atcoder_official's blog

By atcoder_official, history, 7 weeks ago, In English

We will hold AtCoder Beginner Contest 347.

We are looking forward to your participation!

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7 weeks ago, # |
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Good luck to everyone!

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7 weeks ago, # |
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GLHF.

And let's have a guess!

Good Round :
Just so so Round :
Bad Round :

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7 weeks ago, # |
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I hope I can stop dropping rated this round!

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7 weeks ago, # |
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GL&HF.Hope my rating do not drop

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7 weeks ago, # |
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My submission for problem C gives WA for only 1 testcase — LINK.

I can't figure out what's wrong in this,any help is appreciated.

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    7 weeks ago, # ^ |
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    same issue can anyone help

    include

    include

    include

    using namespace std;

    int main() { ios::sync_with_stdio(false); cin.tie(nullptr);

    int N;
    long long A, B;
    cin >> N >> A >> B;
    
    vector<long long> D(N);
    for (int i = 0; i < N; ++i) {
        cin >> D[i];
    }
    
    long long week_length = A + B;
    
    // Calculate the minimum and maximum days later
    vector<long long> temp;
    for (int i = 0; i < N; ++i) {
        long long days_later = D[i] % week_length;
        if (days_later == 0) {
            days_later = week_length;
        }
        temp.push_back(days_later);
    }
    sort(temp.begin(), temp.end());
    long long mini = temp[0];
    long long maxi = temp[N - 1];
    
    // Check if the difference between mini and maxi is less than or equal to A
    bool flag = (maxi - mini < A);
    
    if (flag) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
    
    return 0;

    } this is my code

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    7 weeks ago, # ^ |
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    same issue dude, I missed both C and D today by just 1 test case each, which is the same test case as what you got wrong. I have no idea what could go wrong, spent 1 hr into debugging but of no use :(

    Someone please help.

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    7 weeks ago, # ^ |
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    I think that the problem with your solution is that you didn't take into account that, even after first taking the remainder modulo $$$(A + B)$$$, the array can still be shifted for better adjustment.

    To illustrate this with an example, try the hack:

    2 5 5

    1 7

    The correct answer should be "Yes", but your solution outputs "No". The reason is, after taking the remainder, the array is still $$$[1, 7]$$$, but this gap of length $$$6 > A = 5$$$ between $$$1$$$ and $$$7$$$ means we can still shift the dates so that the first plan is on Day 5 and the second plan is on Day 11, which still works.

    I reckon that anyone else failing to pass one or two test cases might be having this problem, too.

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      7 weeks ago, # ^ |
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      i also failed the same TC

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      7 weeks ago, # ^ |
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      Thanks,I have never thouht of that and WA on 01_test_28.txt

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    7 weeks ago, # ^ |
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    Spoiler

    Answer should be yes.

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was able to solve till E, what's wrong with 1 testcase in C? can anybody tell that testcase?

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C was annoying

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7 weeks ago, # |
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Problem F is almost same with https://www.luogu.com.cn/problem/P3625

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problem C ????????????

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https://atcoder.jp/contests/abc347/submissions/51878189
where is this code going wrong ?? can anyone help . Thank you in advance

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what was the last test case in problem D, I spent more than half hour but was unable to figure out that test case

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7 weeks ago, # |
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D can be solved using DP.

AC Code

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My approach to problem C was I calculated all days[i]%(a + b) then find the maximum and minimum in this array if then length is greater than a then its no otherwise yes but it did fail 1 test case IDK can somebody explain why is this approach wrong ! much appreciated

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    7 weeks ago, # ^ |
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    Do you know what the problem is with your solution because I have the same problem.

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7 weeks ago, # |
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can somebody tell me why c is failing? I am checking if this below condition fails then it should be no: if(ar[i]%(a+b)>0 && ar[i]%(a+b)<=a) https://atcoder.jp/contests/abc347/submissions/51823167

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7 weeks ago, # |
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You can actually pass F without considering those two cases:

if (M <= i && i + M < N - M + 1) // three squares arranged vertically; fix the center square
    chmax(ans, upper_left[i - M].back() + sum[i][j] + lower_right[i + M].front());
if (M <= j && j + M < N - M + 1) // three squares arranged horizontally; fix the center square
    chmax(ans, upper_left.back()[j - M] + sum[i][j] + lower_right.front()[j + M]);

If you comment those in the std you can still get an AC. Weak tests.

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    7 weeks ago, # ^ |
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    You need to apply one more check, that the numbers you have constrcuted have popcounts equal to a and b respectively, As for both num1 and num2 too you have constraint of being <2^60

    Think of a case when a = 60 b = 60 and you have popcount of c = 60

    In this case you will greedilly keep 30 1's in num1 and other 30 1's in num2, This will satisfy xor condition, but you don't have 60 popcounts in num1 and num2

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I must say problem C is very perfect

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apologize for my impulse.

This comment was deleted.

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    7 weeks ago, # ^ |
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    an accepted submission

    data:

    4 1
    25 77 63 69 
    31 57 46 14 
    22 74 67 5 
    58 80 22 62 
    

    ans: 80+77+74=231

    my output is right while I did't pass the problem. The mentioned submission got wrong output.

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    7 weeks ago, # ^ |
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    In addition, the problem F is also the same as a problem in APIO2009. We found that the solution of that problem cannot be accepted in this. The problem link in a Chinese site.

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    7 weeks ago, # ^ |
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    I've just known that there's a problem of APIO which is almost the same as this one. I passed that problem.

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    7 weeks ago, # ^ |
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    ...well, I hacked my submission just now. maybe the tests of F are weak but correct.

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Good contest. Got 1 of WAs in C because got used to printing "YES" in cfs and atcoder requires 'Yes' :p

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My pC also got 1 subtask WA at 01_test_28.txt ; Wondering what's the problem of my idea ; Actually I don't even think we need O(NlogN) to solve this ; Can't we just simply calculate the max and min value of D_i%(A+B) for all elements in D ; And consider the "max-min < A" one as "Yes" case?

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7 weeks ago, # |
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What was C? I kept getting WA on 1 testcase, also I feel D>E.

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My Solution for E !!

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

int main(){
    int n , q;
    cin >> n >> q;
    vector<ll> queries(q+10,0) , vis(n+10,0) , ans(n+10,0) , marked(n+10,0) , sum(q+10,0);

    ll unique = 0;
    for(int i = 1; i <= q;i++){
        cin >> queries[i];
        ll x = queries[i];
        if(vis[x]) vis[x] = 0 , unique--;
        else vis[x] = 1 , unique++;
        sum[i] =  unique;
    }

    for(int i = 1; i <= q; i++) sum[i] += sum[i-1];

    for(int i = 1 ; i <= q; i++){
        ll x = queries[i];
        if(marked[x]){
            ans[x] += sum[i-1];
            marked[x] = 0;
        }
        else{
            ans[x] -= sum[i-1];
            marked[x] = 1;
        }
    }

    for(int i = 1 ; i <= n ; i++) if(marked[i]) ans[i] += sum[q];
    for(int i = 1 ; i <= n ; i++) cout << ans[i] << " ";

    return 0;
}



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Can someone help me understand why my solution for C is wrong. Below is what I did.

for (int D : plans) {
    int rem = D % (A + B);
    if (!(1 <= rem && rem <= A)) {
            return "No";
    }
}
return "Yes";
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    7 weeks ago, # ^ |
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    i also did same, People are sorting the array, i think we didnt get the question. Can somebody explained what is the expectation of the problem?

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    7 weeks ago, # ^ |
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    For this test case

    2 5 2
    6 7
    

    The answer is Yes but you output No

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      7 weeks ago, # ^ |
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      Understood. So basically the start day need not be the 1st day itself. Thanks.

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I really Liked the Problem E,

My solution to E, using two sets

E Solution
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Can anyone Give Counter Test for C https://atcoder.jp/contests/abc347/submissions/51861296 it is just falling on 2 test case

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7 weeks ago, # |
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a possible hack for problem F:

5 2
57 67 65 27 55 
46 20 63 8 71 
44 56 44 11 14 
47 23 46 90 56 
36 30 21 76 12 

the answer is 619.

(i1, i2, i3, i4, i5, i6) = (1, 2, 3, 1, 4, 4).

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    7 weeks ago, # ^ |
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    hey sorry to ask, can you please check my submission and let me know what i am doing wrong

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      7 weeks ago, # ^ |
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      only checking the maximum and minimum values is not enough.

      your solution is far from the right one. suggest you do read the Editorial.

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        7 weeks ago, # ^ |
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        thanks man also don't stay alone

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Can someone help me what's wrong with my E solution? It's AC for some tests but logic looks correct https://atcoder.jp/contests/abc347/submissions/51888633

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Can someone explain me the C's editorial. I didnt get why we are subtracting ei+1- ei?

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Can someone please help me find out where this code is failing for D? Submission Link

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I think the testdata of F is kind of weak.
The solution mentioned 6 situations, but I didn't check the 1st and 4th (three submatrices on [left, mid, right], [top, mid, bottom]), still got AC.

My submission
(the array dp2[][] is useless)

A hack data which my code can't pass:

Input:
3 1
2 2 2
0 0 0
0 0 0

Answer: 6
Output: 4

Update: Here's what happened in my code.

ll dp1[4][maxn][maxn]; // LU LD RU RD

ll ans = -0x3f3f3f3f3f3f3f3fll;
for (int i = 1; i < n; ++i) {
  for (int j = 1; j < n; ++j) {
    // LU RU D
    ans = std::max(ans, dp1[0][i][j] + dp1[2][i][j + 1] + dp1[1][i + 1][n]);
    // LD RD U
    ans = std::max(ans, dp1[1][i + 1][j] + dp1[3][i + 1][j + 1] + dp1[0][i][n]);
    // LU LD R
    ans = std::max(ans, dp1[0][i][j] + dp1[1][i + 1][j] + dp1[2][n][j + 1]);
    // RU RD L
    ans = std::max(ans, dp1[2][i][j + 1] + dp1[3][i + 1][j + 1] + dp1[0][n][j]);
    // I didn't check L M R and U M D
  }
}
std::cout << ans << '\n';
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Need help in debugging my code for problem D. My code fails on the last test case. Please help me in debugging, I'm not able to find any mistake. My Submission for D

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Problem c was very annoying

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Spoiler

Can someone help me figure out why my code for problem C is failing , there's only 1 testcase that it didn't pass and I can't find a counter test case for my code.
According to my logic we reduce A[i] to A[i]%(a+b) at first , then we check for each Di how much we need to add in order for its mod with a+b to lie within [1,a] , based on this we'll get edges of an interval , if the interval is overlapping for all values of Di then I print yes , else no.

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I also received 1 WA in C, but then I encountered an edge case where we can also create a cyclic schedule (which I hadn't considered during the contest).

1 WA Code
AC Code
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    7 weeks ago, # ^ |
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    what's a cyclic schedule?

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      7 weeks ago, # ^ |
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      I mean you can imagine that as the subarray of plans which should have smallest length and fits in holiday schedule (You can make any day as starting point and see).

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        7 weeks ago, # ^ |
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        how to visualize total — plan[i] + plan[i — 1]?

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          7 weeks ago, # ^ |
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          Take an example a = 4, b = 3
          Total Days: 1 2 3 4 5 6 7
          so you can have holiday of length 4 (4 days)
          Plans = 1 2 6

          now as per my code
          - you can say 1 as your first day of holiday so 1 2 3 4 here 6 is not included so it will not work
          - now 2 as your first day of holiday so 2 3 4 5 this will also not work
          - now 6 as your first day so 6 7 1 2, it is covering all plans so it is possible to do plan on this holiday window.

          I hope you got the point.

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Why I didn't take a look at PE... It turns out that PE was much easier than PC.

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Mr. Takahashi, could you please add my ranking when I click on the Show favs only to filter the ranking list . So that we can compare with our good friends. Just like the Codeforces.

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7 weeks ago, # |
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.

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5 weeks ago, # |
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why checking only if(d[i - 1] + a + b - d[i] + 1 <= a){ flag = 1; } is sufficent ?

what about the days at left of i-1 and to the right of i ?

how we will make sure that are also satisfied