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";

temp = a;
a = b;
b = temp;

If making additional variable temp is an overhead, in-place swap algorithms can be used. They are actually typical space-time trade-offs (if you want to avoid memory/space usage, a bit more processing/time is needed).

If variables that have to be swapped are numbers, swapping can be done by using addition and subtraction in the next way:

a = a + b; //a+=b
b = a - b;
a = a - b; //a-=b

Any other variable types can be swapped using XOR swap algorithm in the next way:

a = a XOR b; //a^=b
b = b XOR a; //b^=a
a = a XOR b; //a^=b

These examples are interesting when dealing with value types, with reference types, it’s a bit different.

program reversed;
uses wincrt;
var
a,b:string;
i,s,t:integer;
begin 
  readln(a);
  readln(b);
  while length(a)<length(a) do
    a:=a+'0';
  while length(b)<length(b) do
    b:=b+'0';
  t:=0;
  for i:=1 to length(b) do
  begin
    s:=ord(a[i])+ord(b[i])-2*ord('0')+t;
    a[i]:=chr(ord('0')+(s mod 10));
    t:=s div 10;
  end;
  if t<>0 then
    a:=a+'1';
  while a[length(a)]='0' do
    delete(a,length(a),1);
  while a[1]='0' do
    delete(a,1,1);
  writeln(a);
end.

Solution in example input is 10,83. There are 3 x 2 and 2 x sqrt(2), and remember two meters at the ends, so 4*2+2*sqrt(2)=10.83.

More about Euclidean distance on wiki.

Pascal solution:

program herding_frosh;
uses wincrt;
type
  tarray=record
  x,y:integer;
  end;
var
n,i:integer;
b:array[1..100] of real;
function euclidean_distance(x1,y1,x2,y2:integer):real;
begin
  euclidean_distance:=sqrt(sqr(x1-x2)+sqr(y1-y2));
end;
function calculate:real;
var
  i,n:integer;
  r:real;
  a:array[0..100] of tarray;
begin
r:=0;
readln(n);
with a[0] do
  begin
   x:=0;
   y:=0;
  end;
for i:=1 to n do
readln(a[i].x,a[i].y);
for i:=1 to n do
  begin
    r:=r+euclidean_distance(a[i].x,a[i].y,a[i-1].x,a[i-1].y);
  end;
r:=r+euclidean_distance(a[n].x,a[n].y,a[0].x,a[0].y)+2;
calculate:=r;
end;

begin
readln(n);
for i:=1 to n do
begin
   readln;
   b[i]:=calculate;
end;
for i:=1 to n do
writeln(b[i]:9:2);
end.

TPW code:

program dec_to_bin;
uses wincrt;
var
  a:array[1..100] of 0..1;
  n,k,i:integer;
  label top;
begin
top:
writeln('Enter decimal number !');
readln(n);
k:=0;
while n>0 do
begin
k:=k+1;
a[k]:=n mod 2 ;
n:=n div 2;
end;
writeln('Binary number is : ');
for i:=k downto 1 do
write(a[i]);
writeln;
writeln('Press Enter if you want 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:

195  
591   
___
786   

786
687
___
1,473

1,473

This method leads to palindromes in a few steps for almost all of the integers. But there are interesting exceptions. 196 is the first number for which no palindrome has been found. It has never been proven, however, that no such palindrome exists. The task is to write a program that takes a given number and gives the resulting palindrome (if one exists) and the number of iterations it took to find it. It can be assumed that all the numbers used as test data will terminate in an answer with less than 1,000 iterations (additions), and yield a palindrome that is not greater than 4,294,967,295.
Input
The first line will contain an integer N (0 < N ≤ 100), giving the number of test cases, while the next N lines each contain a single integer P whose palindrome you are to compute.
Output
For each of the N integers, print a line giving the minimum number of iterations to find the palindrome, a single space, and then the resulting palindrome itself.
Sample Input
3
195
265
750
Sample Output
4 9339
5 45254
3 6666
Pascal code:

program reverse_and_add;
uses wincrt;
var
n,steps,p,i:longint;
a:array[1..100] of longint;
function reverse(k:longint):longint;
var
  p,i,g,f:longint;
  a:array[1..100] of 0..9;
begin
  g:=1;
  i:=0;
  f:=0;
  repeat
    begin
     i:=i+1;
     a[i]:=k mod 10;
     k:=k div 10;
    end;
  until k=0;
  for p:=1 to i-1 do
   g:=g*10;
   for p:=1 to i do
     begin
      f:=f+g*a[p];
      g:=g div 10;
     end;
     reverse:=f;
end;
function add(k:longint):longint;
var
  i:integer;
begin
  i:=0;
  repeat
    begin
      k:=k+reverse(k);
      i:=i+1;
    end;
  until reverse(k)=k;
  steps:=i;
  add:=k;
end;
begin
readln(n);
for i:=1 to n do
readln(a[i]);
for i:=1 to n do
   begin
     p:=add(a[i]);
     writeln(steps,' ',p);
   end;
end.

C++ code:

#include <iostream>
#include <cstdio>

using namespace std;

int steps,p,i;
int a [100];

int reverse(int k)
{
  int p,i,g,f;
  int a[100];
  g=1;
  i=0;
  f=0;
  do
    {
     i++;
     a[i] = k % 10;
     k = k / 10;
    }
  while (!k == 0);
  for (p=1;p<=i-1;p++)
   g=g*10;
    for (p=1;p<=i;p++)
     {
      f=f+g*a[p];
      g=g / 10;
      }
     return f;
}
int add(int k)
{
  int i;
  i=0;
  while (reverse(k)>k | reverse(k)<k)
    {
      k = k + reverse(k);
      i++;
    }
  
  steps = i;
  return k;
}
int main()
{
    int n;
    cout<<"enter number of test cases"<<endl;
cin>>n;
for (i=1;i<=n;i++)
cin>>a[i];
for (i=1;i<=n;i++)
   {
     p=add(a[i]);
     cout<<steps<<" "<<p<<endl;
   }
   system("pause");
}

program theorem_153;
uses wincrt;
var
n,i,k:longint;
begin
repeat
readln(n);
until n mod 3 = 0;
i:=1;
repeat
begin
   k:=n;
   n:=0;
   repeat
    begin
      n:=n+(k mod 10)*(k mod 10)*(k mod 10);
      k:=k div 10;
    end;
   until k=0;
   i:=i+1;
end;
until n=153;
writeln(i);
end.

C++:

# include <iostream>
# include <cstdio>

using namespace std;

int main ()
{
  int n,i,k;
cin>>n;
i=1;
while (n>153 | n<153)
{
   k=n;
   n=0;
   while (k > 0)
    {
      n=n+(k % 10)*(k % 10)*(k % 10);
      k=k / 10;
    }
   i++;
}
cout<<i<<endl;
system("pause");
}