I'm coding a little program but it won't print out the right math answer, it just stays at 993.
The Code:
program _Gul_;
uses crt;
var a: integer;
var b: integer;
begin
writeln('1000 - 7?');
a := 1000;
b := a - 7;
while a > 0 do
begin
writeln (a - 7, ' - 7?');
delay(120);
a := a - 7;
writeln (b)
if a = 6 then
break;
end;
writeln('я гуль')
end
I don't quite know why it is not working. I defined "b" and made a command that prints it out and the output is just:
You never update the value in b. In point of fact, b is not necessary to your program at all. Your printing strategy is also more complicated than it needs to be. Print a minus 7, then do the subtraction and print it. This prevents the program telling you the rest of 6 - 7 is 6.
program _Gul_;
uses
crt;
var
a: integer;
begin
a := 1000;
while a > 0 do
begin
writeln (a, ' - 7?');
delay(120);
a := a - 7;
writeln (a);
if a = 6 then
break;
end;
writeln('я гуль')
end.
Related
I have a program that reads a N number of integers and push them into an array, then I need to multiply the odd numbers from the array.
Program p2;
type
tab = array[1..10] of integer;
var
a, c : tab;
n, i, prod : integer;
begin
writeln('n=');
readln(n);
prod := 1;
writeln('Enter array numbers:');
for i := 1 to n do
read(a[i]);
if (a[i] mod 2 = 1) then
prod := a[i] * prod;
writeln('The produs of the odd numbers is: ',prod);
end.
For example when you enter the n as 5, and the numbers: 1, 2, 3, 4, 5;
The result of multiplication of odd numbers should be 15. Can someone help me to fix it working properly.
First off, whitespace and indentation are not terribly significant in Pascal, but they can be your ally in understanding what your program is doing, so let's make your code easier to read.
Program p2;
type
tab = array[1..10] of integer;
var
a, c : tab;
n, i, prod : integer;
begin
writeln('n=');
readln(n);
prod := 1;
writeln('Enter array numbers:');
for i := 1 to n do
read(a[i]);
if a[i] mod 2 = 1 then
prod := a[i] * prod;
writeln('The produs of the odd numbers is: ', prod);
end.
With indentation, it should be apparent what's happening. Your conditional only runs once, rather than each time through your loop. As suggested in the comments, begin and end neatly solve this problem by creating a block where the conditional is checked each time the loop runs.
program p2;
type
tab = array[1 .. 10] of integer;
var
a, c : tab;
n, i, prod : integer;
begin
writeln('n=');
readln(n);
prod := 1;
writeln('Enter array numbers:');
for i := 1 to n do
begin
read(a[i]);
if a[i] mod 2 = 1 then
prod := a[i] * prod;
end;
writeln('The produs of the odd numbers is: ', prod);
end.
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.
My question is how can I read some number of integers that user enters on standard input, and place them in array.However I don't how many numbers user will enter and i can't ask him that? User enters numbers in one line.
Okay i have just one more answer i would like to add.Thanks all for your help this is code written based on suggestions.I added a line for writting array backwards just for you can see that it has readed well.
program backo;
var niz:array [1..100] of integer;
n, i:integer;
begin
i:=1;
writeln('enter elements of array');
read(niz[i]);
while not eoln do
begin
i:=i+1;
read(niz[i]);
end;
for n:=i downto 1 do
writeln(niz[n]);
end.
Ok, based on comments there is three ways demonstrated:
program readmultiint;
{$mode objfpc}{$H+}
uses
StrUtils;
const
CMaxValues = 3;
var
s: string;
darr: array of Integer;
sarr: array [0..CMaxValues-1] of Integer;
i, cnt: Integer;
begin
// Dynamic array using WordCount
Writeln('Enter values:');
Readln(s);
cnt := WordCount(s, StdWordDelims);
SetLength(darr, cnt); // Allocate room for values
for i := 0 to cnt - 1 do
Val(ExtractWord(i + 1, s, StdWordDelims), darr[i]);
for i in darr do
Writeln(i);
// Dynamic array usin EOLN
SetLength(darr, 0);
Writeln('Enter values:');
while not eoln do
begin
SetLength(darr, Length(darr) + 1); // Expand array for next value
Read(darr[High(darr)]);
end;
Readln; // Read <Enter> itself
for i in darr do
Writeln(i);
// Static array
cnt := 0;
Writeln('Enter values:');
while (not eoln) and (cnt < CMaxValues) do // Reads not more then CMaxValues values
begin
Read(sarr[cnt]);
Inc(cnt);
end;
Readln; // Read <Enter> itself
for i := 0 to cnt-1 do
Writeln(sarr[i]);
end.
Feel free to use one of them or provide your own :)
PS: Some readings:
Dynamic arrays
Val procedure
for-in loop
You play a game with 100 opponents. The game has k rounds. Every round you can eliminate some opponents (always atleast 1). You are rewarded for eliminating them.
The reward is: 100.000 * '# of eliminated opponents' / '# of opponents' <= in integers (rounded down)
I want to eliminate the opponents in a way, that gets me the largest amount of money possible.
Example game:
number of rounds = 3
first round we eliminate 50 opponents, so we get 100.000 * 50 / 100 = +50.000
second round we eliminate 30, so we get 100.000 * 30 / 50 = +60.000
last round we eliminate last 20 opponents, so we get 100.000 * 20 / 20 = +100.000
so the total winnings are: 210.000
I tried to write up something, but I don't think it's the most effective way to do it?
Program EliminationGame;
var
selectedHistory : array [1..10] of integer;
opponentCount,roundCount : integer;
maxOpponents,numberSelected : integer;
totalMoney : integer;
i : integer;
begin
totalMoney := 0;
maxOpponents := 100;
opponentCount := maxOpponents;
roundCount := 3; {test value}
for i:=1 to roundCount do begin
if (i = roundCount) then begin
numberSelected := opponentCount;
end else begin
numberSelected := floor(opponentCount / roundCount);
end;
selectedHistory[i] := numberSelected;
totalMoney := floor(totalMoney + (numberSelected / opponentCount * 100000));
opponentCount := opponentCount - numberSelected;
end;
writeln('Total money won:');
writeln(totalMoney);
writeln('Amount selected in rounds:');
for i:= 0 to Length(selectedHistory) do
write(selectedHistory[i],' ');
end.
Also it seems that floor function does not exist in pascal?
It seems the question has a maths answer that can be calculated in advance. As #Anton said it was obvious that the number of points given during the third round did not depend upon the number of eliminated enemies. So the third round should eliminate 1 enemy.
So We get the following function for a thre-round game.
f(x)=100000x/100+100000(99-x)/(100-x)+100000*1/1, where x- the number
of enemies eleminated at first round.
if we find the extrema (local maximum of the function) it appears equal to 90. That means the decision is the following: the first round eliminates 90 the second - 9, the third - 1 enemy.
Of course, for consideration: 90=100-sqrt(100).
In other words: the Pascal decision of the task is to loop a variable from 1 to 99 and see the maximum of this function. X-will be the answer.
program Project1;
var
x, xmax: byte;
MaxRes, tmp: real;
begin
xmax := 0;
MaxRes := 0;
for x := 1 to 99 do
begin
tmp := 100000 * x / 100 + 100000*(99 - x) / (100 - x) + 100000 * 1 / 1;
if tmp > MaxRes then
begin
MaxRes := tmp;
xmax := x;
end;
end;
writeln(xmax);
readln;
end.
The general decision for other number of enemies and rounds (using recursion) is the following (Delphi dialect):
program Project1;
{$APPTYPE CONSOLE}
{$R *.res}
Uses System.SysUtils;
var
s: string;
function Part(RemainingEnemies: byte; Depth: byte;
var OutputString: string): real;
var
i: byte;
tmp, MaxRes: real;
imax: byte;
DaughterString: string;
begin
OutputString := '';
if Depth = 0 then
exit(0);
imax := 0;
MaxRes := 0;
for i := 1 to RemainingEnemies - Depth + 1 do
begin
tmp := i / RemainingEnemies * 100000 + Part(RemainingEnemies - i, Depth - 1,
DaughterString);
if tmp > MaxRes then
begin
MaxRes := tmp;
imax := i;
OutputString := inttostr(imax) + ' ' + DaughterString;
end;
end;
result := MaxRes;
end;
begin
writeln(Part(100, 3, s):10:1);//first parameter-Enemies count,
//2-Number of rounds,
//3-output for eliminated enemies counter
writeln(s);
readln;
end.
This problem can be solved with a dynamic approach.
F(round,number_of_opponents_remained):
res = 0
opp // number_of_opponents_remained
for i in [1 opp]
res = max(res, opp/100 + F(round-1,opp - i) )
return res
I should say this not the complete solution and you add some details about it, and I am just giving you an idea. You should add some details such as base case and checking if opp>0 and some other details. The complexity of this algorithm is O(100*k).