program Hello;
var
a,b,c,x,d: integer;
x1,x2: real;
begin
readln(a,b,c);
if a = 0 then
begin
if b = 0 then
begin
if c = 0 then
begin
writeln('11');
end
else
writeln('21');
end;
end
else
writeln('31');
end;
end
else
d := b^2 - 4*a*c;
if d < 0 then
begin
writeln('Нет Вещественных корней!');
end
else
x1 := (-b + sqrt(d))/(2*a);
x2 := (-b - sqrt(d))/(2*a);
writeln('Первый Корень:' + x1 + ' ' + 'Второй Корень:' + x2);
end;
end;
end.
The reason for this is that your begins and ends are not balanced; disregarding the opening begin and closing end. for the program's syntax to be correct, you should have equal numbers of each, but you have 4 begins and 8 ends.
Obviously, your code is to compute the solutions of a quadratic equation. What I think you should do is to adjust the layout of your code so that it reflects that and then correctly the begins and ends. In particular, your program is trying to detect whether any of a, b and d is zero and, if so, write a diagnostic message, otherwise calculate the roots by the usual formula.
Unfortunately, your begins and ends do not reflect that. Either the whole of the block starting d := ... needs to be executed or none of it does, so the else on the line before needs to be followed by a begin, as in
else begin
d := b*b - 4*a*c; //b^2 - 4*a*c;
if d < 0 then begin
writeln('Нет Вещественных корней!');
end
else begin
x1 := (-b + sqrt(d))/(2*a);
x2 := (-b - sqrt(d))/(2*a);
// writeln('Первый Корень:' + x1 + ' ' + 'Второй Корень:' + x2);
writeln('Первый Корень:', x1, ' Второй Корень:' , x2);
end;
end;
(You don't say which Pascal compiler you are using, but the above fixes two points which are flagged as errors in FreePascal.
If you need more help than that, please ask in a comment.
Btw, there are some grammatical constructs in Pascal implementations where an end can appear without a matching preceding begin such as case ... of ...end.
Related
can you say to me, where is the error. I wanna write in console the content of the a[i] in every for run. The output could be 6 digits. There is compiler Alert:
/usr/bin/ld.bfd: warning: link.res contains output sections; did you forget -T?
...Program finished with exit code 0
program Hello;
type aha = array [1..6] of integer;
var
a: aha;
n,x,i: integer;
begin
a[1]:=2;
a[2]:=6;
a[3]:=4;
a[4]:=2;
a[5]:=4;
a[6]:=3;
n:=6;
x:=a[1];
for i:=2 to n do
begin
{
if (a[i-1]>= x) then
begin
a[i]:=a[i] - x div 2;
end;
else
begin
a[i]:=a[i] + x;
x:= x + mod x(a[i] + 1);
end;
writeln (a[i]);
}
end;
end.
I'm learning algorithms and I'm trying to make an algorithm that extracts numbers lets say n in [1..100] from a string. Hopefully I get an easier algorithm.
I tried the following :
procedure ReadQuery(var t : tab); // t is an array of Integer.
var
x,v,e : Integer;
inputs : String;
begin
//readln(inputs);
inputs:='1 2 3';
j:= 1;
// make sure that there is one space between two integers
repeat
x:= pos(' ', inputs); // position of the space
delete(inputs, x, 1)
until (x = 0);
x:= pos(' ', inputs); // position of the space
while x <> 0 do
begin
x:= pos(' ', inputs); //(1) '1_2_3' (2) '2_3'
val(copy(inputs, 1, x-1), v, e); // v = value | e = error pos
t[j]:=v;
delete(inputs, 1, x); //(1) '2_3' (2) '3'
j:=j+1; //(1) j = 2 (2) j = 3
//writeln(v);
end;
//j:=j+1; // <--- The mistake were simply here.
val(inputs, v, e);
t[j]:=v;
//writeln(v);
end;
I get this result ( resolved ) :
1
2
0
3
expected :
1
2
3
PS : I'm not very advanced, so excuse me for reducing you to basics.
Thanks for everyone who is trying to share knowledge.
Your code is rather inefficient and it also doesn't work for strings containing numbers in general.
A standard and performant approach would be like this:
type
TIntArr = array of Integer;
function GetNumbers(const S: string): TIntArr;
const
AllocStep = 1024;
Digits = ['0'..'9'];
var
i: Integer;
InNumber: Boolean;
NumStartPos: Integer;
NumCount: Integer;
procedure Add(Value: Integer);
begin
if NumCount = Length(Result) then
SetLength(Result, Length(Result) + AllocStep);
Result[NumCount] := Value;
Inc(NumCount);
end;
begin
InNumber := False;
NumCount := 0;
for i := 1 to S.Length do
if not InNumber then
begin
if S[i] in Digits then
begin
NumStartPos := i;
InNumber := True;
end;
end
else
begin
if not (S[i] in Digits) then
begin
Add(StrToInt(Copy(S, NumStartPos, i - NumStartPos)));
InNumber := False;
end;
end;
if InNumber then
Add(StrToInt(Copy(S, NumStartPos)));
SetLength(Result, NumCount);
end;
This code is intentionally written in a somewhat old-fashioned Pascal way. If you are using a modern version of Delphi, you wouldn't write it like this. (Instead, you'd use a TList<Integer> and make a few other adjustments.)
Try with the following inputs:
521 cats, 432 dogs, and 1487 rabbits
1 2 3 4 5000 star 6000
alpha1beta2gamma3delta
a1024b2048cdef32
a1b2c3
32h50s
5020
012 123!
horses
(empty string)
Make sure you fully understand the algorithm! Run it on paper a few times, line by line.
I've beeng having some trouble with this code... I need to create an algorithm which makes the user input a number (X), and then the program calculates the sum of all the odd numbers below (x).
This what I've tried so far, but can't really wrap my head around the logic behind it:
Program odd_numbers;
Var
Num, Limite, Soma: integer;
Begin;
Soma := 0;
Writeln('Choose a limit:');
Readln(Limite);
While (Limite / 2 > 0) do
Begin;
Soma := ((Num < Limite) mod 2 > 0);
Writeln('The sum of odd numbers from 0 to ', Limite, ' é ', Soma);
End;
if (Limite mod 2 = 0) then
Begin;
Soma := ((Num < Limite) mod 2 = 0);
Writeln('The sum of odd numbers from 0 to ', Limite, ' é ', Soma);
End;
End.
*PS: Been writing the code with variables in Portuguese, so don't mind the variables appearing weird to understand. *
I see that everyone is happily looping, but this is not necessary. This is a simple arithmetic sequence, and the sum can be calculated without a loop.
Just think of the following:
1 + 3 = 2 * (1 + 3) / 2 = 2 * 2 = 4 ; limits 3 and 4
1 + 3 + 5 = 3 * (1 + 5) / 2 = 3 * 3 = 9 ; limits 5 and 6
1 + 3 + 5 + 7 = 4 * (1 + 7) / 2 = 4 * 4 = 16 ; limits 7 and 8
1 + 3 + 5 + 7 + 9 = 5 * (1 + 9) / 2 = 5 * 5 = 25 ; limits 9 and 10
1 + 3 + 5 + 7 + 9 + 11 = 6 * (1 + 11) / 2 = 6 * 6 = 36 ; limits 11 and 12
But not only that, you'll see that it is in fact always a perfect square: Sqr((n+1) div 2).
So just calculate:
program odd_numbers;
var
Num, Limite, Soma: Integer;
begin
Write('Choose a limit: ');
Readln(Limite);
Num := (Limite + 1) div 2;
Soma := Num * Num;
Writeln('The sum of odd numbers from 0 to ', Limite, ' is ', Soma);
end.
Looks a little simpler than what the others propose.
The loop While (Limite / 2 > 0) do ... uses real arithmetic and not integer arithmetic. I guess you mean While (Limite div 2 > 0) do ... And you should change Limite in the loop otherwise you get stuck because the exit condition can never be reached.
After you have asked the user to enter a number, Limite, you need to keep that unchanged, because you need it in the final message. You also need a loop where you go through all numbers from Limite towards 0.
You started with a while loop which is ok, you are just missing the loop control variable. That is a variable that eventually gets a terminating value which then stops the loop. Use for example the Num variable you already have declared. You can use the same variable to investigate the numbers between user input and 0, for being odd values.
num := limite-1; // give num a start value based on user input (-1 because of "... numbers below (x)")
while num > 0 do // stop the loop when 0 is reached
begin
// here you investigate if `num` is a odd number (e.g. using `mod` operator or
// possibly your pascal has a built in `function Odd(value: integer): boolean;`)
// and add it to `Soma` if it is
num := num - 1;// decrement num at every iteration
end;
Finally you need to consider changes to the above, to handle negative input from the user.
To test if an integer is an odd value, you could use following function:
function IsOdd( value : Integer) : Boolean;
begin
IsOdd := (value mod 2) <> 0;
end;
Many pascal compilers have a built-in function called Odd(), which you could use.
A while loop works well to solve this problem. If you start with lowest odd number above zero, i.e. one and continue upwards so long we do not exceed the limit value we have a simple start:
function GetOddSumBelowX( X : Integer) : Integer;
var
i,sum: Integer;
begin
i := 1; // Start with first odd number
sum := 0;
while (i < X) do begin // as long as i less than X, loop
if IsOdd(i) then begin
sum := sum + i; // add to sum
end;
i := i + 1; // Increment i
end;
GetOddSumBelowX := sum;
end;
Now, that was simple enough. Next step to simplify the loop is to increment the i variable by two instead, just to jump between all odd numbers:
function GetOddSumBelowX( X : Integer) : Integer;
var
i,sum: Integer;
begin
i := 1; // Start with first odd number
sum := 0;
while (i < X) do begin // as long as i less than X, loop
sum := sum + i; // add to sum
i := i + 2; // Increment to next odd number
end;
GetOddSumBelowX := sum;
end;
i was trying to make something like this
when input : 5
it will print
A B C D E
input : 10
print
A B C D E
J I H G F
input : 15
print
A B C D E
J I H G F
K L M N O
input : 20
A B C D E
J I H G F
K L M N O
T S R Q P
and so on...
here is my code i create
declare
angka number := '&Angka';
i number := trunc(angka/5);
p number := 65;
a number := 1;
b number := 1;
begin
while a <= b loop
if mod(i,2) = 1 then
a := 5;
for b in 1..5 loop
p := p + a
dbms_output.put( chr(p) || ' ' );
a := a - 1;
end loop;
p := p + 5;
else
a := 1;
for b in 1..5 loop
p := p + a
dbms_output.put( chr(p) || ' ' );
a := a + 1;
end loop;
end loop;
dbms_output.put_line(' ');
end;
/
but i was still confused it's still didn't work
and about dbms_output.put_line vs dbms_output.put can someone explain this ? because i was trying print using dbms_output.put it's didn't show.. i don't know why
Thanks
Firstly, the line p := p + a has not been terminated by semi-colon. Ideally, the PL/SQL anonymous block shouldn't compile at first place.
Secondly, with PUT procedure, you haven't completed the line yet. It needs GET_LINES to retrieve an array of lines from the buffer.
There was a similar question, Is dbms_output.put() being buffered differently from dbms_output.put_line()?
You have some problems in your code. I don't believe that you can execute exactly this code. Propably, you forgot to copy some parts of it.
First of all, syntax errors:
declare
angka number := '&Angka';
i number := trunc(angka/5);
p number := 65;
a number := 1;
b number := 1;
begin
while a <= b loop
if mod(i,2) = 1 then
a := 5;
for b in 1..5 loop
p := p + a -- ";" missed
dbms_output.put( chr(p) || ' ' );
a := a - 1;
end loop;
p := p + 5;
else
a := 1;
for b in 1..5 loop
p := p + a -- ";" missed
dbms_output.put( chr(p) || ' ' );
a := a + 1;
end loop;
-- here you missed "end if"
end loop;
dbms_output.put_line(' ');
end;
/
Also you don't need your outer loop ("while a <= b loop"), because its condition always is true and code execution will never ends. And last - when you declare
for b in 1..5 loop
oracle creates here new variable with name "b", and inside the loop previously declared b is not visible. Try to execute this:
declare
b number := 111;
begin
for b in 1..5 loop
dbms_output.put_line(b);
end loop;
dbms_output.put_line(b);
end;
/
You will get:
1
2
3
4
5
111
If you correct these errors, your code will work as you want.
I need a code to go from decimal to binary numbers, but my program shows them invert, example: needs to show 1011000 but it brings out 0001101.
+ I cant use massives and array in this program.
var
x,y,i:longint;
BEGIN
readln(y);
repeat
x:= y mod 2;
y:= y div 2;
write(x);
until y = 0;
END.
I think you can use recursion function. For example:
procedure dec2bin(y)
BEGIN
x := y mod 2;
y := y mod 2;
if y > 1 then
dec2bin(y)
end
write(x)
END
BEGIN
readln(y);
dec2bin(y)
END.
I'm not sure in correct syntax because I working with Pascal long time ago. But I think you can understand my idea and make this.
This should answer what you are looking for
function decimalToBinary(a:LongInt):String;
var d:Integer;
str:String;
Begin
str:='';
while a>0 do begin
d:=a mod 2;
str:=concat(IntToStr(d),str);
a:=a div 2;
end;
decimalToBinary:=str;
End;
This is my code. It works!
program convert;
var
number : integer;
procedure dec2bin(x : integer);
begin
// general case
if (x > 1) then dec2bin(x div 2);
// to print the result
if (x mod 2 = 0) then write('0')
else write('1');
end;
begin
write('Decimal: '); readln(number);
write('Binary : ');
dec2bin(number);
end.