I have a little problem. I have written a program which asks for user for a code which contains 11 digits. I defined it as string but now I would like to use every digit from this code individually and make an equation.
for example if code is 37605030299 i need to do equation:
(1*3 + 2*7 + 3*6 + 4*0 + 5*5 + 6*0 + 7*3 + 8*0 + 9*2 + 1*9) / 11
and find out what's the MOD.
This is a calculation for an ISBN check digit.
Use a loop instead. (I'm only showing the total value and check digit calculation - you need to get the user input first into a variable named UserISBN yourself.)
function AddCheckDigit(const UserISBN: string): string;
var
i, Sum: Integer;
CheckDigit: Integer;
LastCharValue: string;
begin
Assert(Length(UserISBN) = 10, 'Invalid ISBN number.');
Sum := 0;
for i := 1 to 10 do
Sum := Sum + (Ord(UserISBN[i]) * i);
{ Calculate the check digit }
CheckDigit := 11 - (Sum mod 11);
{ Determine check digit character value }
if CheckDigit = 10 then
LastCharValue := 'X'
else
LastCharValue := IntToStr(CheckDigit);
{ Add to string for full ISBN Number }
Result := UserISBN + LastCharValue;
end;
Related
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;
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;
For an app I'm working on, I need to process an array of numbers and return a new array such that the sum of the elements are as close as possible to a target sum. This is similar to the coin-counting problem, with two differences:
Each element of the new array has to come from the input array (i.e. no repetition/duplication)
The algorithm should stop when it finds an array whose sum falls within X of the target number (e.g., given [10, 12, 15, 23, 26], a target of 35, and a sigma of 5, a result of [10, 12, 15] (sum 37) is OK but a result of [15, 26] (sum 41) is not.
I was considering the following algorithm (in pseudocode) but I doubt that this is the best way to do it.
function (array, goal, sigma)
var A = []
for each element E in array
if (E + (sum of rest of A) < goal +/- sigma)
A.push(E)
return A
For what it's worth, the language I'm using is Javascript. Any advice is much appreciated!
This is not intended as the best answer possible, just maybe something that will work well enough. All remarks/input is welcome.
Also, this is taking into mind the answers from the comments, that the input is length of songs (usually 100 - 600), the length of the input array is between 5 to 50 and the goal is anywhere between 100 to 7200.
The idea:
Start with finding the average value of the input, and then work out a guess on the number of input values you're going to need. Lets say that comes out x.
Order your input.
Take the first x-1 values and substitute the smallest one with the any other to get to your goal (somewhere in the range). If none exist, find a number so you're still lower than the goal.
Repeat step #3 using backtracking or something like that. Maybe limit the number of trials you're gonna spend there.
x++ and go back to step #3.
I would use some kind of divide and conquer and a recursive implementation. Here is a prototype in Smalltalk
SequenceableCollection>>subsetOfSum: s plusOrMinus: d
"check if a singleton matches"
self do: [:v | (v between: s - d and: s + d) ifTrue: [^{v}]].
"nope, engage recursion with a smaller collection"
self keysAndValuesDo: [:i :v |
| sub |
sub := (self copyWithoutIndex: i) subsetOfSum: s-v plusOrMinus: d.
sub isNil ifFalse: [^sub copyWith: v]].
"none found"
^nil
Using like this:
#(10 12 15 23 26) subsetOfSum: 62 plusOrMinus: 3.
gives:
#(23 15 12 10)
With limited input this problem is good candidate for dynamic programming with time complexity O((Sum + Sigma) * ArrayLength)
Delphi code:
function FindCombination(const A: array of Integer; Sum, Sigma: Integer): string;
var
Sums: array of Integer;
Value, idx: Integer;
begin
Result := '';
SetLength(Sums, Sum + Sigma + 1); //zero-initialized array
Sums[0] := 1; //just non-zero
for Value in A do begin
idx := Sum + Sigma;
while idx >= Value do begin
if Sums[idx - Value] <> 0 then begin //(idx-Value) sum can be formed from array]
Sums[idx] := Value; //value is included in this sum
if idx >= Sum - Sigma then begin //bingo!
while idx > 0 do begin //unwind and extract all values for this sum
Result := Result + IntToStr(Sums[idx]) + ' ';
idx := idx - Sums[idx];
end;
Exit;
end;
end;
Dec(idx); //idx--
end;
end;
end;
Here's one commented algorithm in JavaScript:
var arr = [9, 12, 20, 23, 26];
var target = 35;
var sigma = 5;
var n = arr.length;
// sort the numbers in ascending order
arr.sort(function(a,b){return a-b;});
// initialize the recursion
var stack = [[0,0,[]]];
while (stack[0] !== undefined){
var params = stack.pop();
var i = params[0]; // index
var s = params[1]; // sum so far
var r = params[2]; // accumulating list of numbers
// if the sum is within range, output sum
if (s >= target - sigma && s <= target + sigma){
console.log(r);
break;
// since the numbers are sorted, if the current
// number makes the sum too large, abandon this thread
} else if (s + arr[i] > target + sigma){
continue;
}
// there are still enough numbers left to skip this one
if (i < n - 1){
stack.push([i + 1,s,r]);
}
// there are still enough numbers left to add this one
if (i < n){
_r = r.slice();
_r.push(arr[i]);
stack.push([i + 1,s + arr[i],_r]);
}
}
/* [9,23] */
I would like to ask if anybody can give a hand in solving the following issue: How should I use the random function in Pascal in order to generate a random combination of digits that are already initialized (I mean that I have given values to four variables and I want via the random function to create a random combination of these four digits).
Thanks in advance!
Rossi
var digits : array[0..3] of integer = (10,20,30,40);
i : integer;
begin
Randomize; // initialize the random generator. Only once per program
for i:=0 to 50 do
Writeln(digits[random(4)]);
end.
The Writeln line draws a number 0<=x<4 so 0..3, and looks it up in the digits array, then writes it to console output. It is draws 50 random numbers and then quits.
var
randomnumber,i:integer;
number:array[0..3] of integer;
begin
randomize;
for i:= 0 to 3 do
begin
readln(number[i]);
end;
randomnumber:= (number[random(4)] * 1000) + (number[random(4)] * 100) + (number[random(4)] * 10) + (number[random(4)] * 1);
writeln(randomnumber);
end.
I hope this could help.
But the given initial value should be between 0 to 9.
If you want that the output contains each digit only once, then you would need to stored the digits which have already been chosen in a set to prevent them from being chosen again.
const
digits: array [0..3] of integer = (1, 3, 5, 7);
var
i, n, total: integer;
stored: set of integer;
begin
Randomize;
stored:= [];
total:= 0;
for i:= 1 to 4 do
begin
repeat
n:= random (4);
until not (n in stored);
stored:= stored + [n];
total:= total * 10 + digits[n];
end;
writeln (total)
end.
Example :
a = 1
b = 2
c = 3
..
..
z = 26
aa = 27
ab = 28
how to convert another string into an integer? for example i want to convert 'lmao' to an integer. please help me :) thank you.
in pascal :)
To convert ordinary base-10 strings into numbers, you take each character from left to right, convert it to its numeric value (between 0 and 9) and add it to the total you already have (which you initialize to zero). If there are more characters following the one you just processed, then multiply the total by 10. Repeat until you run out of characters.
For example, the number 374 is 3×102 + 7×101 + 4×100. Another way of writing that, which more closely models the conversion algorithm I described above, is (((3)×10+7)×10+4.
You can adapt that to handle any string of characters, not just numeric characters. Instead of 10, the base is 26, so multiply by that. And instead of digits, the characters are a through z. Your example string would be evaluated like this: (((l)×26+m)×26+a)×26+o. Substitute numbers for those letters, and you get 219,742.
Here's some code to do it. It doesn't check for errors; it assumes that the string will only contain valid characters and that the string won't represent a number that's too big to fit in an Integer variable.
function SpecialStrToInt(const s: string): Integer;
var
i: Integer;
subtotal: Integer;
c: Char;
charval: Integer;
begin
subtotal := 0;
for i := 1 to Length(s) do begin
c := s[i];
charval := Ord(c) - Ord('a') + 1;
subtotal := subtotal * 26;
subtotal := subtotal + charval;
end;
SpecialStrToInt := subtotal;
end;
An oddity about your format is that there's no way to represent zero.