III Concurso de Preparación ACM-ICPC México Abril 2012

Analysis for problems of the competition ...

Problem-Setter: Ing. Yonny Mondelo HernándezInstitution: Universidad de las Ciencias Informáticas (CUBA)~ Contact-Mail: ymondelo @ uci.cu ~ COJ-Account: ymondelo20

Sites:Universidad Vasco de Quiroga (UVAQ)Universidad Panamericana (UP) – BonaterraBenemérita Universidad Autónoma de Puebla (BUAP)Instituto Tecnológico de Estudios Superiores (ITES) de ZamoraInstituto Tecnológico y de Estudios Superiores de Monterrey (ITESM) – Querétaro

III Concurso de Preparación ACM-ICPC México Abril 2012

Problem: 1793 - Substring Changes

Two cases for check... consider the sample

T = "ACM" and E = "ICPC".

1 - You hold the first letter of T, with every position of E,

and check with S finishing with the last valid intersection

between T and E.

2 - You hold the first letter of E, with every position of T,

and similarly check with S finishing the last valid

Intersection between T and E.

III Concurso de Preparación ACM-ICPC México Abril 2012

Problem: 1793 - Substring Changes

For the sample there are 6 cases:

ACM---

--ICPC

ACM--

-ICPC

ACM-

ICPC

-ACM

ICPC

--ACM

ICPC-

---ACM

ICPC--

Need 4 changes...

Need 4 changes...

Need 3 changes...

Need 4 changes...

Need 4 changes...

Need 3 changes...

III Concurso de Preparación ACM-ICPC México Abril 2012

1794 - Who Committed the Crime

We call people who said the suspect committed a murder

accused and the people who say suspect did not commit

murder justifying. Consider the number of suspected k.

Suppose that he had committed murder, and calculate

how much time people have told the truth. People who

have told the truth - these are people who accused him,

and the people who did not justify it.

Thus, their number is:

III Concurso de Preparación ACM-ICPC México Abril 2012

1794 - Who Committed the Crime

The number of people who have accused him, plus the

number of people who are acquitted anyone, and minus the

number of people who are acquitted him. After these steps,

in O(1) calculate for each suspect the number of people who

speak the truth in that, if he was a criminal, and this number

is m, then this person may be the culprit. And once we have

found a list of possible offenders, then it is easy for each

person to determine whether he could tell the truth or not.

III Concurso de Preparación ACM-ICPC México Abril 2012

1795 - Dealing with the K-th Pair of Elements

To begin with, I want to note that this problem has caused

multiple questions, although the subject was clearly and

unambiguously says what to do.

First sort the array S. If all the numbers are different, then

the obvious answer - a pair (S [k / n], S [k % n]), where

the operation % means the remainder of the module.

III Concurso de Preparación ACM-ICPC México Abril 2012

1795 - Dealing with the K-th Pair of Elements

However, if the array is a duplicate of, then the situation

becomes a bit trickier, consider an example: let the data

array (1, 1, 2, 2), then the pair recorded as follows:

(1, 1), (1, 1), ( 1, 1), (1, 1), (1, 2), (1, 2), (1, 2), (1, 2),

(2, 1), (2, 1), (2, 1), (2, 1), (2, 2), (2, 2), (2, 2), (2, 2),

each pair is taken into account and is not going away,

always treated exactly n^2 pairs of numbers.

III Concurso de Preparación ACM-ICPC México Abril 2012

1795 - Dealing with the K-th Pair of Elements

What can I do in this case ?

III Concurso de Preparación ACM-ICPC México Abril 2012

1795 - Dealing with the K-th Pair of Elements

It is still true that the first number - this is S [k / n].

A second number - this is S [(k-n · cnt) / num], where cnt:

- number of elements in the array is strictly smaller than

S [k / n], and the num: - number of elements in the array,

which is equal to S [k / n].

This formula is easy to see by considering a few examples

of repetitive elements. In a pinch, you can write a trivial

solution for the O (n^2·log(n)) and make sure that you

have correctly come up with a formula.

III Concurso de Preparación ACM-ICPC México Abril 2012

1800 - Maximum Subset of Array

You count the number of positive numbers, negative

numbers and zeros.

- IF there are positive numbers, the maximum sum

will be the sum of all positive numbers. - Else IF there

are at least one zero, the maximum sum will be 0.

- Other case, the maximum sum will be the max number.

But, what with the count of possible solutions???

Well, you only should thinks about the number of zeros :)

III Concurso de Preparación ACM-ICPC México Abril 2012

1801 - ASCII Area

Scan the picture top-to-bottom, or left-to right...

--- Count the number of "\" and "/":

◦ If even, we're outside the polygon

◦ If odd, we're inside the polygon

Area = (number of "\" and "/") / 2 + number of "." inside...

III Concurso de Preparación ACM-ICPC México Abril 2012

1802 - Binary Encoding

- Find the smallest n, such that: m <= 2^n

- Now k = 2^n – m, is the number of “unused”

codes compared to binary encoding

◦ k is exactly the max number of codes with n-1 bits

So, to get the answer write

◦ For i in [0, k-1] write binary encoding of i with n-1 bits.

◦ For i in [k, m-1] write binary encoding of (i + k) with n bits.

III Concurso de Preparación ACM-ICPC México Abril 2012

1803 - The Grille

Pretty easy, wasn’t it?

III Concurso de Preparación ACM-ICPC México Abril 2012

1804 - Stack Machine Executor

Straightforward Simulation.

Beware of Integer Overflow (MUL).

III Concurso de Preparación ACM-ICPC México Abril 2012

Analysis for problems of the competition ...

Problem-Setter: Ing. Yonny Mondelo HernándezInstitution: Universidad de las Ciencias Informáticas (CUBA)~ Contact-Mail: ymondelo @ uci.cu ~ COJ-Account: ymondelo20

Sites:Universidad Vasco de Quiroga (UVAQ)Universidad Panamericana (UP) – BonaterraBenemérita Universidad Autónoma de Puebla (BUAP)Instituto Tecnológico de Estudios Superiores (ITES) de ZamoraInstituto Tecnológico y de Estudios Superiores de Monterrey (ITESM) – Querétaro