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Educational Codeforces Round 82 Editorial
By awoo, history, 5 years ago, translation, In English

1303A - Erasing Zeroes

Idea: Roms

Tutorial
Solution (Roms)

1303B - National Project

Idea: adedalic

Tutorial
Solution (adedalic)

1303C - Perfect Keyboard

Idea: Roms

Tutorial
Solution (Ne0n25)

1303D - Fill The Bag

Idea: Roms

Tutorial
Solution (Roms)

1303E - Erase Subsequences

Idea: adedalic

Tutorial
Solution (adedalic)

1303F - Number of Components

Idea: Neon

Tutorial
Solution (Ne0n25)

1303G - Sum of Prefix Sums

Idea: Neon

Tutorial
Solution (BledDest)
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5 years ago, # |
  Vote: I like it -47 Vote: I do not like it

Why this post doesn't have any comments?

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5 years ago, # |
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can someone please explain a little bit about the proof of algorithm of problem F? Why considering all the delete queries in the last doesn't affect the final results?

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    5 years ago, # ^ |
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    because the delete queries are after the add queries.

    when considering color $$$t$$$,all the add queries is the queries with $$$c_i=t$$$ ,then for sure,they are consquent,and so for sure all the delete queries are after the add queries

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      5 years ago, # ^ |
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      But this is true for a particular color right? Like for a color c what ideally should have been done is to perform add query and then immediately after that perform it's delete queries but in code above they are doing all add first and then all delete query.

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        5 years ago, # ^ |
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        No...for a certain color c and all the queries with ci=c,the queries are consquent,so they changed some of the cells into color c.There isn't any delete query between them.After that,some queries may change the color of some cells which were color c,these are the delete queries.

        Like this example:

        2 2 4
1 1 1
1 2 1
2 1 2
1 1 2

        For color 1,the add queries are the 1st and the 2nd while the delete query is the 4th.

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

          Sorry but I think I didn't convey my doubt properly. I am talking about this part of code ..

          forn(i, clrs) recalc(add[i], +1);
forn(i, clrs) recalc(del[i], -1);

          Here we are processing all the delete queries at the end (even after the add queries of same cell i.e. even after add query of (x, y) from c1 to c2).

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5 years ago, # |
  Vote: I like it +11 Vote: I do not like it

In problem D

If in binary representation of n the i-th bit is equal to 1 and we have at most one box of size 2^i,

I think this should be at least?

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5 years ago, # |
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can someone please explain me about problem B why TotalG = ceil(needG / g) * (g + b) , i don't understand that clearly =)) ????

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    5 years ago, # ^ |
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    every (g + b) days have exactly (g) days good, we call it a block. So if we need (needG) days good, we must have ceil(needG / g) blocks, thus TotalG = ceil(needG / g) * (g + b).

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5 years ago, # |
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What is the time complexity for G? nlgnlgn?

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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +5 Vote: I do not like it

    Yes. You process each node O(lg n) times using centroid decomposition and each time you do it, you add 2 lines and you make 2 queries for the best path in O(lg n). Total is O(n lg^2 n).

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5 years ago, # |
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Has anyone used adjacency list to solve C ? If so could you share your idea/implementation thanks

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    5 years ago, # ^ |
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    Here is my submission — 70873445
    Just store the letter as an undirected graph and make sure that every vertex has degree 1 or 2.
    Then loop through all the vertex and find the vertex which has degree 1 because that would be the start/end of the graph.

    Now go for dfs, if you find a cycle in the graph, that means it's not possible, otherwise dfs will give you the starting letters of the keyboard.

    Then just add the remaining alphabets back in the string.

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      5 years ago, # ^ |
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      your code is very messy didnt understand a thing ... or may be I m big big noob

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

The problem statement of F says $$$1 \le c_i \le max(1000,\lceil \frac{2⋅10^6}{nm} \rceil)$$$.

Should not that be $$$min$$$ instead of $$$max$$$?

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5 years ago, # |
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Can someone help me understand why my code for G is TLE on test 33? I've read the editorial and I think I am doing the same thing.

https://codeforces.net/contest/1303/submission/71252477

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

    I had also struggled from test 33, and finally made it accepted.

    I don't know what exactly is test 33, but this test generator code would be helpful. It takes about 20 sec on N=150000 on your code.

    #include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <random>
#include <ctime>
#include <queue>
#include <cmath>

using namespace std;

inline int randint(int a, int b) {
	return rand()%(b-a+1) + a;
}

int main() {
	int n;
	//cin >> n;
	n = 150000;
	
	cout << n << endl;
	
	srand(time(0));

	vector<int> perm(n+1, 0);
	for (int i=1; i<=n; i++) perm[i] = i;
	for (int it=n*10; it--; ) {
		int a = randint(1, n);
		int b = randint(1, n);
		swap(perm[a], perm[b]);
	}

	// sqrt tree, binary tree
	int sq = sqrt(n);
	// int sq = 2; // binary
	queue<int> qu;
	qu.push(1);
	int p = 2;
	while (p <= n) {
		int a = qu.front(); qu.pop();
		for (int i=0; i<sq; i++) {
			cout << perm[a] << ' ' << perm[p++] << endl;
			if (p == n+1) break;
			qu.push(p-1);
		}
	}
	/* */
	
	for (int i=1; i<=n; i++) {
		cout << randint(1, 1000000) << ' ';
	}
	cout << endl;
}

    Use the output of the generator code as the input.

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      5 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      For me, it was a problem from cache. I had used so many random-accessing on arrays, and it became x8 faster when I fixed into a cache-friendly code.

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        5 years ago, # ^ |
        ← Rev. 2   Vote: I like it +8 Vote: I do not like it

        Thanks I may have a chance to fix it now. All my own generated samples ran in <2 seconds

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5 years ago, # |
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can someone explain the solution for problem b

i think it gives right answer for 5 1 5 but wrong for 10 1 10

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5 years ago, # |
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Typo in D, case 2, "most" should be "least".

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5 years ago, # |
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How to proof that D solution is optimal(finds min number of divides)?

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    22 months ago, # ^ |
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    we should consider the lowerest bit first , if we don't have the bit , than look for the nearest high bit and decompose it.

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5 years ago, # |
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Hi awoo! Why the scheduled codeforces educational round 83 was deleted?

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5 years ago, # |
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For problem G, how can we get all the "first parts" and "second parts" efficiently?

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5 years ago, # |
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I think there is a mistake in the tests in the B problem. MY solution as well as the tutorial both fail at the same place in the second test. Here is my code: (https://codeforces.net/problemset/submission/1303/79669445)(https://codeforces.net/problemset/submission/1303/79668860). Could someone tell me if there is a mistake

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    5 years ago, # ^ |
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    When you calculate minN = round(n / 2), n is truncated before round is called because it's an integer. So n always rounds down instead of rounding up. You can fix this by doing minN = (n+1) / 2 instead.

    Also, it's a little arrogant to assume that the tests are wrong if your code doesn't get AC. Especially if nearly 6,000 others have solved without any issues. So test your code a little on some cases first, before assuming the writers messed up.

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5 years ago, # |
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80808954 Which case am i missing? (Problem B)

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    4 years ago, # ^ |
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    you have to take the case when req is divisible by g than you dont need to add the last b days because you have already finished your work and dont and instead of subtracting g from req and than doing other operations better try to do it like this

    ans -> the number of days that we need to calculate

    int n; // length of road
int nn = (n+1)/2;
long long req= nn/g;
if(nn%g == 0)
{
   ans = req* g + (req- 1)*b;
}
else
{
   ans = req* g + req* b + nn % g;
}
ans = max(ans,n);
cout << ans << "\n";
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      4 years ago, # ^ |
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      You mean a case like this:

      1
12 3 8

      It outputs : 

      14
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        4 years ago, # ^ |
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        No just take the example of 16 3 8 as input Your code will give 16 as output But the correct answer will be 24

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4 years ago, # |
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In problem F, can someone please explain the intuition behind deleting cells. How is it similar to adding in reverse order? awoo

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4 years ago, # |
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in D how do we go about the optimality?

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4 years ago, # |
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Alternate dp for E. Let $$$dp[len_{s}][len_{a}]$$$ be the max $$$len_{b}$$$ such that the prefix of $$$s$$$ of length $$$len_{s}$$$ contains non-intersecting subsequences of $$$a$$$ and $$$b$$$ of $$$len_{a}$$$ and $$$len_{b}$$$. Set it to -1 if the prefix does not contain a subsequence of length $$$len_{a}$$$. With this approach we don't have to calculate the array $$$nxt$$$ or make the observation in the editorial. Code 97250299

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4 years ago, # |
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How is solution for D optimal?

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3 years ago, # |
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Erasing Zeroes //my solution

include<stdio.h>

include<string.h>

main(){ int t; scanf("%d",&t); char str[101]; while(t--){ scanf("%s",&str); int poslast1,posfirst1,c=0; int l=strlen(str); for(int i=0;i<l;i++){ if(str[i]=='1') posfirst1=i; break; } for(int j=l-1;j>=0;j--){ if(str[j]=='1') poslast1=j; break; } for(int k=posfirst1;k<=poslast1;k++){ if(str[k]=='0') c++; } printf("%d\n",c); } return 0; }

//what's problem in this

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18 months ago, # |
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In problem G, the value of a path between any two nodes can also be queried using binary lifting. (When you have to compute for upward path, just do it like normal binary lifting. When you have to compute for downward path, "invert" its coinciding upward path using trivial relation between prefix sum and suffix sum)

Implementation

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