HackerRank::Algorithms::Warmup::Angry Professor

  We just need to input N number and find the number of numbers which are less or equal than 0. And then compare the result to the given integer K.

HackerRank::Algorithms::Warmup::Lonely Integer

  To keep it short, if we XOR( exclusive OR) all the numbers together, the result will be our answer.
 For not getting too deep into the math proofs, we need to prove 4 properties
1. a^a=0 ( any number XORed with itself is 0)
2. a^0=a (any number XORed with 0 is equal to himself)
3. (a^b)^c=a^(b^c) (the associative rule)
4. a^b=b^a (the commutative rule)
 All those rules are really very obvious to prove.
So, the only thing we need to do is to input all the numbers, XOR them and output the result.

HackerRank::Algorithms::Warmup::The Love-Letter Mystery

  No matter what, we need to change every "pair" to be equal ( by saying pair we understand the letters which need to be equal to one another to form a palindrome like the first and the last, second and the pre-last). Which means if we have a pair of letters A and B and we need to change them in order for them to be equal, we need to do at least absolute (B-A) moves.

HackerRank::Algorithms::Warmup::Maximizing XOR

   The constraints: the interval is at most 10^3 which means that there are at most 10^6 pairs of numbers, so manually checking them and choosing the maximum one is sure going to be accepted.

HackerRank::Algorithms::Warmup::Utopian Tree

   The constraints are very low, so just implementing the cycle of the growth is enough to get accepted.

HackerRank::Algorithms::Warmup::Solve me Second

   The continuation of the A+B task, except this time you need to know loops as well.

HackerRank::Algorithms::Warmup::Solve me First

   The first task of almost any online judge is the A+B task. Not much to tell here.

       

ACM 1106. Two Teams / Две команды

Here is a little suggestion. Let us start a wave from any node. That chosen node we will send to the first team, then all of his unmarked neighbors will be sent to the second team, and then the neighbors of all the nodes form the previous to the first team and so on. If we have a forest ( where there are sub graphs not connected to one another) then we should do this for all the sub graphs. Then we need to check whether our result is a valid solution or not. If it is not, then there are no other ways of making the teams. 

ACM 1139. City Blocks / Городские кварталы

 First let us answer the following question : how many coordinates will the helicopter pass which have a natural coordinates. We need to look at the geometrical illustration.

Have a look at the picture below.

So, this is our coordinate system and the points (0,0) and (9,6) are connected, what are all the coordinates which have natural coordinates and the red line passes over them those are (0,0) (3,2) (6,4) (9, 6) . I think you already see the pattern here but let's try proving it. So we can build a right angle triangle with the points (0,0) (9,0) and (9,6) , we will call that triangle 1. And let's build another one with the points (0,0) (0,6) (6,4) we will call that triangle 2. So, if we compare the triangles 1 and 2 we can say that they are similar (due to equal angles) which means that their sides are   multiples. A Conclusion  all the multiples of the pair (9,6) (to the one which we connected the initial (0,0) point) are vertexes of such triangles, => are the points which have natural coordinates which our line crosses. Let's have a little case which is (0,0) to (2,3). (2,3) do not have any multiples so there is no point with natural coordinates on the way. which means that the line crosses all the available vertical lines and all the available horizontal lines separately. The total number of those is 2+3-1=4 which gives us the first idea of the answer which is N+M-1. If there are natural coordinated points then our task is divided into sub tasks, in out case we have 3 different sub-tasks each has the same answer. For finding the number of natural coordinates on the way we need to find the GCD (greatest common divisor). In this case it would be (M/GCD+N/GCD-1) * GCD which gives the answer of M+N-GCD.

ACM 1638. Bookworm/ Книжный червь

 This task has a difficulty of only 84  and yet almost 2/3 of the total submissions are not accepted. I liked this one really. It is a very tricky task. Let's start with examining the input. Some people say that it is wrong, it looks wrong at first but eventually you will understand. So here is the input.

10 1 1 2

We Expect the answer to be 22, because we expect that the from the first page of book 1 to the last page of book 2 is a total distance of 2 covers and 2 volumes. But it's not correct because we need to imagine the book shelf in our minds and see for ourselves. I made a little picture to illustrate my point below. Have a quick look.

I think now everything is settled. As you can see it takes only 2 covers to travel from the beginning of the first book to the end of the second book which buildup our generaly formula  (end-start)*2*cover + (end-start-1)*volume. 

But there are 2 more tricky cases. It is not guaranteed that the starting number is always less then the ending number. If they are equal then the worm passes through the whole volume and if the second one is smaller than the first one then the worm passes though 2 more additional volumes.