We have 490 students
We have a capacity of 20 students per classroom
Ie we have 490/20 = 24.5 ie 25 sections
First I want to arrange students in alphabetical order
Secondly I want to make a table containing
Id_classroom and id_student attributes
My Problem is in id_classroom and how to populate it
How can I automatically give this
Students from 1 to 20 === id_classroom = 1
Students from 21 to 40 === id_classroom = 2
and so on
thank you in advance
In your SQL RESQUEST, the function FLOOR is your friend.
What it does is reduce your number to the largest integer less than or equal to the expression.
Let's say you have a student id 46.
The operation would look like this:
46-1= 45 //index starts at 1 instead of 0.
45/20=2.25
Floor(2.25)=2
2+1=3
So your SQL would have something as such in it:
id_classroom=FLOOR((id_student-1)/20)+1
Haven't tested this, I just thought something up :)
EDIT: Just noticed this was in SQL not C# forum. Not sure if still useful.
TotalStudents = 490;
MaxCap = 20;
RoomID = 0;
Dictionary<RoomID, MaxCap>() StudentsAndRooms = new Dictionary();
void PopulateRoom(Dictionary<int,int> myD)
{
myD = StudentsAndRooms;
for(int i = 0; i < TotalStudents; i++)
{
if(myD[RoomID].Value.Count <= MaxCap)
{
myD[RoomID] = myD[RoomID +1]
Debug.Log("Added Room");
}
else if(myD[RoomID].Value.Count != MaxCap)
{
myD[RoomID] +=1;
Debug.Log("Added Student");
}
else
{
Debug.Log("Something went wrong!");
}
}
void Start()
{
PopulateRoom(StudentsAndRooms);
}
Related
I can't figure out how to solve the following problem: there is a number n. Output the numbers to the console in order separated by a space, but so that the next digit in the iteration is output as many times as it is a digit, and at the same time so that there are no more than n digits in the output. Сan anyone suggest the correct algorithm?
example: have n = 7, need print (1 2 2 3 3 3 4) in kotlin
what i try:
var n = 7
var count = 1
var i = 1
for (count in 1..n) {
for (i in 1..count) {
print(count)
}
}
}
var n = 11
var count = 1
var i = 1
var size = 0
// loop# for naming a loop in kotlin and inside another loop we can break or continue from outer loop
loop# for (count in 1..n) {
for (i in 1..count) {
print(count)
size++
if (size == n){
break#loop
}
}
}
You can use "#" for naming loops and if you want to break from that loop, you can use this syntax in kotlin. It worked for me.
For kotlin labeled break you can look at this reference: link
var count = 1
var n = 7
for(count in 1..n) {
print(count.toString().repeat(count))
}
count.toString() converts an integer to a string, .repeat() function repeats count times the string.
In case you need to add a space between each number, you can add the following:
print(" ")
Using generateSequence and Collections functions:
val n = 7
println(generateSequence(1) {it + 1}
.flatMap{e -> List(e){e}}
.take(n)
.joinToString(" "))
Your example is correct, You have to put a space between the printing
you can follow the code from this link
Kotlin lang code snippet
or the following code snippet
fun main() {
var n = 7
var count = 1
var i = 1
for (count in 1..n) {
for (i in 1..count) {
print(count)
print(' ')
}
}
}
For completeness, here's another approach where you write your own sequence function which produces individual values on demand (instead of creating intermediate lists)
sequence {
var digit = 1
while (true) {
for (i in 1..digit) yield(digit)
digit++
}
}.take(7)
.joinToString(" ")
.run(::print)
Not a big deal in this situation, but good to know!
I have a 2D array that absolutely will not return the values I need. I start off with this array:
var userdata:Array = new Array(new Array(1000),new Array(4))
Then I try to set all values to 0, with this:
this.onLoad()
{
for (i = 0; i < 1000; i++)
{
for (j = 0; j < 4; j++)
{
userdata[i][j] = 0
trace(userdata[i][j])
}
}
}
This trace returns 8 0s and then a giant amount of "undefined"s. I can't figure out why this would be. I try something like this as well:
userdata[5][0] = 0
trace(userdata[5][0])
It still returns "undefined". Can anyone help with this?
To understand why you got only 8 "zeros" and many undefined values, let's start by your array declaration :
var userdata:Array = new Array(new Array(1000),new Array(4));
Here you should understand that you have created an array with only 2 cells ( that's why userdata[5][0] is undefined ) : the 1st cell is an array of 1000 elements and the 2nd one is an array of 4 elements, and that's why you can only set 8 items ( 2 x 4 ) : the 4th first items from the 1000 of the the 1st cell + the the 4th first items from the 4 of the 2nd cell.
Let's return to your question, you want create a multidimensional array of 1000 rows and 4 columns. To start, we create an array of 1000 rows (cells) :
var a:Array = [1000]; // you can write it : new Array(1000);
Then, we create 4 columns for every row, and set values like this :
var i:Number, j:Number;
for (i = 0; i < 1000; i++)
{
// create the 4 columns
a[i] = [4]; // you can write it : a[i] = new Array(4);
for (j = 0; j < 4; j++)
{
a[i][j] = 0;
}
}
Then we can verify our array :
trace(a[0][0]); // gives : 0
trace(a[255][2]); // gives : 0
trace(a[255][5]); // gives : undefined, because we have only 4 columns
trace(a[1500][0]); // gives : undefined, because we have only 1000 rows
Hope that can help.
i have a problem with the below code. i need to get the number of parents that it's value = to the sum of it's 2 children. example if parent value = 10 and its children are 2 and 8.
then i have to count this parent as 1. i need to check for all nodes in the tree.
this is what i tried to do: could you please advise:
int BinaryTree::numberOfSum (){
return numberOfSumImpl (root);
}
int BinaryTree::numberOfSumImpl (BTNode *rootNode, int el){
if(rootNode ==0) return 0;
int count=0;
if(rootNode->hasTwoChildren() || rootNode->isLeaf())
else if{
if (rootNode->info==el) return count=1;
return count + numberOfSumImpl(el,rootNode->left) + numberOfSumImpl(el,rootNode->right);
}
}
Many Thanks,
There seem to be a few problems with the call to numberOfSumImpl, as in the main function it is only called with one parameter, and in the recursive function it is called with two parameters, but with the order inversed. This code does not even compile!
Besides, the condition rootNode->hasTwoChildren() || rootNode->isLeaf()) looks strange and, having its body empty even more strange.
You did not post the definition of the BTNode, but looks like you are looking for something as
int BinaryTree::numberOfSumImpl (BTNode* p) {
if (not p) return 0;
int count;
if (p->hasTwoChildren() and p->info == p->left->info + p->right->info) {
count = 1;
} else {
count = 0;
}
return count + numberOfSumImpl(p->left) + numberOfSumImpl(p->right);
}
And this function would possibly be static.
I am a newbie in programming and i start with Objective C as my first language.
I am messing around with some books and tutorials, at last programing a calculator...
Everything fine and i am getting into (programming makes really fun)
Now i am asking myself how I could translate arabic numbers to chinese numbers
(e.g. arabic 4 is in chinese 四 and 8 is 八 which means 四 + 四 = 八
The chinese number system is kind of different than arabic they have signs for 100, 1000, 10000 and ja kind of twisted, which screws up my brain ... anyway do anybody have some advice, hints, tips or solutions how i can tell the computer how to work with this numbers, or even how to calculate with them?
I think everything is possible so i wont ask "If its even possible?"
Considering the Chinese numerical system (Mandarin) as described by wikipedia http://en.wikipedia.org/wiki/Chinese_numerals, where for instance:
45 is interpreted as [4] [10] [5] and written 四十五
114 is interpreted as [1] [100] [1] [10] [4] and written 一百一十四
So the trick is to decompose a number as powers of 10:
x = c(k)*10^k + ... + c(1)*10 + c(0)
where k is the largest power of 10 that divides x such that the quotient is at least 1. In the 2nd example above, 114 = 1*10^2 + 1*10 + 4.
This x = c(k)*10^k + ... + c(1)*10 + c(0) becomes [c(k)][10^k]...[c(1)][10][c(0)]. In the 2nd example again, 114 = [1] [100] [1] [10] [4].
Then map each number within bracket to the corresponding sinogram:
0 = 〇
1 = 一
2 = 二
3 = 三
4 = 四
5 = 五
6 = 六
7 = 七
8 = 八
9 = 九
10 = 十
100 = 百
1000 = 千
10000 = 万
As long as you keep track of the [c(k)][10^k]...[c(1)][10][c(0)] form, it's easy to convert to an integer that the computer can handle or to the corresponding Chinese numeral. So it's this [c(k)][10^k]...[c(1)][10][c(0)] form that I'd store in an integer array of size k+2.
I'm not familiar with Objective-C, thus I can't help you with a solution for iOS.
Nonetheless, following is the Java code for Android...
I assumed it might help you, as well as it helped me.
double text2double(String text) {
String[] units = new String[] { "〇", "一", "二", "三", "四",
"五", "六", "七", "八", "九"};
String[] scales = new String[] { "十", "百", "千", "万",
"亿" };
HashMap<String, ScaleIncrementPair> numWord = new HashMap<String, ScaleIncrementPair>();
for (int i = 0; i < units.length; i++) {
numWord.put(units[i], new ScaleIncrementPair(1, i));
}
numWord.put("零", new ScaleIncrementPair(1, 0));
numWord.put("两", new ScaleIncrementPair(1, 2));
for (int i = 0; i < scales.length; i++) {
numWord.put(scales[i], new ScaleIncrementPair(Math.pow(10, (i + 1)), 0));
}
double current = 0;
double result = 0;
for (char character : text.toCharArray()) {
ScaleIncrementPair scaleIncrement = numWord.get(String.valueOf(character));
current = current * scaleIncrement.scale + scaleIncrement.increment;
if (scaleIncrement.scale > 10) {
result += current;
current = 0;
}
}
return result + current;
}
class ScaleIncrementPair {
public double scale;
public int increment;
public ScaleIncrementPair(double s, int i) {
scale = s;
increment = i;
}
}
You can make use of NSNumberFormatter.
Like below code, firstly get NSNumber from chinese characters, then combine them.
func getNumber(fromText text: String) -> NSNumber? {
let locale = Locale(identifier: "zh_Hans_CN")
let numberFormatter = NumberFormatter()
numberFormatter.locale = locale
numberFormatter.numberStyle = .spellOut
guard let number = numberFormatter.number(from: text) else { return nil }
print(number)
return number
}
How many possible combinations of the variables a,b,c,d,e are possible if I know that:
a+b+c+d+e = 500
and that they are all integers and >= 0, so I know they are finite.
#Torlack, #Jason Cohen: Recursion is a bad idea here, because there are "overlapping subproblems." I.e., If you choose a as 1 and b as 2, then you have 3 variables left that should add up to 497; you arrive at the same subproblem by choosing a as 2 and b as 1. (The number of such coincidences explodes as the numbers grow.)
The traditional way to attack such a problem is dynamic programming: build a table bottom-up of the solutions to the sub-problems (starting with "how many combinations of 1 variable add up to 0?") then building up through iteration (the solution to "how many combinations of n variables add up to k?" is the sum of the solutions to "how many combinations of n-1 variables add up to j?" with 0 <= j <= k).
public static long getCombos( int n, int sum ) {
// tab[i][j] is how many combinations of (i+1) vars add up to j
long[][] tab = new long[n][sum+1];
// # of combos of 1 var for any sum is 1
for( int j=0; j < tab[0].length; ++j ) {
tab[0][j] = 1;
}
for( int i=1; i < tab.length; ++i ) {
for( int j=0; j < tab[i].length; ++j ) {
// # combos of (i+1) vars adding up to j is the sum of the #
// of combos of i vars adding up to k, for all 0 <= k <= j
// (choosing i vars forces the choice of the (i+1)st).
tab[i][j] = 0;
for( int k=0; k <= j; ++k ) {
tab[i][j] += tab[i-1][k];
}
}
}
return tab[n-1][sum];
}
$ time java Combos
2656615626
real 0m0.151s
user 0m0.120s
sys 0m0.012s
The answer to your question is 2656615626.
Here's the code that generates the answer:
public static long getNumCombinations( int summands, int sum )
{
if ( summands <= 1 )
return 1;
long combos = 0;
for ( int a = 0 ; a <= sum ; a++ )
combos += getNumCombinations( summands-1, sum-a );
return combos;
}
In your case, summands is 5 and sum is 500.
Note that this code is slow. If you need speed, cache the results from summand,sum pairs.
I'm assuming you want numbers >=0. If you want >0, replace the loop initialization with a = 1 and the loop condition with a < sum. I'm also assuming you want permutations (e.g. 1+2+3+4+5 plus 2+1+3+4+5 etc). You could change the for-loop if you wanted a >= b >= c >= d >= e.
I solved this problem for my dad a couple months ago...extend for your use. These tend to be one time problems so I didn't go for the most reusable...
a+b+c+d = sum
i = number of combinations
for (a=0;a<=sum;a++)
{
for (b = 0; b <= (sum - a); b++)
{
for (c = 0; c <= (sum - a - b); c++)
{
//d = sum - a - b - c;
i++
}
}
}
This would actually be a good question to ask on an interview as it is simple enough that you could write up on a white board, but complex enough that it might trip someone up if they don't think carefully enough about it. Also, you can also for two different answers which cause the implementation to be quite different.
Order Matters
If the order matters then any solution needs to allow for zero to appear for any of the variables; thus, the most straight forward solution would be as follows:
public class Combos {
public static void main() {
long counter = 0;
for (int a = 0; a <= 500; a++) {
for (int b = 0; b <= (500 - a); b++) {
for (int c = 0; c <= (500 - a - b); c++) {
for (int d = 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2656615626.
Order Does Not Matter
If the order does not matter then the solution is not that much harder as you just need to make sure that zero isn't possible unless sum has already been found.
public class Combos {
public static void main() {
long counter = 0;
for (int a = 1; a <= 500; a++) {
for (int b = (a != 500) ? 1 : 0; b <= (500 - a); b++) {
for (int c = (a + b != 500) ? 1 : 0; c <= (500 - a - b); c++) {
for (int d = (a + b + c != 500) ? 1 : 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2573155876.
One way of looking at the problem is as follows:
First, a can be any value from 0 to 500. Then if follows that b+c+d+e = 500-a. This reduces the problem by one variable. Recurse until done.
For example, if a is 500, then b+c+d+e=0 which means that for the case of a = 500, there is only one combination of values for b,c,d and e.
If a is 300, then b+c+d+e=200, which is in fact the same problem as the original problem, just reduced by one variable.
Note: As Chris points out, this is a horrible way of actually trying to solve the problem.
link text
If they are a real numbers then infinite ... otherwise it is a bit trickier.
(OK, for any computer representation of a real number there would be a finite count ... but it would be big!)
It has general formulae, if
a + b + c + d = N
Then number of non-negative integral solution will be C(N + number_of_variable - 1, N)
#Chris Conway answer is correct. I have tested with a simple code that is suitable for smaller sums.
long counter = 0;
int sum=25;
for (int a = 0; a <= sum; a++) {
for (int b = 0; b <= sum ; b++) {
for (int c = 0; c <= sum; c++) {
for (int d = 0; d <= sum; d++) {
for (int e = 0; e <= sum; e++) {
if ((a+b+c+d+e)==sum) counter=counter+1L;
}
}
}
}
}
System.out.println("counter e "+counter);
The answer in math is 504!/(500! * 4!).
Formally, for x1+x2+...xk=n, the number of combination of nonnegative number x1,...xk is the binomial coefficient: (k-1)-combination out of a set containing (n+k-1) elements.
The intuition is to choose (k-1) points from (n+k-1) points and use the number of points between two chosen points to represent a number in x1,..xk.
Sorry about the poor math edition for my fist time answering Stack Overflow.
Just a test for code block
Just a test for code block
Just a test for code block
Including negatives? Infinite.
Including only positives? In this case they wouldn't be called "integers", but "naturals", instead. In this case... I can't really solve this, I wish I could, but my math is too rusty. There is probably some crazy integral way to solve this. I can give some pointers for the math skilled around.
being x the end result,
the range of a would be from 0 to x,
the range of b would be from 0 to (x - a),
the range of c would be from 0 to (x - a - b),
and so forth until the e.
The answer is the sum of all those possibilities.
I am trying to find some more direct formula on Google, but I am really low on my Google-Fu today...