The first lecture of the ACM Aleppo CPC training. The local contest of ICPC. This lecture will help you get started in programming contests word with the lower bound techniques. The lectures focus on the C++ programming language and the STL library to solve programming problems.

1. ACM Aleppo CPC Training
Part 1 C++ Programming Concepts
By Ahmad Bashar Eter

2. About the contest and the training
• What is the ACM ICPC?
• Who this contest is for?
• What will I get from this contest?

3. About the contest and the training
• What should I do to win in this contest?
• Who this training is for?
• How far I will reach in programming contest word after this part is
finished?

4. The Problems
Each problem has:
• Problem description
• Input description
• Output description
• Sample input (1 or more)
• Sample output (1 or more)

5. Judge Response
• Accepted
• Wrong Answer
• Time Limit Exceeded
• Runtime Error
• Compilation Error
• Presentation Error

6. Example 1
HQ9+ is a joke programming language which has only four one-character instructions:
"H" prints "Hello, World!",
"Q" prints the source code of the program itself,
"9" prints the lyrics of "99 Bottles of Beer" song,
"+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of
the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing
this program will produce any output.
Input
The input will consist of a single line p which will give a program in HQ9+. String p will
contain between 1 and 100 characters, inclusive. ASCII-code of each character of p will
be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Solve at: http://codeforces.com/problemset/problem/133/A

7. Self Training resources
• First of all Google and Wikipedia.
• To ask a question go to www.stackoverflow.com
• C++ and STL libraries reference: www.cplusplus.com
• C++ tutorials: www.learncpp.com
• Online Judges:
Code Forces
UVA
A2OJ
Top Coder
SPOJ
Hacker Earth

8. Self Training resources
Books:
1. Programming Challenges
2. Competitive Programming 3
Tutorials:
Topcoder tutorials:
Hacker Earth Code Monk
GeeksForGeeks.org

9. C++ Review

10. Variables & Data types
• To define a variable: [TypeName] [VariableName];
• Integer data types:
Note: int data type is 2 byte signed on some compilers and 4 on other.
Type name Size Range
char 1 byte signed -128 to 127
unsigned char 1 byte unsigned 0 to 255
short 2 byte signed -32,768 to 32,767
unsigned short 2 byte unsigned 0 to 65,535
long 4 byte signed -2,147,483,648 to 2,147,483,647
unsigned long 4 byte unsigned 0 to 4,294,967,295
long long 8 byte signed
-9,223,372,036,854,775,808 to
9,223,372,036,854,775,807
unsigned long long 8 byte unsigned 0 to 18,446,744,073,709,551,615

11. Variables & Data types
• char type is also used to represent alphabets.
• Floating point data types (Real values):
• Boolean data type (true or false state variable): bool its size is 1 byte.
Type name Size Range Precision
float 4 bytes ±1.18 x 10-38 to ±3.4 x 1038 6-9 significant digits, typically 7
double 8 bytes ±2.23 x 10-308 to ±1.80 x 10308 15-18 significant digits, typically 16

12. Operators
• Arithmetic
Operator Symbol Form Operation
Assignment = x = y Assign value y to x
Addition + x + y x plus y
Subtraction - x - y x minus y
Multiplication * x * y x multiplied by y
Division / x / y x divided by y
Addition assignment =+ x += y Add y to x
Subtraction assignment -= x -= y Subtract y from x
Multiplication assignment =* x *= y Multiply x by y
Division assignment =/ x /= y Divide x by y
Modulus assignment =% x %= y Put the remainder of x / y in x

13. Operators
• Arithmetic
• For the complete list of operators precedence go to:
http://www.learncpp.com/cpp-tutorial/31-precedence-and-
associativity/
Operator Symbol Form Operation
Prefix increment (pre-increment) ++ ++x Increment x, then evaluate x
Prefix decrement (pre-decrement) –– ––x Decrement x, then evaluate x
Postfix increment (post-increment) ++ x++ Evaluate x, then increment x
Postfix decrement (post-decrement) –– x–– Evaluate x, then decrement x
Prefix increment (pre-increment) ++ ++x Increment x, then evaluate x

14. Operators
• Relational operators (comparisons)
Operator Symbol Form Operation
Greater than > x > y true if x is greater than y, false otherwise
Less than < x < y true if x is less than y, false otherwise
Greater than or equals >= x >= y true if x is greater than or equal to y, false otherwise
Less than or equals <= x <= y true if x is less than or equal to y, false otherwise
Equality == x == y true if x equals y, false otherwise
Inequality =! x != y true if x does not equal y, false otherwise
Operator Symbol Form Operation
Greater than > x > y true if x is greater than y, false otherwise
Less than < x < y true if x is less than y, false otherwise
Greater than or equals >= x >= y true if x is greater than or equal to y, false otherwise

15. Operators
• Logical operators
• And there are a lot of other operators we will discuss them when
we will approach them.
Operator Symbol Form Operation
Logical NOT ! !x true if x is false, or false if x is true
Logical AND && x && y true if both x and y are true, false otherwise
Logical OR || x || y true if either x or y are true, false otherwise

16. Flow control statements
• If statement.
• Switch statement.
• While loop statement.
• For loop statement.
• Break and continue.

17. Flow control: if statement
• Syntax: If (Conditional expression)
one expression or {expressions block}
else if (Conditional expression)
one expression or {expressions block}
else if (Conditional expression)
one expression or {expressions block}
.
.
.
else if (Conditional expression)
one expression or {expressions block}
else
one expression or {expressions block}

18. Flow control: switch statement
• Syntax: switch(integer expression)
{
case (case value):
[break];
case (case value):
[break];
case (case value):
[break];
default:
[break];
}

19. Flow control: switch statement
• Used over if statement if we have a lot of equal condition if
statement (it is much faster than if statement).
• Note: we can leave out the brake statement if we want some cases
to do the same statements.

20. Flow control: while loop statement
• Syntax: while (Conditional expression)
one expression or {expressions block}
• Anther form: do
one expression or {expressions block}
while (Conditional expression)
• Used when we don’t know how many times we will loop.
• Note: we can use it to do infant loop if the condition will never be
false (ex. 10>0 , true).

21. Flow control: for loop statement
• Syntax
for(loop initialization; loop condition; increment expression)
one expression or {expressions block}
• Used when we know exactly how many times we will loop.
• We can omit the loop initialization , loop condition or increment
expression if we want to.
• Note: we can use it to do infant loop if the loop condition will never
be false (ex. 10>0 , true) or if we leave out the loop condition .

22. Flow control: break & continue
• If we want to skip the running iteration of the loop, we can use the
continue statement.
• If we want to terminate all the work of the loop, we can use the
break statement.

23. IO methods
• In C++ if we want to output something on the screen, we can use
cout stream.
• If we want to get an input from what the user is typing on the
screen, we can use cin stream.
• In general, cin cout streams are slow in programing contests.
• To get faster IO methods you may use the C programing language
methods printf and scanf.
• Or you may put this statement before any cin or cout statement:
ios::sync_with_stdio (false);
• This statement turns off the synchronization between the C++ IO
and C IO, so once you use it you can’t use printf and scanf.

24. Example 2: A. Choosing Teams
The Saratov State University Olympiad Programmers Training Center (SSU
OPTC) has n students. For each student you know the number of times
he/she has participated in the ACM ICPC world programming
championship. According to the ACM ICPC rules, each person can
participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in
the world championship. Each team must consist of exactly three
people, at that, any person cannot be a member of two or more teams.
What maximum number of teams can the head make if he wants each
team to participate in the world championship with the same members
at least k times?
Solve at:

25. Example 2: A. Choosing Teams
Input:
The first line contains two integers, n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 5). The
next line contains n integers: y1, y2, ..., yn (0 ≤ yi ≤ 5), where yi shows the
number of times the i-th person participated in the ACM ICPC world
championship.
Output:
Print a single number — the answer to the problem.
Solve at:

26. Example 2: A. Choosing Teams
1. #include <iostream>
2. using namespace std;
3. int main() {
4. ios::sync_with_stdio(false);
5. int teamscount = 0, invidualcount = 0, n, k, x;
6. cin >> n >> k;
7. for (int i = 0; i < n; i++) {
8. cin >> x;
9. if (k + x <= 5) {
10. invidualcount++;
11. if (invidualcount == 3) {
12. invidualcount = 0;
13. teamscount++;
14. }
15. }
16. }
17. cout << teamscount << endl;
18. return 0;
19.}
Solve at:

27. Example 3: The 3n + 1 problem
Background
Problems in Computer Science are often classified as belonging to a certain
class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be
analyzing a property of an algorithm whose classification is not known for all
possible inputs.
The Problem
Consider the following algorithm:
1. input n
2. print n
3. if n = 1 then STOP
4. if n is odd then tex2html_wrap_inline44
5. else tex2html_wrap_inline46
6. GOTO 2
Solve at:

28. Example 3: The 3n + 1 problem
Given the input 22, the following sequence of numbers will be printed 22 11 34
17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed)
for any integral input value. Despite the simplicity of the algorithm, it is
unknown whether this conjecture is true. It has been verified, however, for all
integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers
than this.)
Given an input n, it is possible to determine the number of numbers printed
(including the 1). For a given n this is called the cycle-length of n. In the
example above, the cycle length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length
over all numbers between i and j.
Solve at:

29. Example 3: The 3n + 1 problem
The Input
The input will consist of a series of pairs of integers i and j, one pair of integers
per line. All integers will be less than 1,000,000 and greater than 0.
You should process all pairs of integers and for each pair determine the
maximum cycle length over all integers between and including i and j.
You can assume that no operation overflows a 32-bit integer.
The Output
For each pair of input integers i and j you should output i, j, and the maximum
cycle length for integers between and including i and j. These three numbers
should be separated by at least one space with all three numbers on one line
and with one line of output for each line of input. The integers i and j must
appear in the output in the same order in which they appeared in the input
and should be followed by the maximum cycle length (on the same line).
Solve at:

30. Example 3: The 3n + 1 problem
1. #include <iostream>
2. using namespace std;
3. int main() {
4. ios::sync_with_stdio(false);
5. int n, m;
6. while (cin >> n >> m) {
7. int maxCycle = 0;
8. int start,end;
9. if (n > m) {
10. start = m;
11. end = n;
12. }
13. else {
14. start = n;
15. end = m;
16. }
Solve at:
17. for(int i = start;i <= end; i++){
18. int Cycle = 1, x = i;
19. while (x > 1){
20. if (x % 2)
21. x = x * 3 + 1;
22. else
23. x = x / 2;
24. Cycle++;
25. }
26. if (Cycle > maxCycle)
27. maxCycle = Cycle;
28. }
29. cout<<n<<" "<<m<<" "<<maxCycle<<endl;
30. }
31. return 0;
32.}

31. Example 4: B. Present from Lena
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a
present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The
largest digit n should be located in the centre. The digits should decrease as they approach
the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Solve at:

32. Example 4: B. Present from Lena
Your task is to determine the way the handkerchief will look like by the
given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of
spaces before the first digit on each line. Every two adjacent digits in the
same line should be separated by exactly one space. There should be no
spaces after the last digit at the end of each line.
Solve at:

33. Other Exercises
Try these good exercises and remember the more you exercise, the
better you will become ;).
• Codeforces A. Nearly Lucky Number
• Codeforces A. I_love_%username%
• Codeforces A. Lunch Rush
• UVA 00278 – Chess
• UVA 10976 - Fractions Again?!

34. Thank you :) :) :)