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.
Related
I have my bubble sorting algorithm which works correctly but I want to set it up so it prints each line in the process of the final output(19 lines).I have tried almost everything, but it doesn't print correctly:
program Bubble_Sort;
const N = 20;
var
d : array[1..N] of integer;
var
i,j,x : integer;
begin
randomize;
for i := 1 to N do d[i] := random(100);
writeln('Before sorting:'); writeln;
for i := 1 to N do write(d[i], ' ');
writeln;
for j := 1 to N - 1 do
for i := 1 to N - 1 do
write(d[i], ' ');
if d[i] > d[i+1] then
begin
x := d[i]; d[i] := d[i+1]; d[i+1] := x;
end;
writeln('After sorting:'); writeln;
for i := 1 to N do write(d[i], ' ');
writeln;
end.
The outer loop in the center of your code, the for j ... loop runs for each bubble iteration. That is where you want to output the state of the sorting. Because you thus have more than one statement within that for j ... loop, you must also add a begin .. end pair:
for j := 1 to N - 1 do
begin
//one round of sorting
//display result so far
end;
The sorting is ok as you have it, except when you added the write(d[i], ' '); presumably to output the sort result for one iteration, you changed the execution order to become totally wrong.
Remove the write(d[i], ' '); from where it is now.
To display the sorting result after each iteration add a new for k ... loop and a writeln;
for k := 1 to N do
write(d[k], ' ');
writeln;
Final sorting and progress display should be structured like:
for j := 1 to N - 1 do
begin
for i := 1 to N - 1 do
// one round of sorting
for k := 1 to N - 1 do
// output result of one sorting round
end;
Cant obtain a correct answer it sums incorectly and multiplicates too, i must use repeat when solving this problem, i suppose the problem is in sum:=i+i but i dont know how to solve it
program SAD;
uses crt;
var a, i, sum, prod: integer;
begin
clrscr;
sum:=0;
prod:=0;
{Sum}
repeat
for i:=1 to 26 do
if i mod 4 = 0 then sum:=i+i;
until i = 26;
{Multiplication}
repeat
for a:=1 to 26 do
if a mod 4 = 0 then prod:=a*a;
until a = 26;
writeln('Suma numerelor divizate la 4 este:', sum);
writeln('Produsul numerelor divizate la 4 este:', prod);
end.
I think the instruction "use repeat" probably intends that you should avoid using for as well.
There are a few errors in your code:
In the sum loop, you should add i to sum, not to itself.
In the prod loop, since you set prod to zero at the start, it will stay as zero because zero times anything is zero. So you need to adjust the logic of your prod calculation so that if prod is zero, when the mod 4 condition is satisfied, you set prod to the current value of a, otherwise you multiply it by a.
Here is some code which fixes the above points and avoids the use of for.
program Sad;
uses crt;
var
a, i, sum, prod: integer;
begin
clrscr;
sum:=0;
prod:=0;
{Sum}
i := 0;
repeat
inc(i);
if (i mod 4) = 0 then
sum := sum + i;
until i = 26;
{Multiplication}
a :=0;
repeat
inc(a);
if a mod 4 = 0 then begin
if prod = 0 then
prod := a
else
prod := prod * a;
end;
until a = 26;
writeln('Suma numerelor divizate la 4 este:', sum);
writeln('Produsul numerelor divizate la 4 este:', prod);
readln;
end.
Part of the program I have checks if an input number is a perfect number. We're supposed to find a solution that runs in O(sqrt(n)). The rest of my program runs in constant time, but this function is holding me back.
function Perfect(x: integer): boolean;
var
i: integer;
sum: integer=0;
begin
for i := 1 to x-1 do
if (x mod i = 0) then
sum := sum + i;
if sum = x then
exit(true)
else
exit(false);
end;
This runs in O(n) time, and I need to cut it down to O(sqrt(n)) time.
These are the options I've come up with:
(1) Find a way to make the for loop go from 1 to sqrt(x)...
(2) Find a way to check for a perfect number that doesn't use a for loop...
Any suggestions? I appreciate any hints, tips, instruction, etc. :)
You need to iterate the cycle not for i := 1 to x-1 but for i := 2 to trunc(sqrt(x)).
The highest integer divisor is x but we do not take it in into account when looking for perfect numbers. We increment sum by 1 instead (or initialize it with 1 - not 0).
The code if (x mod i = 0) then sum := sum + i; for this purpose can be converted to:
if (x mod i = 0) then
begin
sum := sum + i;
sum := sum + (x div i);
end;
And so we get the following code:
function Perfect(x: integer): boolean;
var
i: integer;
sum: integer = 1;
sqrtx: integer;
begin
sqrtx := trunc(sqrt(x));
i := 2;
while i <= sqrtx do
begin
if (x mod i = 0) then
begin
sum := sum + i;
sum := sum + (x div i) // you can also compare i and x div i
//to avoid adding the same number twice
//for example when x = 4 both 2 and 4 div 2 will be added
end;
inc(i);
end;
if sum = x then
exit(true)
else
exit(false);
end;
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)
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.