Good editorial!

Code for D is messed up a bit.

Mike Mirzayanov thanks for awesome editorial

Question D is a great problem,but why is the code for D so messy？

Hey, can anyone tell me why did I get a MLE on D with submission 166572453? I precomputed every segment that can be painted in one step, however there can at most be 100*10 such segments for every test case. If every segment is 3 integers (12 bytes) then they use at most 12000 bytes of memory. Lets say that I copy all segments once into final, then I use at most 2*12000=24000 bytes per test case. This is way below the 256 MB memory limit, so why did my solution get MLE? Thanks.

How to solve question D with DP? Can you tell me about it and share your code? If you can help me, I will thank you very much.

Indented solution of problem D from editorial:

#include<bits/stdc++.h>
#define len(s) (int)s.size()
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
int ans = 0;
bool ok = true;

void Find(int a, int b, string &t, vector<string>&str, vector<pair<int, int>>&match){
    int Max = 0, id = -1, pos = -1;
    for(int i = a; i <= b; i++){
        for(int j = 0; j < len(str); j++){
            string s = str[j];
            if(i + len(s) > len(t) || i + len(s) <= b)continue;
            if(t.substr(i, len(s)) == s) {
                if(i + len(s) > Max){
                    Max = i + len(s);
                    id = j;
                    pos = i;
                }
            }
        }
    }
    if(id == -1) {
        ok = false; return;
    }
    else {
        match.emplace_back(id, pos);
        ans++;
        if(Max == len(t)) return;
        else Find(max(pos + 1, b +1), Max, t, str, match);
    }
}

void solve(){
    ans = 0;
    ok = true;
    string t;
    cin >> t;
    int n;
    cin >> n;

    vector<string>s(n);
    vector<pair<int, int>>match;

    forn(i, n) {
        cin >> s[i];
    }

    Find(0, 0, t, s, match);
    if(!ok) cout << "-1\n";
    else{
        cout << ans << endl;
        for(auto &p : match) cout << p.first + 1 << ' ' << p.second + 1 << endl;
    }
}

int main(){
    int q;
    cin >> q;
    while(q--){
        solve();
    }
    return 0;
}

"When you color a letter in red again, it stays red."

Missed this line during contest does this make the question similar to Jump game? ->where we can jump from index to any index in range (i,i+a[i])

Can someone help me out with my wa for E? https://codeforces.com/problemset/submission/1714/166799175

Approach:

• I move the highest number to a remainder of 8
• Everything else can either reach the number or not, and I set conditions for it.

For i in (0, n - 1)
Let k = a[i] / 10, l = a[max] / 10
    if remainder = 1, 2, 4, 8 then (l - k) % 2 = 0
    if remainder = 3, 6, 7, 9 then (l - k) % 2 = 1 
    if remainder = 5, a[i] + 5 == a[max]
    if remainder = 0, a[i] == a[max]
    else return NO

The code for F: Build a Tree and That is it does not really clarify things mentioned in tutorial rather it has checks and implementations that makes it seem like the code to be working on a different idea. Can anyone either explain the tutorial or the code ? Thanks.

My Editorial for the problems I solved

It was a very good contest D was interesting but i think you should reorder the problem like this B C A E D G F

I read the solution "Thus, we will apply the operation to each element of the array until its remainder modulo 10 becomes, for example, 2, and then check that the array does not contain both remainders 2 and 12 modulo 20." I don't know why do we need to check that "the aray does not contain both remainders 2 and 12 modulo 20". Why do we have to do that. Can you explain for me. Thanks

How to solve C using dp ??

Problem D and E were very nice But I think problem E is much easier than D.

Code for $$$Problem D$$$ should be posted like this :

#include<bits/stdc++.h>
#define len(s) (int)s.size()
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std; 
int ans = 0; 
bool ok = true;

void Find(int a, int b, string &t, vector&str, vector<pair<int, int>>&match)
{ 
    int Max = 0, id = -1, pos = -1; 
    for(int i = a; i <= b; i++)
    { 
        for(int j = 0; j < len(str); j++)
        { 
            string s = str[j]; 
            if (i + len(s) > len(t) || i + len(s) <= b) continue; 
            if (t.substr(i, len(s)) == s) 
            { 
                if (i + len(s) > Max){ 
                    Max = i + len(s); id = j; pos = i; 
                } 
            } 
        } 
        if(id == -1) { ok = false; return; } 
        else
        { 
            match.emplace_back(id, pos); 
            ans++; 
            if (Max == len(t)) return; 
            else Find(max(pos + 1, b +1), Max, t, str, match); 
        } 
    }

void solve()
{ 
    ans = 0; ok = true;

    string t;
    cin >> t;
    int n;
    cin >> n;

    vector<string>s(n);
    vector<pair<int, int>>match;

    forn(i, n) {
        cin >> s[i];
    }

    Find(0, 0, t, s, match);
    if(!ok) cout << "-1\n";
    else{
        cout << ans << endl;
        for(auto &p : match) cout << p.first + 1 << ' ' << p.second + 1 << endl;
    }
}

int main(){ 
    int q; 
    cin >> q; 
    while(q--){ 
        solve(); 
    } 
    return 0; 
}

I think Vladosiya has forgotten to write </spoiler> at the end.

Does anybody have E wa3 ,only one line of code was different between the two submissions Submissions id(wa3) : 166896597 Submissions id(ac) : 166900206 I'd like to know what the problem is. Could you please take a look at it for me，thanks！！！！！

G is not that hard compared to F, D

agree?

Problem D can be also done with DP approach: 166555787

It took some time for me but I solved it finally :)

I started by stating out the different series that are being made and what commonalities they have. I deduced that there are two disjoint sets (if you exclude the multiples of 5 of which there are several disjoint sets.) that you get.

I called the set {1,2,4,8,16,22,24,...,96} for evenSet. And the set {3, 6, 12, 14, 18, 26, ..., 98} for oddSet.

The trick is to know that it's only the last two digits in every number which denotes which set it belongs to. For example 111 is equivalent to 11 in the series since they can converge.

What about prime numbers like 17? For them, we can see that 17 + 17 % 10 = 24 is a member of the evenSet. This is true for all such numbers, we can just check if either the number or the next number of the series of the current number is in the set we're interested to check.

What about multiples of 5? For them, we know that {5, 10} is one set, {15,20} another. The set cannot increase and thus we only need to worry about checking if all numbers are equal to the latter in the set. i.e. if a % 10 == 5, then we should look for a + 5. But note that we will check for all other numbers b, if b == a or b+5==a.

Once we have this, the solution should be quite straightforward:

1. Generate the evenSet and the oddSet.

2. Check if there are any multiples of 5, if so then we know that all of them should be in the same multiples set.

3. If there were not any multiples of 5, then check if a or a + a%10 is in the evenSet or the oddSet. (Only take the last two digits of the number!). If there are already any in the oddSet, then since the sets are disjoint, then we know there are no way to make them equal. Similarly, we check in the oddSet and if there are any in the evenSet.

Took a bit of time but well worth it, learned a lot. Can probably be cleaned up a lot:

import java.util.HashSet;
import java.util.Scanner;

public class E {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        int t = input.nextInt();

        while (t > 0) {
            solution(input);
            t--;
        }
    }
    public static void solution(Scanner input) {
        int n = input.nextInt();
        int[] a = new int[n];
        String result = "Yes";
        boolean odds = false;
        boolean evens = false;
        boolean fives = false;

        HashSet<Integer> evenSet = new HashSet<>();
        HashSet<Integer> oddSet = new HashSet<>();

        for (int i = 0; i < n; i++) {
            a[i] = input.nextInt();
        }

        // 1. Generate even sets: {1,2,4,8} etc. A member or the member + member % 10 counts.
        int firstEven = 1;
        while (firstEven < 100) {
            evenSet.add(firstEven);
            firstEven += firstEven % 10;
        }

        int firstOdd = 3;
        while (firstOdd < 100) {
            oddSet.add(firstOdd);
            firstOdd += firstOdd % 10;
        }

        // 2. Check if multiple of 5.
        int target = 0;
        for (int e : a) {
            if (e % 10 == 5) {
                if (fives && (e + 5) != target) {
                    result = "No";
                    break;
                }

                if (target == 0) {
                    target = e + 5;
                }
                fives = true;
            }

            if (e % 10 == 0) {
                if (fives && e != target) {
                    result = "No";
                    break;
                }

                if (target == 0) {
                    target = e;
                }
                fives = true;
            }
        }

        // 3. Check in odd and even sets if there are no fives.
        if (!fives) {
            for (int e : a) {
                String numStr = String.valueOf(e);
                int num;

                if (numStr.length() > 2) {
                    num = Integer.parseInt(numStr.substring(numStr.length() - 2));
                } else {
                    num = Integer.parseInt(numStr);
                }

                String nextNumStr = String.valueOf(num + num % 10);
                int nextNum;

                if (nextNumStr.length() > 2) {
                    nextNum = Integer.parseInt(nextNumStr.substring(nextNumStr.length() - 2));
                } else {
                    nextNum = Integer.parseInt(nextNumStr);
                }

                if (evenSet.contains(num) || evenSet.contains(nextNum)) {
                    if (odds) {
                        result = "No";
                        break;
                    }
                    evens = true;
                }

                if (oddSet.contains(num) || oddSet.contains(nextNum)) {
                    if (evens) {
                        result = "No";
                        break;
                    }
                    odds = true;
                }

            }
        }

        System.out.println(result);

    }
}

In problem E, why flag12 and flag2 should not be true for the answer to be YES?

i coded D iteratively, but its giving me MLE, can someone please tell me why ?

iterative D soln

Why I got TLE ? What's wrong with my code[submission:166546367]

Is there a way to solve G using binary lifting