First place: BLU TEAM 5 problems solved
Second place: /dev/null 4 problems solved
Third place: NibbleSoft 2 problems solved
Category: Uva Problems
Good news everyone finally we get a lab for the “advanced training course”.
Every Thursday (Jueves) from 3 pm to 6 pm. In the CISCO Lab in the CSI building (1 floor).
This week Maximum Sum.
Just to remind you that the advance class was change it to Thursday (JUEVES) from 3pm to 6pm. And the begginers class is saturday (SABADO) from 9am to 12pm.
Problem description , input and output:
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:
Problem description, input and output:
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.
Problem Description, Input and Output:
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.
Problem Description, Input and Output:
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:
Problem description, Inputs and Outputs:
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:
Sorry for all the time i haven’t post anything new, but the work didn’t let me do anything.
i’ll try to post more often from now on, so be patient and wait for more problems explained.
Problem description, Input and Output:
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).
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