10
Balta B 1 15
Melyna M 2 15
Zalia Z 3 12
Raduona R 4 10
Geltona G 5 10
Violetine V 6 12
Pilka P 7 10
Oranzine O 8 12
Alyvuogiu A 9 12
Juoda J 10 10
3 5
Andrius B 4 P 7 R 4 B 1 V 6
Tomas V 6 A 9 B 6 O 8 P 2
Evelina R 4 P 7 R 4 P 7 B 1
program Spalvotos_korteles;
type Spalvos = record
SPav : string[15]; // SpalvosPavadinimas
SNr, SSk : integer; // SpalvosNumeris, SpalvosSkaicius
SI : char; // SpalvuIndeksas
end;
Mokiniai = record
V : string[15]; // Mokinio vardas
MI : char; // Mokinio istrauktas indeksas
Mnr, TR, NR : integer; // Mokinio uzrasytas numeris TeisingiRasymai, NeteisingiRasymai
end;
Mas = array[1..100] of Spalvos;
Mas1 = array[1..100] of Mokiniai;
Mas2 = array[1..100] of char;
Mas3 = array[1..100] of integer;
var n, Q, MokSk, MokT : integer;
S : Mas;
M : Mas1;
VI : Mas2; // Visi Indeksai
VNr, Istraukta, TeisingiRasymai, Nepanaudota : Mas3; // VNr - Visi Numeriai
procedure Nuskaitymas;
var df : text;
Qq, i, j, z, ii : integer;
begin
Qq:=1;
assign(df,'duom.txt');
reset(df);
readln(df, n);
for i:= 1 to n do
readln(df, S[i].SPav, S[i].SI, S[i].SNr, S[i].SSk);
readln(df, MokSk, MokT);
for j := 1 to MokSk do
begin
read(df,M[j].V);
for z := 1 to MokT do
begin
read(df, M[z].MI, M[z].Mnr);
VI[Qq] := M[z].MI;
VNr[Qq] := M[z].Mnr;
Qq:=Qq+1;
for ii := 1 to n do
if (M[z].MI = S[ii].SI) and (M[z].Mnr = S[ii].SNr) then M[j].TR := M[j].TR+1;
end;
end;
Q:=Qq-1;
close(df);
end;
procedure Uzrasymai_ant_korteliu;
var i, j : integer;
begin
for i:= 1 to n do
begin
for j:= 1 to Q do
begin
if S[i].SI = VI[j] then
Istraukta[i]:=Istraukta[i]+1;
if (S[i].SI = VI[j]) and (S[i].SNr = VNr[j]) then
TeisingiRasymai[i]:= TeisingiRasymai[i]+1;
end;
Nepanaudota[i]:= S[i].SSk - Istraukta[i];
end;
end;
procedure Rezultatas;
var i, j : integer;
Rf : text;
begin
assign(rf,'SpalvuRezultas.Txt');
rewrite(rf);
for i := 1 to MokSk do
writeln(rf,M[i].V, M[i].TR);
writeln(rf);
for j := 1 to n do
writeln(rf,S[j].SPav, Istraukta[j], TeisingiRasymai[j], Nepanaudota[j]);
close(rf);
end;
begin
Nuskaitymas;
Uzrasymai_ant_korteliu;
Rezultatas;
end.
My goal is to read the 3 last roads and look how many numbers by the symbols are correct (the correct ones are symbol & first number from the columns) but when I try to read then I get the 106error wrong numeric format. I somehow understand that the problem because of the char symbols but I have no clue how to fix it. Could someone help me?
The main idea is to find out how the program interprets your read statement in your program.
Let ch, ch1, ch2 ... be a char and int, int1, int2 ... be an integer.
INPUT:
P 1 C 2
execute read(ch, int)
RESULT:
ch = P
int = 1
It works perfectly ok.
Let's expand this to read four elements.
INPUT:
P 1 C 2
execute read(ch1, int1, ch2, int2)
RESULT:
ch1 = P
int1 = 1
ch2 = C
int2 = 2
However,
INPUT:
P 1 C 2
execute read(ch1, int1); read(ch2, int2);
RESULT:
ch1 = P
int1 = 1
ERROR 106 when execute read(ch2, int2);
Why?
Let's see the following illustration.
reading ch1
1234567
P 1 C 2
^
The first read(ch1, int1) asks the program to read a char and then a integer, separated by SPACE. The program first read a char and move the pointer to position 2 and then knows it is a SPACE so move on read the next int which move the pointer to position 3.
reading int1
1234567
P 1 C 2
^
So it reads 1 into int1. And move the pointer to the next position. Then the first read is finished.
1234567
P 1 C 2
^
What's going on? Second read will read the char on position 4 which is a SPACE. It won't give out any warning as it is really a character. After that the position of pointer should be on position 5.
1234567
P 1 C 2
^
The program will not skip position 5 as it is not a SPACE. Therefore, 'C' is read into int2. As 'C' is not an integer, the program gives out a RUNTIME ERROR 106.
It seems that readln(df, S[i].SPav, S[i].SI, S[i].SNr, S[i].SSk); and read(df,M[j].V);read(df, M[z].MI, M[z].Mnr); have the same logic, but in the execution, they do not.
Moreover, you have to go to the next line after you have read all elements on a row by readln(df);.
FYI, below is the edited code:
program post;
type
Spalvos = record
SPav: string[15]; // SpalvosPavadinimas
SNr, SSk: integer; // SpalvosNumeris, SpalvosSkaicius
SI: char; // SpalvuIndeksas
end;
Mokiniai = record
V: string[15]; // Mokinio vardas
MI: char; // Mokinio istrauktas indeksas
Mnr, TR, NR: integer;
// Mokinio uzrasytas numeris TeisingiRasymai, NeteisingiRasymai
end;
Mas = array[1..100] of Spalvos;
Mas1 = array[1..100] of Mokiniai;
Mas2 = array[1..100] of char;
Mas3 = array[1..100] of integer;
var
n, Q, MokSk, MokT: integer;
S: Mas;
M: Mas1;
VI: Mas2; // Visi Indeksai
VNr, Istraukta, TeisingiRasymai, Nepanaudota: Mas3; // VNr - Visi Numeriai
temp: char;
procedure Nuskaitymas;
var
df: Text;
Qq, i, j, z, ii: integer;
begin
Qq := 1;
Assign(df, 'duom.txt');
reset(df);
readln(df, n);
for i := 1 to n do
readln(df, S[i].SPav, S[i].SI, S[i].SNr, S[i].SSk);
readln(df, MokSk, MokT);
for j := 1 to MokSk do
begin
Read(df, M[j].V);
for z := 1 to MokT do
begin
Read(df, M[z].MI, M[z].Mnr, temp);
writeln(M[z].MI, M[z].Mnr);
VI[Qq] := M[z].MI;
VNr[Qq] := M[z].Mnr;
Qq := Qq + 1;
for ii := 1 to n do
if (M[z].MI = S[ii].SI) and (M[z].Mnr = S[ii].SNr) then
M[j].TR := M[j].TR + 1;
end;
readln(df);
end;
Q := Qq - 1;
Close(df);
end;
procedure Uzrasymai_ant_korteliu;
var
i, j: integer;
begin
for i := 1 to n do
begin
for j := 1 to Q do
begin
if S[i].SI = VI[j] then
Istraukta[i] := Istraukta[i] + 1;
if (S[i].SI = VI[j]) and (S[i].SNr = VNr[j]) then
TeisingiRasymai[i] := TeisingiRasymai[i] + 1;
end;
Nepanaudota[i] := S[i].SSk - Istraukta[i];
end;
end;
procedure Rezultatas;
var
i, j: integer;
Rf: Text;
begin
Assign(rf, 'SpalvuRezultas.Txt');
rewrite(rf);
for i := 1 to MokSk do
writeln(rf, M[i].V, M[i].TR);
writeln(rf);
for j := 1 to n do
writeln(rf, S[j].SPav, Istraukta[j], TeisingiRasymai[j], Nepanaudota[j]);
Close(rf);
end;
begin
Nuskaitymas;
Uzrasymai_ant_korteliu;
Rezultatas;
end.
Thank you for your answer, but I already found out about char reading spaces too and I fixed it. Don't know if it is the best way but it works.
readln(df, MokSk, MokT);
for j := 1 to MokSk do
begin
read(df,M[j].V);
for z := 1 to MokT-1 do
begin
read(df, MI, Mnr, Tuscias); // M[z].MI P M[z].Mnr O
VI[Qq] := MI;
VNr[Qq] := Mnr;
Qq:=Qq+1;
for ii := 1 to n do
if (MI = S[ii].SI) and (Mnr = S[ii].SNr) then M[j].TR := M[j].TR+1;
end;
read(df, MI, Mnr);
VI[Qq] := MI;
VNr[Qq] := Mnr;
Qq:=Qq+1;
for ii := 1 to n do
if (MI = S[ii].SI) and (Mnr = S[ii].SNr) then M[j].TR := M[j].TR+1;
readln(df);
end;
Andrius B 4 P 7 R 4 B 1 V 6
Tomas V 6 A 9 B 6 O 8 P 2
Evelina R 4 P 7 R 4 P 7 B 1
I don't know any other way to read this program, so I made it like this.
P.s. I tried your way before but it didn't work because after the last pair there's no space and the program makes an error out of it. But thanks for your help kind sir :)
Related
I have a task to print each number from the input alternately, firstly numbers with even indexes, then numbers with odd indexes. I have solved it, but only for one line of numbers, but I have to read n lines of numbers.
Expected input:
2
3 5 7 2
4 2 1 4 3
Expected output:
7 5 2
1 3 2 4
Where 2 is number of lines, 3 and 4 are numbers of numbers, 5, 7, 2 and 2, 1 , 4, 3 are these numbers.
Program numbers;
Uses crt;
var k,n,x,i,j: integer;
var tab : Array[1..1000] of integer;
var tab1 : Array[1..1000] of integer;
var tab2 : Array[1..1000] of integer;
begin
clrscr;
readln(k);
for i:=1 to k do
begin
read(n);
for j:=1 to n do
begin
read(tab[j]);
if(j mod 2 = 0) then
tab1[j]:=tab[j]
else
begin
tab2[j]:=tab[j];
end;
end;
end;
for j:=1 to n do
if tab1[j]<>0 then write(tab1[j], ' ');
for j:=1 to n do
if tab2[j]<>0 then write(tab2[j], ' ');
end.
Let's clean up the formatting, and use a record to keep track of each "line" of input.
program numbers;
uses
crt;
type
TLine = record
count : integer;
numbers : array[1..1000] of integer
end;
var
numLines, i, j : integer;
lines : Array[1..1000] of TLine;
begin
clrscr;
readln(numLines);
for i := 1 to numLines do
begin
read(lines[i].count);
for j := 1 to lines[i].count do
read(lines[i].numbers[j])
end
end.
We can read each line in. Now, how do we print the odd and even indices together? Well, we could do math on each index, or we could just increment by 2 instead of 1 using a while loop.
program numbers;
uses
crt;
type
TLine = record
count : integer;
numbers : array[1..1000] of integer
end;
var
numLines, i, j : integer;
lines : Array[1..1000] of TLine;
begin
clrscr;
readln(numLines);
// Read in lines.
for i := 1 to numLines do
begin
read(lines[i].count);
for j := 1 to lines[i].count do
read(lines[i].numbers[j])
end;
// Print out lines.
for i := 1 to numLines do
begin
j := 1;
while j <= lines[i].count do
begin
write(lines[i].numbers[j], ' ');
j := j + 2
end;
j := 2;
while j <= lines[i].count do
begin
write(lines[i].numbers[j], ' ');
j := j + 2
end;
writeln
end
end.
Now if we run this:
2
3 4 5 6
4 6 2 4 1
4 6 5
6 4 2 1
One thing we can note is that the following loop is the same for both odd and even indexes, except for the start index.
while j <= lines[i].count do
begin
write(lines[i].numbers[j], ' ');
j := j + 2
end;
This is a perfect place to use a procedure. Let's call it PrintEveryOther and have it take an index to start from and a line to print.
program numbers;
uses
crt;
type
TLine = record
count : integer;
numbers : array[1..1000] of integer
end;
var
numLines, i, j : integer;
lines : Array[1..1000] of TLine;
procedure PrintEveryOther(start : integer; line :TLine);
var
i : integer;
begin
i := start;
while i <= line.count do
begin
write(line.numbers[i], ' ');
i := i + 2
end
end;
begin
clrscr;
readln(numLines);
for i := 1 to numLines do
begin
read(lines[i].count);
for j := 1 to lines[i].count do
read(lines[i].numbers[j])
end;
for i := 1 to numLines do
begin
PrintEveryOther(1, lines[i]);
PrintEveryOther(2, lines[i]);
writeln
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 need an algorithm to print all possible sums of a number (partitions).
For example: for 5 I want to print:
1+1+1+1+1
1+1+1+2
1+1+3
1+2+2
1+4
2+3
5
I am writing my code in Pascal. So far I have this:
Program Partition;
Var
pole :Array [0..100] of integer;
n :integer;
{functions and procedures}
function Minimum(a, b :integer): integer;
Begin
if (a > b) then Minimum := b
else Minimum := a;
End;
procedure Rozloz(cislo, i :integer);
Var
j, soucet :integer;
Begin
soucet := 0;
if (cislo = 0) then
begin
for j := i - 1 downto 1 do
begin
soucet := soucet + pole[j];
if (soucet <> n) then
Write(pole[j], '+')
else Write(pole[j]);
end;
soucet := 0;
Writeln()
end
else
begin
for j := 1 to Minimum(cislo, pole[i - 1]) do
begin
pole[i] := j;
Rozloz(cislo - j, i + 1);
end;
end;
End;
{functions and procedures}
{Main program}
Begin
Read(n);
pole[0] := 101;
Rozloz(n, 1);
Readln;
End.
It works good but instead of output I want I get this:
1+1+1+1+1
2+1+1+1
2+2+1
3+1+1
3+2
4+1
5
I can't figure out how to print it in right way. Thank you for help
EDIT: changing for j:=i-1 downto 1 to for j:=1 to i-1 solves one problem. But my output is still this: (1+1+1+1+1) (2+1+1+1) (2+2+1) (3+1+1) (3+2) (4+1) (5) but it should be: (1+1+1+1+1) (1+1+1+2) (1+1+3) (1+2+2) (1+4) (2+3) (5) Main problem is with the 5th and the 6th element. They should be in the opposite order.
I won't attempt Pascal, but here is pseudocode for a solution that prints things in the order that you want.
procedure print_partition(partition);
print "("
print partition.join("+")
print ") "
procedure finish_and_print_all_partitions(partition, i, n):
for j in (i..(n/2)):
partition.append(j)
finish_and_print_all_partitions(partition, j, n-j)
partition.pop()
partition.append(n)
print_partition(partition)
partition.pop()
procedure print_all_partitions(n):
finish_and_print_all_partitions([], 1, n)
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.
Before i must say this : Please, excuse me for my bad english...
I'm student.My teacher gave me problem in pascal for my course work...
I must write program that calculates 2^n for big values of n...I've wrote but there is a problem...My program returns 0 for values of n that bigger than 30...My code is below...Please help me:::Thanks beforehand...
function control(a: integer): boolean;
var
b: boolean;
begin
if (a >= 10) then b := true
else b := false;
control := b;
end;
const
n = 200000000;
var
a: array[1..n] of integer;
i, j, c, t, rsayi: longint; k: string;
begin
writeln('2^n');
write('n=');
read(k);
a[1] := 1;
rsayi := 1;
val(k, t, c);
for i := 1 to t do
for j := 1 to t div 2 do
begin
a[j] := a[j] * 2;
end;
for i := 1 to t div 2 do
begin
if control(a[j]) = true then
begin
a[j + 1] := a[j + 1] + (a[j] div 10);
a[j] := a[j] mod 10;
rsayi := rsayi + 1;
end;
end;
for j := rsayi downto 1 do write(a[j]);
end.
The first (nested) loop boils down to "t" multiplications by 2 on every single element of a.
30 multiplications by two is as far as you can go with a 32-bit integer (2^31-1 of positive values, so 2^31 is out of reach)
So the first loop doesn't work, and you probably have to rethink your strategy.
Here is a quick and dirty program to compute all 2^n up to some given, possibly large, n. The program repeatedly doubles the number in array a, which is stored in base 10; with lower digit in a[1]. Notice it's not particularly fast, so it would not be wise to use it for n = 200000000.
program powers;
const
n = 2000; { largest power to compute }
m = 700; { length of array, should be at least log(2)*n }
var
a: array[1 .. m] of integer;
carry, s, p, i, j: integer;
begin
p := 1;
a[1] := 1;
for i := 1 to n do
begin
carry := 0;
for j := 1 to p do
begin
s := 2*a[j] + carry;
if s >= 10 then
begin
carry := 1;
a[j] := s - 10
end
else
begin
carry := 0;
a[j] := s
end
end;
if carry > 0 then
begin
p := p + 1;
a[p] := 1
end;
write(i, ': ');
for j := p downto 1 do
write(a[j]);
writeln
end
end.