Description
To win the Powerball lottery (an extremely unlikely event so don’t waste your time) you have to pick six numbers correctly. The first five numbers are drawn from a drum containing 53 balls and the sixth is drawn from a drum containing 42 balls. The chances of doing this are 1 in 120,526,770. Write a program to generate a set of Powerball numbers by utilizing the choice function in Python’s random module.

Input
Ask the user how many sets of Powerball numbers he or she would like.

Output
The program will print each set of Powerball numbers in numeric order.

Sample session
Official (but fruitless) Powerball number generator

How many sets of numbers? 3

Your numbers: 3 12 14 26 47 Powerball: 2
Your numbers: 1 4 31 34 51 Powerball: 17
Your numbers: 10 12 49 50 53 Powerball: 35

from random import randint
def BubbleSort(l):
    e=0
    for i in range(len(l)-1):
        for j in range(len(l)-1):
            if l[j]>l[j+1]:
                e=l[j+1]
                l[j+1]=l[j]
                l[j]=e
    return l
def MyPrint(L,pb):
        BubbleSort(L)
        print "Your numbers: %d %d %d %d %d Powerball: %d"%(L[0],L[1],L[2],L[3],L[4],pb)

print "Official (but fruitless) Powerball number generator"
No=int(raw_input("How many sets of numbers? "))
for i in range(No):
    numbers=[]
    powerball=randint(1,42)
    i=0
    while (i!=5):
        numbers.append(randint(1,53))
        i+=1
    for i in range(len(numbers)-1):
        for j in range(len(numbers)-1):
            if i==j:
                j+=1
            if numbers[i]==numbers[j] or powerball==numbers[j]:
                numbers[j]=randint(1,53)
                j-=1
    MyPrint(numbers,powerball)

Few notes:
After you choose how many sets of numbers you want, the first ‘while’ loop gives some random numbers to our powerball list. After that, the 2 ‘for’ loops check if we have double numbers in our list. Because after we draw one ball, that one cannot be chosen again and it is out of the game.

Then the process repeats for the rest of the sets.

Pascal solution:

program even_nums_sum;
var i,n,s:integer;
begin
    readln(n);
    s:=0;
    for i:=1 to n do
    begin
        if i mod 2 = 0 then
        begin
            s:=s+i;
        end;
    end;
    writeln(s);
    readln();
end.

This task is interesting because different bitwise operators can be used while writing an algorithm in C++:

int a = 86, b = 57, product = 0;
while (b)
{
if (b & 1)
product += a;
a <<= 1;
b >>= 1;
}
cout << product;
cout << "\n";

First of all “while (b)” actually means “while (b!=0)”. (b & 1) is actually a check if number is odd. And << and >> are shifting operators which effectively give the result of integer multiplication and division respectively.

The solution can be written even more short by using for loop and ternary operator:

int a,b,product = 0;
for(;b;product+=(b&1?a:0), b>>=1, a<<=1);
cout << product;
cout << "\n";

<?php
function getDistance($a_lat,$a_lng,$b_lat,$b_lng,$mi=false){
    var $radius;
    ($mi ? $radius=10253 : $radius=6371);
    $a_lat = deg2rad($a_lat);
    $a_lng = deg2rad($a_lng);
    $b_lat = deg2rad($b_lat);
    $b_lng = deg2rad($b_lng);
    if($a_lat==$b_lat && $a_lng==$b_lng) return 0;
    if ((sin($b_lat)*sin($a_lat ) + cos($b_lat)*cos($a_lat )*cos($b_lng-$a_lng))>1)
        return $radius * acos(1));
     return $radius * acos(sin($b_lat)*sin($a_lat)+cos($b_lat)*cos($a_lat)*cos($b_lng-$a_lng)));
}
?>

Euclid Algorithm that does the same thing (more here).
Except Euclid Algorithm, Binary Algorithm is used, more on wiki.
C++ code:

# include <iostream>
# include <cstdio>

using namespace std;

int gcd(int u,int v)
{
     int s;
     if (u == 0 || v == 0)
       return u | v;
     for (s = 0; ((u | v) & 1) == 0; ++s) {
         u >>= 1;
         v >>= 1;
     }
     while ((u & 1) == 0)
       u >>= 1;
     do {
         while ((v & 1) == 0) 
           v >>= 1;
         if (u < v) {
             v -= u;
         } else {
             int diff = u - v;
             u = v;
             v = diff;
         }
         v >>= 1;
     } while (v != 0);
     return u<<s;
}

int main()
{
    int a,b;
    cin>>a;
    cin>>b;
    cout<<gcd(a,b)<<endl;
    system("pause");
}

Here is the solution in TPW:

program integral;
uses wincrt;
const dx = 0.0001;
var
  i,a,b,p:real;
  label top;
begin
top:
writeln('Enter start of segment of Ox !');
readln(a);
repeat
begin
writeln('Enter end of segment of Ox !');
readln(b);
end;
until b>a;
i:=a+dx;
p:=0;
repeat
begin
  p:=p+abs(dx*sin(i)*ln(i));
  i:=i+dx;
end;
until i>b;
writeln('Solution is ',p:9:4);
writeln('Press Enter to try again !');
readln;
clrscr;
goto top;
end.

5 div 2 = 2
5 mod 2 = 1

If you get the modulus of a number when divided with 10, you would get last digit, than you can divide it by 10 and do modulus again till you reach zero.

Here is the implementation in TPW:

program cut_digits_out_of_number;
uses wincrt;
var
  n:longint;
  i,j:integer;
  a:array [1..100] of integer;
begin
writeln('Program for cutting digits');
writeln('out of an integer number !');
writeln('Enter integer number !');
readln(n);
i:=0;
repeat
begin
   i:=i+1;
   a[i]:=n mod 10;
   n:=n div 10;  
end;
until n=0;
writeln('Digits are : ');
for j:=i downto 1 do
writeln(a[j]);
end.

But if you want to use function like in Delphi strtoint(), you can make it in Pascal. Like i made it. And i used string cutting and i made that function, so solution looks like this:

program cut_digits_out_of_number;
uses wincrt;
var
  n:string;
  i:integer;
  a:array [1..100] of integer;
function numcheck(s: string): boolean;
var
  x:integer;
begin
numcheck:=true;
for x:=1 to length(s) do
  if (ord(s[x])<48) or (ord(s[x])>57) then
   if s[x]<>'' then
    numcheck:=false;
end;
function str_to_int(s:string):integer;
var
  check:integer;
  i:integer;
begin
   if numcheck(s) then
    val(s,i,check);
   str_to_int:=i;
end;
begin
writeln('Program for cutting digits');
writeln('out of an integer number !');
writeln('Enter integer number !');
readln(n);
for i:=1 to length(n) do
begin
a[i]:=str_to_int(n[i]);
end;
for i:=1 to length(n) do
writeln(a[i]);
end.

Or you can use val procedure, explained in Help:

TPW code:

program matrix_read_write;
uses wincrt;
var
  a:array[1..100,1..100] of integer;
  i,j,m,n:integer;
begin
writeln('Enter length of matrix !');
readln(n);
writeln('Enter height of matrix !');
readln(m);
for i:=1 to n do
  for j:=1 to m do
    read(a[i,j]);
for i:=1 to n do
  begin
    writeln;
  for j:=1 to m do
    write(a[i,j],' ');
  end;
end.

program calendar;
uses wincrt;
type
  months=1..12;
  dayinweek =0..6;
var
  days:array[months] of integer;
  month,year:integer;

  function leapyear(year:integer):boolean;
  begin
  if (year mod 100)=0 then
  leapyear :=(year mod 400) = 0
  else
  leapyear := (year mod 4) = 0
  end;

  function firstjanuary(year:integer): dayinweek;
  begin
  firstjanuary := (year+(year-1) div 4-(year-1) div 100 +(year-1) div 400 ) mod 7
  end;

  procedure setdays (year:integer);
  begin
  days[1]:=31;
  if leapyear(year) then
  days[2]:=29
  else
  days[2]:=28;
  days[3]:=31;
  days[4]:=30;
  days[5]:=31;
  days[6]:=30;
  days[7]:=31;
  days[8]:=31;
  days[9]:=30;
  days[10]:=31;
  days[11]:=30;
  days[12]:=31;
  end;

  function firstdayinmonth (month,year:integer): dayinweek;
  var
  i,brdana:integer;
  begin
  setdays (year);
  brdana :=0;
  for i:=1 to month-1 do
  brdana:= brdana + days[i];
  firstdayinmonth := (firstjanuary (year) + brdana) mod 7;
  end;

  procedure writing (month,year :integer );
  var
  i,n:integer;
  begin
  writeln;
  writeln('==================================');
  write('Calendar for ',month,'. month ');
  writeln (year,'. year');
  writeln ('----------------------------------');
  writeln (' Sun Mon Tue Wed Thu Fri Sat');
  n:=firstdayinmonth(month,year);
  for i:=0 to n-1 do
  write ('    ');
  for i:=1 to days[month] do
  begin
  write (i:4);
  n:=n+1;
  if (n mod 7) =0 then
  writeln;
  end;
  writeln;
  writeln;
  end;

  begin
  repeat
  write (' Enter month : ');
  readln (month);
  until (month>=1) and (month<=12);
  repeat
  write (' and year : ');
  readln (year)
  until (year>=0) and (year<= 4000);
  writing (month, year);
  Writeln;
  end.

p:=p+abs(sin(i)*dx);

And input is segment between a and b. Parameter b has to be bigger than a.
For example, if pi=3.141, if you enter segment from 0 to pi, you’ll get solution 2, because that area is 2, as you can see in the picture:

Code in TPW:

program find_surface_of_function_sinx;
uses wincrt;
const dx = 0.0001; 
var
  i,a,b,p:real;
begin
writeln('Program for finding area in segment of sin() function');
writeln('Enter start of segment of Ox');
readln(a);
repeat
begin
writeln('Enter end of segment of Ox');
readln(b);
end;
until b>a;
i:=a+dx;
p:=0;
repeat
begin
  p:=p+abs(sin(i)*dx);
  i:=i+dx;
end;
until i>b;
writeln('Solution is ',p:9:4);
end.