**Problem A. Chat room**

Solution is greedy algorithm. The first thing we do is find in our string the first letter 'h'. Then we find letter 'e' which is righter that found 'h'. If we find the whole word 'hello' in such way, obliviously, answer is YES.

Now let's prove that if answer exists, we find it. Let see on position of the 'h' in right answer. If we move it to the first 'h' in our string, nothing changes. But now we can say that our greedy algorithm correctly Now let's do such with the second letter, and so on.

We have greedy algorithm with work time O(n), where n - length of the input.

Now let's prove that if answer exists, we find it. Let see on position of the 'h' in right answer. If we move it to the first 'h' in our string, nothing changes. But now we can say that our greedy algorithm correctly Now let's do such with the second letter, and so on.

We have greedy algorithm with work time O(n), where n - length of the input.

**Problem B. Coins**

This problem also have greedy solution: we run down all number from 2 and greater and while denomination of the last added to answer coin is divisible by current number, we divide and increase answer.

You can prove correctness, if you take a look at prime factorizations of coins' denominations. In each next denomination each prime have less or equal power than the current one (it's equivalent to '

You can prove correctness, if you take a look at prime factorizations of coins' denominations. In each next denomination each prime have less or equal power than the current one (it's equivalent to '

*a*divide*b*'). Obliviously, if summary degree of primes decreases at least two (for example, we had number*a*=*x*·*y*·*z*(where*y*,*z*> 1, and the current number is*b*=*x*), then we can add one more coin with demonitaion*a*>*c*=*x*·*y*>*b*. So, in the optimal answer sum of degrees decreasing at exactly one. Our greedy algorithm do exactly what it need.**Problem C. Trees.**

The first thing we notice - beautiful sequence is can be determined with any member. The next thing - at least one tree will remain the same height. Prove: let's fix height of the first tree, and correct heights of all other ones. Obliviously, they all remain positive.

Solution with work time

This solution can be optimized to linear solution: if we don't touch some tree, we know the first element of sequence. Let's count for each possible element amount of trees, which have required height. It can be done with linear loop and the 'increment' operation on array. After that we just find value of the first element, for which amount of 'good' trees is maximal and output

Solution with work time

*O*(*n*^{2}): we run down which tree we will fix, determine required height of the first tree and then relax answer.This solution can be optimized to linear solution: if we don't touch some tree, we know the first element of sequence. Let's count for each possible element amount of trees, which have required height. It can be done with linear loop and the 'increment' operation on array. After that we just find value of the first element, for which amount of 'good' trees is maximal and output

*n*-*x*, where*x*is this amount.

Problem D. Calendar

Problem D. Calendar

We know, that all lines of calendar should have equal length, so we can find this length. It's just , where

We know one line, let's do similary with the second line and so on.

*suml*- summary length of all strings. Now let take a look at string, which will be the first one in our calendar. Obliviously, it's profitably to take string with maximal*s*+*d*(where*s*is our string and*d*- is character from input). Such string is unique otherwise we have two equal strings in input (as*d*is not contained in any string); Of cause, you should remember that if you fix the first string in a line, you fix length of the second one - it's required to have at least one such. Great, now we know the first string in our calendar. Now let's determine the second one. We know it's length, so we need just to take minimal string with such length.We know one line, let's do similary with the second line and so on.

**Problem E. Expression**

Under construction. The main idea is 3D dynamic with O(10

^{2}) transition.
I hope you complete this with write part E , good luck man :D

Good Job Man

And necroposting is not a good job

When can you give us the solution for problem E ?

Can anyone give ideas about the DP solution for E.

Can you say what is test case #6 of problem A

tymbzjyqhymedasloqbq

how you get this test case?

by submitting wrong solution

I'm 8 years late, but 58C - Trees has weak testcases. some solutions pass although they allow some tree heights to become non positive.

Case:

`10`

`1 1 1 2 3 3 2 1 1 1`

Answer should be 8, (correct sequence:

`1 2 3 4 5 5 4 3 2 1`

)Wrong answer is 4, (Assumes the correct sequence is:

`-1 0 1 2 3 3 2 1 0 -1`

)Wrong submission: 51244937

Correct submission: 51245002

Div2c used very odd concept. Can anyone suggest me some more questions using similar concepts ??

So when we can get the solution of E?

88215616 i don't know why my code is giving wrong answer on test case#8 plz somebody help me out.

binod

Please don't post irrelevant comments here. This is an actual community trying to help each other

OK.

can somebody help me with the Solution of A .. I didnt get it!! C++ code will also help!

Thanks!

You are welcome.

Could not understand the solution can you help?

thanks

Problem Cneeds stronger test cases, my submission 100033643(pardon the spaghetti code)got accepted although it's based on faulty logic exposed by this test case for example:The correct output is

`5`

and not`7`

as my wrong solution.****I don't know what's wrong with this code?****

## include <bits/stdc++.h>

using namespace std;

int main() { string s; cin>>s;

}

Could anyone please explain the DP for E? The labels in the solutions on here arent very clear to me