awoo's blog

Educational Codeforces Round 148 Editorial
By awoo, history, 20 months ago, translation, In English

1832A - New Palindrome

Idea: BledDest

Tutorial
Solution (Neon)

1832B - Maximum Sum

Idea: BledDest

Tutorial
Solution (awoo)

1832C - Contrast Value

Idea: BledDest

Tutorial
Solution (Neon)

1832D1 - Red-Blue Operations (Easy Version)

Idea: BledDest

Tutorial

1832D2 - Red-Blue Operations (Hard Version)

Idea: BledDest

Tutorial
Solution (awoo)

1832E - Combinatorics Problem

Idea: BledDest

Tutorial
Solution (BledDest)

1832F - Zombies

Idea: BledDest

Tutorial
Solution (awoo)
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Comments Comments (15)
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20 months ago, # |
  Vote: I like it +40 Vote: I do not like it

imbalanceForces

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20 months ago, # |
← Rev. 2   Vote: I like it +11 Vote: I do not like it

In the editorial of $$$E$$$, it is mentioned that $$$c_{i,0}=\sum_{j=1}^{i}{a_j}$$$. Actually, it should be $$$c_{i,0}=\sum_{j=1}^{i+1}{a_j}$$$ (with $$$c_{0,0}=a_1$$$ and $$$c_{n,0}$$$ is not needed).

Reason:

When we say $$$b_i$$$ (at $$$k$$$) $$$=$$$ $$$b_{i-1}$$$ (at $$$k$$$) $$$+$$$ $$$b_{i-1}$$$ (at $$$k-1$$$), this is true only when $$$k>1$$$, because this will cause the last term in both $$$\sum_{j=1}^{j=i}{{i-j \choose k} \cdot a_j}$$$ and $$$\sum_{j=1}^{j=i}{{i-j \choose k-1} \cdot a_j}$$$ to be $$$0$$$ (as $$$0 \choose x$$$ is $$$0$$$ for positive $$$x$$$). However, this is not the case for the $$$2^{nd}$$$ summation when $$$k=1$$$. The last term will not be eliminated as $$${0 \choose 0}=1$$$.

Submission.

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    20 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Fixed this issue, thank you!

    The editorial might take a while to update, but hopefully it will show the new version soon.

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20 months ago, # |
  Vote: I like it 0 Vote: I do not like it

SorrowForces

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20 months ago, # |
← Rev. 5   Vote: I like it 0 Vote: I do not like it

D1

I see applying Operation 2 $$$n$$$ or $$$n-1$$$ times from begin.

then, $$$k-(n-(n+k) ~\text{mod}~ 2)-1$$$ should be $$$k-(n-(n+k) ~\text{mod}~ 2)+1$$$ ?

Or, $$$k-n+1 + (n+k) \text{mod}~2$$$

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20 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it 

Problem B:

I don't understand what is 'k — m' maximums, and I don't know what is k when m is the number of operations. Can anyone explain from me pls ??

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    20 months ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    $$$k$$$ is given in the statement. upd: $$$k$$$ is the total number of operations, $$$m$$$ is the number of operations of the first type (when we delete two minimum elements)

    $$$(k-m)$$$ maximums is $$$(k-m)$$$ greatest elements of the array, i. e. $$$(k-m)$$$ last elements in sorted order.

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20 months ago, # |
  Vote: I like it 0 Vote: I do not like it

For problem F, quadrangle inequality properties also apply to array partition DP. So we can use Knuth's optimization for a second time on $$$dp$$$, which leads an $$$O(n^2)$$$ algorithm.

This is my submission: https://codeforces.net/contest/1832/submission/206185565.

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20 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Hi. can someone please tell me the error in this code which I used without prefix array

#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;


int maxRemoval(vector<int> &v){
    int n = v.size(), sum = 0;
    for(int i=0;i<n-1;++i){
        sum += v[i];
    }
    return sum;
}

int minRemoval(vector<int> &v){
    int n = v.size(), sum = 0;
    for(int i=2;i<n;++i){
        sum += v[i];
    }
    return sum;
}
signed main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t;
    cin >> t;
    while(t--){
    
        int n, k;
        cin >> n >> k;
        vector<int> v(n);
        for(int i=0;i<n;++i){
            cin >> v[i];
        }
        sort(begin(v), end(v));
       
        int ans = 0;
        for(int i=0;i<k;++i){
            int s1 = maxRemoval(v);
            int s2 = minRemoval(v);
        
            ans = max(s1, s2);
            if(s1 < s2){
                v.erase(v.end()-1);
            }
            else{
                v.erase(v.begin());
                v.erase(v.begin());
            }

        }
  
        cout << ans << endl;
  
    }
}

It runs correctly for 1st test case but tells that there is wrong answer on test 2 wrong answer 2459th numbers differ - expected: '8', found: '7' I mean does it really check 2459th number. did the testers even count upto that thing. I doubt it. Link to My Submission

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    20 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I dont think that greede solution is correct. But it also O(n*k) and even it be correct, it gets TLE.

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19 months ago, # |
  Vote: I like it +3 Vote: I do not like it

Hello guys, I am getting TLE in 1832E - Combinatorics Problem. I don't think I have made any logical errors, but I can't seem to understand why it's giving TLE. Please help :( -> 209759092

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    19 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    hey, you are using too many modulo operators in your add and mul. In spite of you having the right complexity, your constant factor is bad. Here is a fixed version of your code 209761902

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      19 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thank you so much man, I didn't know that modulo operation also affected complexity. Got to learn something new. Tysm :)

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11 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

I like the awoo solutions.. He make the solutions like my friend explaining to me.. Thx man++.

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10 months ago, # |
  Vote: I like it 0 Vote: I do not like it

can someone please explain how this test case in problem B maximum sum is giving o/p 46

6 2

15 22 12 10 13 11

please help my code

include <bits/stdc++.h>

using namespace std;

define ll long long

define all(x) x.begin(), x.end()

define pb push_back

define F first

define S second

const int M = 1e9 + 7;

define ll long long

define yes cout << "YES" << endl

define no cout << "NO" << endl

define n1 cout << -1 << endl

int main() { int t; cin >> t; while (t--) { ll int n; cin >> n; int k; cin >> k; vector< ll int> v; for (ll int i = 0; i < n; i++) { ll int x; cin >> x; v.push_back(x); }

    sort(v.begin(), v.end());
    while (k--)
    {

       if (v[0] + v[1] <= v[v.size() - 1])
       {
         //cout << "hi" << endl;

         v.erase(v.begin(), v.begin() + 2);
       }
       else
       {
         //cout << "hello" << endl;

         v.pop_back();
       }
       if(v.size()<=2)
       {
         break;
       }
    }

    ll int ans = 0;
    for (ll int i = 0; i < v.size(); i++)
    {
       ans = ans + v[i];
    }
    cout << ans << endl;
}

}

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