc++.h>

using namespace std;
using i64 = long long int;

void testcase() {
  vector<vector<int>> a(3, vector<int>(3));
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
      cin >> a[i][j];

  vector<int> p(9);
  iota(p.begin(), p.end(), 0);

  int total = 0, favourable = 0;

  do {
    bool ok = 0;
    vector<vector<int>> rows(3);
    vector<vector<int>> columns(3);
    vector<vector<int>> diagonals(2);

    for (int i = 0; i < 9; i++) {
      int x = p[i] / 3, y = p[i] % 3;
      rows[x].emplace_back(a[x][y]);
    }

    for (int i = 0; i < 9; i++) {
      int x = p[i] / 3, y = p[i] % 3;
      columns[y].emplace_back(a[x][y]);
    }

    for (int i = 0; i < 9; i++) {
      int x = p[i] / 3, y = p[i] % 3;
      if ((x == 0 && y == 0) || (x == 1 && y == 1) || (x == 2 && y == 2))
        diagonals[0].emplace_back(a[x][y]);
      else if ((x == 0 && y == 2) || (x == 1 && y == 1) || (x == 2 && y == 0))
        diagonals[1].emplace_back(a[x][y]);
    }

    for (int i = 0; i < 3; i++)
      if (rows[i][0] == rows[i][1] && rows[i][1] != rows[i][2])
        ok = 1;

    for (int i = 0; i < 3; i++)
      if (columns[i][0] == columns[i][1] && columns[i][1] != columns[i][2])
        ok = 1;

    for (int i = 0; i < 2; i++)
      if (diagonals[i][0] == diagonals[i][1] &&
          diagonals[i][1] != diagonals[i][2])
        ok = 1;

    total++;
    favourable += 1 - ok;
  } while (next_permutation(p.begin(), p.end()));

  double ans = 1.0 * favourable / total;
  cout << setprecision(10) << ans << "\n";
}

int main() {

  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
#endif

  int T = 1;
  // cin >> T;
  for (int t = 1; t <= T; t++) {
    testcase();
  }

  return 0;
}

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Imran435

9 месяцев назад, # ^ |

Found the bug!!

Just leaving the comment though in case anyone stumbled upon.

Correct Code

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213976

9 месяцев назад, # |

how to solve D?

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Lovely_CatHxy

9 месяцев назад, # ^ |

Binary Search

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ThePeak

9 месяцев назад, # ^ |

In D, what if the order of the words didn't matter, so would that problem be solved by dp ? How would we implement the solution in that case ?

→ Ответить

__SSB__

9 месяцев назад, # ^ |

D can be solved using binary search, choose l as 0 and r as 1e9.

Then apply binary search, with capacity as (l+r)/2, for the given capacity check if we can fill all words in m or less lines.

finally print value of left.

Basically capacity represents the best values for columns against m rows.

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vikas_s

9 месяцев назад, # ^ |

← Rev. 3 →

include

using namespace std;

bool checkIfPossible(vector&a , long long mid , int m){

int n = a.size();
long long int sum = 0;
int c = 1;

for(int i = 0;i<n;i++){
    sum = sum + a[i] + 1;
    if(sum-1 > mid){
       c += 1;
       sum = a[i]+1;
    }
}
return c <= m;

}

void solve(){ long long n , k; cin>>n>>k; vectora(n); for(int i = 0;i<n;i++){ cin>>a[i]; }

long long low = 0;
long long  high =  1000303000000000 ;

while(low<=high)
{

    long long  mid = (low + high)/2;
    if(checkIfPossible(a , mid , k)){
       high = mid-1;
    }
    else{
       low = mid+1;
    }
}
cout<<low<<"\n";

}

int main() { #ifndef ONLINE_JUDGE freopen("./../input.txt","r",stdin); //file input.txt is opened in reading mode i.e "r" freopen("./../output.txt","w",stdout); //file output.txt is opened in writing mode i.e "w" #endif int t; // cin>>t; t = 1; while(t--){ solve(); } }

For Question D . What is wrong with my solution ? What am i doing wrong ?

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n1k1450

9 месяцев назад, # ^ |

what if single element value is greater than mid?

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vikas_s

9 месяцев назад, # ^ |

Thank you .Yes , that was the mistake

→ Ответить

Mighty_CK

9 месяцев назад, # ^ |

Hey can you repaste the correct code here please,considering the case: if (word_width > capacity) { return false; // Single word cannot fit within capacity }

→ Ответить

vikas_s

9 месяцев назад, # ^ |

include

using namespace std;

bool checkIfPossible(vector&a , long long mid , int m){

int n = a.size();
long long int sum = 0;
int c = 1;

for(int i = 0;i<n;i++){
    if(a[i]>mid){
       return false;
    }
    sum = sum + a[i] + 1;
    if(sum-1 > mid){
       c += 1;
       sum = a[i]+1;
    }
}
return c <= m;

}

void solve(){ long long n , k; cin>>n>>k; vectora(n); for(int i = 0;i<n;i++){ cin>>a[i]; }

long long low = 0;
long long  high =  10003030000000000 ;

while(low<=high)
{

    long long  mid = (low + high)/2;
    if(checkIfPossible(a , mid , k)){
       high = mid-1;
    }
    else{
       low = mid+1;
    }
}
cout<<low<<"\n";

}

→ Ответить

Mighty_CK

8 месяцев назад, # ^ |

Thank You.

→ Ответить

0Fqwq

9 месяцев назад, # ^ |

Consider using a dichotomous answer for this question.

Regarding the proof of monotonicity for this question: since the wider the width of the display, the more width can be displayed in a line, the fewer lines are needed for the display (the total number of words remains the same). So the width of the monitor and the number of lines needed to display it** have a monotonically decreasing relationship**, so you can use dichotomization.

Dichotomize enumerates the width of the display $$$w$$$, and then calculates the minimum number of lines needed to display a word with a width of $$$w$$$, if the number of lines needed is less than or equal to $$$m$$$, then it meets the requirement, otherwise it doesn't meet the requirement.

How is this calculated?

If the width of the display $$$w$$$ is less than the maximum width of the word $$$mw$$$, it means that a word can not be put into the display, obviously does not meet the requirements.

So that $$$sum $$$ that the current line has been used for how much width, $$$cnt $$$ that the current use of how many lines, an enumeration of the length of the word $$$ l_i $$$, if the current width plus a space and $$$l_i $$$ the length of more than $$$ w $$$ that the current line can not be put in the word, the need to open a new line, so that $$$cnt $$$ plus $$$ 1 $$$, and so that $$$sum $$$ equal to $$$ l_i $$$, otherwise $$$sum $$$ plus $$$ l_i $$$ and $$$ l_i $$$ and $$$ l_i $$$, otherwise $$$ sum $$$ plus $$$l_i$$$ and the length of a space.

Time complexity: $$$O(\log n)$$$ for bisection, $$$O(n)$$$ for one calculation, $$$O(n \log n)$$$ for overall time complexity.

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Lovely_CatHxy

9 месяцев назад, # |

First five problems without C, how awkward, and bad problem.

→ Ответить

st_pigeon

9 месяцев назад, # |

+10

Laugh at A, get confused about B, and then give up understanding at C.

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__SSB__

9 месяцев назад, # |

I request Atcoder to take some action against a mass group whose user name starts with "klu". These all students belong from a certain university and have clearly mass copied, this drastically pushed the ranks.

→ Ответить

9 месяцев назад, # ^ |

Even with mass cheating, these losers could solve only the easiest three.

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__SSB__

9 месяцев назад, # ^ |

Even i was able to solve only the easiest 3, it took me ages to realise solving c wasnt worth it, but i clearly dont deserve a rank of 6000 when it should be in range of 3000-4000

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9 месяцев назад, # ^ |

totally agree.

→ Ответить

Lyrically

9 месяцев назад, # ^ |

+16

wtf why do those Indian cheaters start invading AtCoder too?

→ Ответить

Banis

9 месяцев назад, # |

Can someone explain the statement of C?

→ Ответить

no_oneplb

9 месяцев назад, # |

In problem C,Takahashi can see numbers in each cell in 9! different ways and we have to calculate valid permutations.A valid permutation is one in which for each row,column and diagonal which consists of two same values and one different value,he must visit the cell with different value before atleast one of the other two cells with same value. After that , the implementation is pretty much brute force.

But the definition of valid permutation (not disappointed) was not clear at all atleast for me personally, it took me 40 min to understand what question is asking us to do and 7 odd minutes to implement it.

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aditya_108

9 месяцев назад, # ^ |

Same with me.The problem statement should be more clear.

→ Ответить

vantaablackk

9 месяцев назад, # ^ |

I considered the 9! factorials ways of choosing the grid cells. For every such order, I made a 3X3 grid from them, and then checked whether the permutation was valid by considering the 16 possible unique lines. But I'm getting a much lower answer for sample 1. Can you please take a look at what I'm doing wrong? submission

→ Ответить

SXZZ5

8 месяцев назад, # ^ |

What do you mean you checked "whether the permutation was valid" ? Aren't all permutations valid ? They are just a way of picking up the picking up the elements from the grid. I think the real question is whether a permutation is "disappointing" for Takahashi.

→ Ответить

vantaablackk

8 месяцев назад, # ^ |

Yeah, that's what I meant really, it's not valid, it's disappointing.

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ikr

9 месяцев назад, # |

Yeah, C was a total buzz kill today. I wonder how much clearer was the Japanese description (for the Japanese speakers)...

→ Ответить

Include_Z_F_R_

9 месяцев назад, # |

G is even easier than F

→ Ответить

100M_

9 месяцев назад, # |

can someone explain the solution for problem E?

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9 месяцев назад, # |

I think this can be called a ABC319C disaster.

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Lovely_CatHxy

9 месяцев назад, # ^ |

Can't agree more. Many of my classmates give up solving D,E when seeing C.

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stalkkk_19

9 месяцев назад, # ^ |

same i do :(

→ Ответить

Chimpanzee

9 месяцев назад, # |

+15

What is C? Why is C?

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Blog that explains this idea

9 месяцев назад, # |

atcoder_official Editorials when?

→ Ответить

123asdf123

9 месяцев назад, # |

Why there's no English Editorial?

→ Ответить

HexShift

9 месяцев назад, # |

can someone explain the sol of G? i'm thinking about bfs but i can't seem to optimized it.

→ Ответить

StayAlone

9 месяцев назад, # |

When will the English Editorial be released

→ Ответить

Aliyyiakbar

9 месяцев назад, # |

← Rev. 2 →

-8

How to solve G? I used simple BFS with o(n^2*log(n)) solution using sets and got TLE.

→ Ответить

ElonMuskOfficial

9 месяцев назад, # ^ |

-8

Oh typical Turkish guys, $$$n \leq 2*10^5$$$.

→ Ответить

Aliyyiakbar

9 месяцев назад, # ^ |

-8

I'm not turkish.

→ Ответить

ElonMuskOfficial

9 месяцев назад, # ^ |

-8

Azerbaijanis are Turks.

→ Ответить

Aliyyiakbar

9 месяцев назад, # ^ |

They're Azerbaijani Turks.

→ Ответить

diobrando97

9 месяцев назад, # |

Is there any problem related to G? Very thanks!

→ Ответить

NimaAryan

9 месяцев назад, # ^ |

Problem

→ Ответить

italianDefence

9 месяцев назад, # ^ |

Thanks a lot. I was able to solve the problem after reading the blog. Upvoted!

→ Ответить

LetsHaveFun2

9 месяцев назад, # |

Can someone help me understand why my code for problem D doesn't pass all the tests? https://atcoder.jp/contests/abc319/submissions/45406059

→ Ответить

llc5pg

9 месяцев назад, # ^ |

I modified your code a bit. The submission is here: https://atcoder.jp/contests/abc319/submissions/45437738

→ Ответить

LetsHaveFun2

9 месяцев назад, # ^ |

Thanks! Can you help me understand why the modified code works and the original not? As far as I understand you increase the lower bound so there will be less checks during the binary search so it could improve the running time but it shouldn't effect the correctness of the program but for some reason it does.

→ Ответить

ZXG_DZXX

9 месяцев назад, # |

C is the most shit problem

→ Ответить

9 месяцев назад, # |

I have a strange solution for F:

Enumerate over all possible visiting orders of vertices with $$$t_i=2$$$. For each permutation, consider the following greedy strategy:

Find a monster you can reach with the minimum strength. If you can't defeat it, take the reachable medicine with the highest priority (determined by the visiting order).

Repeat this step, until all monsters are cleared or there is no medicine available.

The overall time complexity is $$$\mathcal O(m!\times n\log n)$$$, where $$$m$$$ is the number of medicine vertices. So it can't fit in the time limit, but actually with some simple optimizations it ran in $$$800\mathrm{ms}$$$ (maybe can be hacked?).

However, I got WAx1 (submission). Did I make a mistake in the program or the idea is wrong?

→ Ответить

9 месяцев назад, # ^ |

← Rev. 3 →

I used the same solution as yours.I found I got 3 TLE.So I break when it will up to 2 seconds with No.Then I got 2 WA.So I change to output Yes or No randomly with the one-third probability outputs No,otherwise outputs Yes.Then I pass the task with judging 9 times in all.submission(After 7 times,Oh god,I found I had forgot to add srand(time(0)); at first!)

→ Ответить

9 месяцев назад, # ^ |

Oh,I found another one.Using random_shuffle instead of next_permutation can get a higher probability to pass this task.I tried 3 times,and they all passed.

→ Ответить

9 месяцев назад, # ^ |

Thanks. But my code tried all the permutations, without any randomization. I expected TLE, but why does it get WA? Could you please have a look at my code?

→ Ответить

9 месяцев назад, # ^ |

OK,let me have a look.

→ Ответить

9 месяцев назад, # ^ |

Some optimizations:

use __gnu_pbds::priority_queue instead of std::priority_queue for a better constant factor
if $$$(\text{the current strength})\times (\text{the product of all the available }g_i)\ge (\text{maximum monster strength})$$$, then print Yes.

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9 месяцев назад, # ^ |

I'am in China now,I can't download the testcase.But you can try it.And I found another thing:I'am wrong,I can reach the subtree of a medicine without taking it.But I passed!

→ Ответить

9 месяцев назад, # ^ |

Maybe the test cases are too weak?

And the test cases for ABC319 are't published yet (in fact, the latest public test cases from AtCoder is that of ABC311).

→ Ответить

9 месяцев назад, # ^ |

Oh,that's too bad.

→ Ответить

Tudor67

9 месяцев назад, # ^ |

← Rev. 2 →

I got the same WA x 1 (my submission) for problem F and I would be happy to know what's wrong with my solution.

Are there any special cases or I misunderstood the statement of problem F?

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