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.

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.

Solution in Pascal:

program upper_lower_case;
uses wincrt;
var
  s,k:string;
  label error;
function uppercase(s:string):string;
var
i:integer;
begin
  for i:=1 to length(s) do
    if (s[i]>='a') and (s[i]<='z') then
      dec(s[i],32);
    uppercase:=s;
end;
function lowercase(s: string): string;
var
i:integer;
begin
  for i:=1 to length(s) do
    if (s[i]>='A') and (s[i]<='Z') then
     inc(s[i],32);
    lowercase:=s;
end;
begin
  writeln('Enter string ?');
  readln(s);
  error:
  writeln('Enter "u" for upper or "l" for');
  writeln('lower case string ?');
  readln(k);
  if k='u' then
     writeln(uppercase(s))
  else
    if k='l' then
      writeln(lowercase(s))
    else   
     goto error;
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.

var
input:text;

Next 4 commands are essential:

assign(text_variable,string_location);
reset(text_variable);
rewrite(text_variable);
close(text_variable);

Reset opens a file for reading, and rewrite opens a file for
writing, and assign assigns the file with the text variable,
or connects them, and close closes that connection.
“string_location” represents a string that is location of file.
That string, if the file is in the same dir or folder as .exe
is for example “input.txt” or “input.in”,
and if you want to specify olcation you use standard dir declaration
for example “C:/text/input.txt”.
ATTENTION: File must be created before writing or reading from it.
You can not assign non-existed file.

Next thing after you assign and reset file is reading from it,
you can use read() or readln() functions like

readln(file_variable,variable_list);

Write a program which reads an integer variable from file “input.in”
and writes on screen 2*that_variable.

TPW:

program example1;
uses wincrt;
var
n:integer;
input:text;
begin
assign(input,'input.in');
reset(input);
readln(input,n);
writeln(2*n);
close(input);
end.

ATTENTION : That variable must be in the first line of that file.
For writing to file, write() and writeln() functions the same way as you use
read() and readln() functions, like this:

writeln(file_variable,variable_list);

Make a program which reads 2 numbers from 2 separate lines from file
“input.txt” and writes all numbers between them into file “output.txt”.
If the second number is smaller than first then it writes numbers from
second to first, and if they are equal, then it writes in file
“The numbers are equal.”.

program example2;
uses wincrt;
var
i,a,b:integer;
input,output:text;
begin
assign(input,'input.txt'); reset(input);
assign(output,'output.txt'); rewrite(output);
readln(input,a);
readln(input,b);
if a=b then
  writeln(output,'The numbers are equal.')
else
  if a>b then
   for i:=b to a do
     writeln(output,i)
  else
    if a<b then
      for i:=a to b do
        writeln(output,i);
close(input);
close(output);
end.

You should always close the files on the end.

program separate_words;
uses wincrt;
var
s,locin,locout:string;
input,output:text;
function alphabetcheck(s:string):boolean;
var
  x:integer;
  y:boolean;
begin
  y:=true;
    for x:=1 to length(s) do
      if ((s[x]<'A') or (s[x]>'Z')) and (y=true) then
       if s[x]='' then
         y:=true
       else
         if ((s[x]<'a') or (s[x]>'z')) then
           y := false;
        if y=true then
            alphabetcheck:=true
        else
            alphabetcheck:=false;
end;
function lowercase(s:string):string;
var
  i:integer;
begin
  for i:=1 to length(s) do
    if (s[i]>='A') and (s[i]<='Z') then
      inc(s[i], 32);
    lowercase:=s;
end;
function remove_chars(s:string):string;
var
  i,br:integer;
  g:array[1..10] of char;
begin
br:=0;
  for i:=1 to length(s) do
    if not alphabetcheck(s[i]) then
      remove_chars:=remove_chars(copy(s,1,i-1)+copy(s,i+1,length(s)))
    else
     br:=br+1;
     if br=length(s) then
       remove_chars:=s;    
end;
procedure one_line(s:string);
begin
  if pos(' ',s)=0 then
    writeln(output,lowercase(remove_chars(s)))
  else
    begin
      writeln(output,lowercase(remove_chars(copy(s,1,pos(' ',s)-1))));
      one_line(copy(s,pos(' ',s)+1,length(s)));
    end;
end;
begin
  writeln('Enter location of input file !');
  readln(locin);
  writeln('Enter location of output file !');
  readln(locout);
  assign(input,locin); reset(input);
  assign(output,locout); rewrite(output);
  repeat
    begin
      readln(input,s);
      one_line(s);
    end;
  until eof(input);
  close(input);
  close(output);
  writeln('task completed succesfully !');
end.

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:

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");
}