I have another question as this code does not work in the compilers... sorry for the many questions, I think other questions will follow:
program findMaximum(input,output);
var
number, Maximum: integer;
begin
writeln('Insert a number and finish up with 0').
while number < >0 do
begin
if number> Maximum then
Maximum := number;
readln(number)
end;
if Maximum <> 0 then
writeln('The biggest number is',Maximum)
else
writeln('The input was zero.')
end.
var
number, Maximum: integer;
begin
maximum:=0;
writeln('Insert a number and finish up with 0');
while number < >0 do
begin
readln(number);
if number> Maximum then Maximum := number;
end;
if Maximum <> 0 then
writeln('The biggest number is',Maximum)
else
writeln('The input was zero.')
end.
Related
I have a sample number in a column of oracle table which is binary's 2's complimanet -
e.g 0110001000110111
I want to convert this to normal decimal number in 2's compliment.
Expected output-
Reference link - https://www.rapidtables.com/convert/number/decimal-to-binary.html
You can loop as applying powers of 2 while multiplying by each bit(0 or 1) starting from the right end of your presented value such as
SET SERVEROUTPUT ON
DECLARE
bin_nr VARCHAR2(100) := '0110001000110';
dec_nr NUMBER;
BEGIN
FOR i IN 1..LENGTH(bin_nr)
LOOP
dec_nr := NVL(dec_nr,0) + SUBSTR(bin_nr,-i,1)*(2**(i-1));
END LOOP;
DBMS_OUTPUT.PUT_LINE(dec_nr);
END;
/
which results 3142 as the decimal value.
Demo
For 2s compliment, the most-significant bit represents the sign bit and if that is 1 then you have a negative number:
DECLARE
bin_nr VARCHAR2(100) := '1111001110111010';
sign PLS_INTEGER;
dec_nr PLS_INTEGER;
BEGIN
IF SUBSTR(bin_nr, 1, 1) = '1' THEN
bin_nr := TRANSLATE(bin_nr, '01', '10');
sign := -1;
dec_nr := 1;
ELSE
sign := 1;
dec_nr := 0;
END IF;
FOR i IN 1 .. LENGTH(bin_nr) LOOP
IF SUBSTR(bin_nr, -i, 1) = '1' THEN
dec_nr := dec_nr + POWER(2, i-1);
END IF;
END LOOP;
dec_nr := dec_nr * sign;
DBMS_OUTPUT.PUT_LINE(dec_nr);
END;
/
Outputs -3142
If you are expecting an N-bit binary number as the input (for example, the link in the question expects a 16-bit binary number as an input for 2s compliment) then you should LPAD with zeroes if you have fewer than that many bits.
db<>fiddle here
program Loans(Input,Output);
uses crt;
Var
i,m,l:Integer;
Month:Integer;
LoanID:String;
QUARTERS:array[1..4] of Integer;
max:integer;
min:integer;
maxloc:integer;
minloc:integer;
begin
clrscr;
For i:=1 to 4 do
QUARTERS[m]:=0;
For l:=1 to 8 do
begin
Writeln('Enter LoanID and Month');
Readln(LoanID);
Readln(Month);
if Month <=3 then
QUARTERS[1]:=QUARTERS[1]+1
else if Month <=6 then QUARTERS[2]:=QUARTERS[2]+1
else if Month <=9 then QUARTERS[3]:=QUARTERS[3]+1
else QUARTERS[4]:=QUARTERS[4]+1
end;
max:= 0;
min:= 999;
for m:= 1 to 4 do
begin
if QUARTERS[m]>max then
max:=QUARTERS[m];
maxloc:=m;
end;
begin
if QUARTERS[m]<min then
min:=QUARTERS[m];
minloc:=m;
end;
for m:=1 to 4 do
Writeln('Quarter ',m,' has ',QUARTERS[m],' loans');
Writeln('The maximum number of loans taken is ', max);
Writeln('The maximum number of loans were taken in quarter ', maxloc);
Writeln('The minimum number of loans taken is ', min);
Writeln('The minimum number of loans were taken in quarter ', minloc);`enter code here`
end.
You definitely must format the code properly - then you'll see your mistakes with begin/end
for m:= 1 to 4 do
begin
if QUARTERS[m]>max then begin
max:=QUARTERS[m];
maxloc:=m; //update maxloc here!!!
end; //if
if QUARTERS[m]<min then
begin
min:=QUARTERS[m];
minloc:=m;
end;//if
end;//for m
and you can change all your quarters code
if Month <=3 then
QUARTERS[1]:=QUARTERS[1]+1
else if Month <=6 then QUARTERS[2]:=QUARTERS[2]+1
else if Month <=9 then QUARTERS[3]:=QUARTERS[3]+1
else QUARTERS[4]:=QUARTERS[4]+1
end;
to single line:
Inc(QUARTERS[1 + (Month - 1) div 4]);
The following pl/sql program generates an error on execution on line the sum :=temp*sum; encountered symbol ; when expecting ( . Please explain my mistake.
declare
n number;
temp number;
sum number := 1;
begin
n := &n;
temp := n;
while temp>0 loop
sum := temp*sum;
temp := temp-1;
end loop;
dbms_output.put_line('Factorial of '||n||' is '||sum);
end;
/
Maybe not the answer to your question, but there is no need for PL/SQL here:
select round(exp(sum(ln(level))))
from dual
connect by level <= 5;
where 5 is your number (5!).
Additionally, if you like to operate faster in PL/SQL use pls_integer instead of number.
UPDATE
So according to comments I felt free to test:
create or replace package test_ is
function by_query(num number) return number deterministic;
function by_plsql(num number) return number deterministic;
end test_;
/
create or replace package body test_ is
function by_query(num number) return number deterministic
is
res number;
begin
select round(exp(sum(ln(level))))
into res
from dual
connect by level <= num;
return res;
end;
function by_plsql(num number) return number deterministic
is
n number := 0;
begin
for i in 1..num loop
n := n + ln(i);
end loop;
return round(exp(n));
end;
end test_;
So there are two functions with different content. Test query:
declare
dummy number;
begin
for i in 1..10000 loop
dummy := test_.by_query(5);
end loop;
end;
0.094 sec.
declare
dummy number;
begin
for i in 1..10000 loop
dummy := test_.by_plsql(5);
end loop;
end;
0.094 sec.
You'll say I am cheater and using deterministic keyword but here it is obvious and is needed by logic. If I remove it, the same scripts are working 1.7 sec vs 1.3 sec, so procedure is only a bit faster, there is no even double-win in performance. The totally opposite effect you will get if you use the function in a query so it is a fair trade.
Sum is reserved word in sql. Change variable name like
declare
n number;
temp number;
sum_ number := 1;
begin
n := &n;
temp := n;
while temp>0 loop
sum_ := temp*sum_;
temp := temp-1;
end loop;
dbms_output.put_line('Factorial of '||n||' is '||sum_);
end;
/
declare
n number;
i number;
sum_of_log_10s number;
exponent number;
base number;
begin
n := &n;
i := 1;
sum_of_log_10s := 0;
while i <= n loop
-- do stuff
sum_of_log_10s := sum_of_log_10s + log(10,i);
i := i + 1;
end loop;
dbms_output.put_line('sum of logs = '||sum_of_log_10s);
exponent := floor(sum_of_log_10s);
base := power(10,sum_of_log_10s - exponent);
dbms_output.put_line(n||'! = '||base||' x 10^'||exponent);
end;
I came up with this code that I like even better than #smnbbrv's answer. It's a great way to check the speed of a machine. I've been using a variation of this since my Atari 800
ALTER SESSION FORCE PARALLEL DDL PARALLEL 16;
ALTER SESSION FORCE PARALLEL DML PARALLEL 16;
ALTER SESSION FORCE PARALLEL QUERY PARALLEL 16;
with t as (
select /*+materialize*/
rownum i
from dual connect by rownum < 100000 -- put number to calculate n! here
)
,t1 as (
select /*+parallel(t,16)*/ /*+materialize*/
sum(log(10,i)) logsum
from t
)
select
trunc(power(10,(mod(logsum,1))),3) ||' x 10^'||trim(to_char(floor(logsum),'999,999,999,999')) factorial
-- logsum
from t1
;
-- returns 2.824 x 10^456,568
Here is the simple code for finding factorial of number at run time...
declare
-- it gives the final answer after computation
fac number :=1;
-- given number n
-- taking input from user
n number := &1;
-- start block
begin
-- start while loop
while n > 0 loop
-- multiple with n and decrease n's value
fac:=n*fac;
--dbms_output.put(n||'*');
n:=n-1;
end loop;
-- end loop
-- print result of fac
dbms_output.put_line(fac);
-- end the begin block
end;
I need an algorithm in Delphi to generate partitions for a specified integer value.
Example: for 13 if 5 is specified as the max value for partition it will give 5,5,3; if 4 is specified as max partition value the result should be 4,4,4,1, and so on.
It's simple enough to solve the problem using div and mod. Here's an example program that I don't think needs any further explanation:
program IntegerPartitions;
{$APPTYPE CONSOLE}
function Partitions(const Total, Part: Integer): TArray<Integer>;
var
Count: Integer;
Rem: Integer;
i: Integer;
begin
Assert(Total>0);
Assert(Part>0);
Count := Total div Part;
Rem := Total mod Part;
if Rem=0 then
SetLength(Result, Count)
else
SetLength(Result, Count+1);
for i := 0 to Count-1 do
Result[i] := Part;
if Rem<>0 then
Result[Count] := Rem;
end;
var
Value: Integer;
begin
for Value in Partitions(13, 5) do
Writeln(Value);
Readln;
end.
I write this program with pascal
which ask the user to enter two arrays and constant value which is K
the program muli the K with arrays .
and then save the answer in new array
and do some operation in new array
addition << work well
Subtraction << also work
BUT the problem in Multi << I am trying to ask the user to enter a new array and do Muti but still there is a problem.
ALSO
I want these operation repeated until the user press exit <<< I could not do this options because i am not perfect with pascal .
I would be grateful if you could help me
This is My Code
program BST6;
const maxN=100;maxM=100;
type mat=array[1..maxN,1..maxM]of integer;
var A,B,c:mat;
n,m,l,s,i,j,k:integer;
ch : char;
procedure readMat(var A:mat;var m,n:integer);
begin
for i:=1 to m do
for j:=1 to n do
begin
write('mat[',i,',',j,']=');
readln(A[i,j]);
end;
end;
procedure writeMat(A:mat;m,n:integer);
begin
for i:=1 to m do
begin
for j:=1 to n do
write(a[i,j]:4);
writeln;
end;
end;
function multK(A:mat;k:integer):mat;
begin
for i:=1 to n do
for j:=1 to m do
begin
B[i,j]:= K*A[i,j];
end;
multK:=B;
end;
function minus(A,B:mat):mat;
begin
for i:=1 to m do
for j:=1 to n do
C[i,j]:=A[i,j]-B[i,j];
minus:=C;
end;
function plus(A,B:mat):mat;
begin
for i:=1 to m do
for j:=1 to n do
C[i,j]:=A[i,j]+B[i,j];
plus:=C;
end;
function mult(A,B:mat;m,l,n:integer):mat;
begin
for i:=1 to m do
for j:=1 to n do
for k:=1 to l do
c[i,j]:=c[i,j]+A[i,k]*B[k,j];
mult:=C;
end;
begin
write('input m<=',maxM,'.. m=' );readln(m);
write('input n<=',maxN,'.. n=');readln(n);
readMat(A,m,n);
writeln('input the const K');readln(k);
B:=multK(A,K);
writeln('The matrix A : ');
writeMat(A,m,n);
writeln('The matrix B=K*A : ');
writeMat(B,m,n);
writeln('choose the operation + , - or * ');
readln(ch);
case ch of
'+' : c:=plus(A,B);
'-' : c:=minus(A,B);
'*' : begin
writeln('input m<=',maxM,'input l<=',maxN);readln(m,l);readMat(A,m,l);
writeln('input l<=',maxN);readln(n);readMat(B,l,n);
c:=mult(A,B,m,l,n);
end;
end;
writeMat(c,m,n);
readln;
end.
First of all having global one letter variables which collide with function parameters with the same name is insane.
Why does multK modify the global variable B as a sideeffect?
Why does minus modify the global variable C as a sideeffect?
Why global integers as for index variables?
And mult is even worse: It doesn't only modify C as a sideeffect, but it assumes C contains meaningful values beforehand. I think it needs to initialize C to all zeros beforehand.
My guess is some of your side effects interfere in strange ways. But I don't want to think it through. Refactor your code first. In particular learn how and when to use local variables.