Second place: /dev/null 4 problems solved

Third place: NibbleSoft 2 problems solved

Every Thursday (Jueves) from 3 pm to 6 pm. In the CISCO Lab in the CSI building (1 floor).

This week Maximum Sum.

http://uva.onlinejudge.org/index.php?option=onlinejudge&Itemid=99999999&page=show_problem&category=&problem=668

In this problem we need to change an infix expression to a posfix expresion, the key words are IN and POS.

INfix means that the operator is between the two “numbers”

POSfix means that the operator is after the two “numbers”

Theres plenty of ways to do this, most of them use stacks but there are other ways using arrays to store the things inside the parenthesis either way you do it there just like three diferent cases:

1- When you receive a number (0-9) you printed

2- if you receive a left parenthesis ‘(‘ you start storing inside the stack

3- if you receive the right parenthesis ‘)’ you start ‘poping’ everything form the stack

4- when you receive and operand you stored to  print it after the two “numbers”

5- last but not least with “numbers” i mean a single number or a bunch of them including the operands

Source Code - part2

Source Code - part1

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=2380

In this problem there is some things we can assume with the problem description:

1. The maximun Fibonacci number we need to calculate is the 45th number.

2. The maximun length of the cypher text is 100 characters(not all of them uppercase letters)

3. All of the input numbers are fibonacci (no need to validate them).

4. The maximun output string would be 45 cause there isn’t trailing spaces.

5. You need to check only X amount of numbers where X is the amount of uppercase letters in the cypher text.

6. The output string must be filled with spaces before we add the letters to it.

With all of this in mind the problem is almost finish, now we need to follow just a few more steps.

first we pre-calculate the first 45 fibonacci elements then we store the numbers and ONLY the uppercase letters (you can store the whole line and then store the uppercase letters in another array) with that we need two more things, one, we need to store the amount of upper case letters (a.k.a. the length of the array), second, we need the biggest fibonacci number inside the number we are going to use to put a ” at the end to stop the string (or we can start at the 45th position in the array and going backwards until we find something diferent to a ‘ ‘ (space) and we put a ” in the position next to it.

Source Code

Source Code

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=9&page=show_problem&problem=722

Ok, this is one of the problems that just a few people have solved, and my intension is not to raise that number, is more like to explain how can it be solved and show a path and with that all of you are going to be able to solve it by yourselves.

Well, with all the restrictions and things described in the description of the problem we can assume a lot of things, and that make this problem easier to solve, actually theres just 1 case that is the hardest to solve, the rest is easy.

One way of solving this problem (the one i took) would be considering the graph as a hierarchical tree, each child has one more grade than the father (father => grade 1 ; son => grade 2). and that make it easier to verify the restrictions.

source code

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=29&page=show_problem&problem=1078

This one is not too difficult you need to calculate the sum of the cost above the average price and also the sum of the costs below the average. The number you print is the higher of the two.

like you need to use double data type to use the decimal part of the numbers but you only need 2 digits after the decimal point you need to “round” the number, you can do it this way:

(double)((long)(price – avg)*100)/100

that way you eliminate all the digits after the second spot.

REMEMBER TO OUTPUT THE “$” CHARACTER !!!!

source code

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3051

This one is so easy, first of all when i first read the description very quick i thought it was like a knapsack problem, but then i read it again and realize that you only need it the weight in increasing order and try to grab as many eggs as possible, and then i realize that it says that they already give you the numbers in NON-DECREASING order (thats just another way to say they are in INCREASING order ) so you don’t have to even worry about using qsort or something similar, you just grab the numbers and start putting eggs into your bowl until one of three situation occur:

1) you have P eggs (cause thats the maximum amount of eggs you can have)

2) you have Q gm of eggs already in your bowl (cause thats the maximum weight you can handle)

3) the more obvious one, you ran out of eggs XD

so this can be easily solved by a loop, there are no special cases so you don’t need to worry about that.

Source Code

i’ll try to post more often from now on, so be patient and wait for more problems explained.

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3060

The approach i used with this problem is to consider 2 diferent formulas:

1) when the bottle of milk is more than half full

2) when the bottle of milk is less than half full

like you can see in the first picture the red part is the first case and the white part is the second case. In the picture is also the variables and the degrees of the corners(we only have 3 diferent degrees º(Theta), 90, (90 – º).

the only formula we need is:

a/sin(a) = b/sin(b) = c/sin(c)

with that we can find the missing distance cause we only know L, W, H but we also need the size of the segment where the milk level is.

you need to remember that when Theta is 0 that means you only need to print L*W*H and if Theta is 90 you print 0

P.D. You can find the formulas in the picture also so don’t worry.

Source Code

Visual Description