Good luck to everyone!

GLHF.

And let's have a guess!

Good Round :
Just so so Round :
Bad Round :

I hope I can stop dropping rated this round!

GL&HF.Hope my rating do not drop

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.

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

C was annoying

Problem F is almost same with https://www.luogu.com.cn/problem/P3625

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

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

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

D can be solved using DP.

AC Code

https://atcoder.jp/contests/abc347/submissions/51845269

any help for C test case?

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

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

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.

Does greedy not work for D?

https://atcoder.jp/contests/abc347/submissions/51871701

I must say problem C is very perfect

apologize for my impulse.

This comment was deleted.

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

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?

What was C? I kept getting WA on 1 testcase, also I feel D>E.

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;
}

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";

I really Liked the Problem E,

My solution to E, using two sets

    auto solve = [&]()->void
    {

        int N , Q ;
        cin>>N>>Q ;
        multiset< int > s1 , s2 ;

        for( int i = 0 ; i < N ; i++ )
        {
            s2.insert(0);
        }
        vector<ar>A(N,{0,0});
        int e = 0 ;

        for( int i = 0 ; i < Q ; i++ )
        {
            int k ;
            cin>>k ;
            k--;

            int first = A[k][1];
            int val = A[k][0] ;

            if( first )
            {
                A[k][0] = val + e ;
                s1.erase( s1.find(val) );
                s2.insert( A[k][0] );
                e += s1.size();
            }
            else{

                int put = val - e ;

                s2.erase( s2.find(val) );
                s1.insert(put);
                A[k][0] = put ;
                e += s1.size();
            }
            A[k][1] ^= 1 ;
        }

        for( int i = 0 ; i < N ; i++ )
        {
            int v = A[i][0] ;
            int f = A[i][1] ;

            if(f){
                cout<<(v+e)<<" ";
            }
            else{
                cout<<v<<" ";
            }
        }

        cout<<endl;

    };

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

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).

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

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

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

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';

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

Problem c was very annoying

#include<cstdio>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<cstring>
#include<cmath>
#include<climits>
#include<iostream>
#include<cstdlib>
#include<queue>
using namespace std;

#define int long long
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define print(a) for (auto &x : a)cout<<x<<" ";

void solve() {
    int n,a,b; cin>>n>>a>>b;

    vector<int>A(n);

    for(int i = 0;i<n;i++) {
        int x; cin>>x;
        A[i] = x%(a+b);
    }

    int prel = a+b-A[0]+1, prer = a+b-A[0]+a;

    for(int i = 1;i<n;i++) {
        int curl = a+b-A[i]+1, curr = a+b-A[i]+a;
        if(prel>curr or prer<curl) {
            cout<<"No"<<endl;
            return;
        } 

        prel = max(prel,curl);
        prer = min(prer,curr);
    }
    cout<<"Yes"<<endl;
}

signed main() {
    ios_base::sync_with_stdio(false);cin.tie(0);
    int t = 1; 
    // cin >> t;
    for(int i = 1;i<=t;i++) {
        solve();
    }
}

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.

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).

ll n, a, b;
cin >> n >> a >> b;
vector<ll> plan(n);
ll total = a + b;
for (auto &x: plan)
{
    cin >> x;
    x = x % total;
}

ll mine = *min_element(plan.begin(), plan.end());
ll maxe = *max_element(plan.begin(), plan.end());

cout << (maxe - mine < a ? "Yes" : "No");
return 0;

ll n, a, b;
cin >> n >> a >> b;
vector<ll> plan(n);
ll total = a + b;
for (auto &x: plan)
{
    cin >> x;
    x = x % total;
}

sort(plan.begin(), plan.end());
ll res = INT_MAX;
res = min(res, plan[n - 1] - plan[0]);

for (ll i = 1; i < n; i++)
{
    res = min(res, total - plan[i] + plan[i - 1]);
}

cout << (res < a ? "Yes" : "No");
return 0;

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

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.

.

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