I have used querystring to accept number in my classic asp code, and then the program will output the sum of digits of the number.
<%
dim n,d,sum
sum = 0
d = 0
n = request.querystring("n") //Taking value from querystring
while n<>0
d = n mod 10
response.write("<br>"&d)
sum = sum + d``
n = Cint(n/10)
wend
response.write("<br>Sum of digits of "&request.querystring("n")&" is :
"&sum)
%>
But the problem is it does not show correct answers for few values like
for eg. n=91....then output will be 11
but for n=123...the output will be 6.
pls help.
It might be easier to treat n as a string, which is really a character array. Because a string is a character array we can loop through the array and sum each value.
Dim n, d, sum
sum = 0
d = 0
n = Request.QueryString("n")
If IsNumeric(n) Then
For i = 1 To Len(n)
d = CInt(Mid(n,i,1))
Response.Write("<br />" & d)
sum = sum + d
Next
Response.Write("<br />Sum of digits of " & n & " is : " & sum)
End If
Only have to do one change in the code , i.e. use the operator \ instead of /
when you reduce your number....
i.e. instead of n = CInt(n/10) use n = n\10
Therefore the code will work fine as butter:
<%
dim n,sum,d
sum = 0
n = request.querystring("n") 'taking value from querystring
while n <> 0
d = n mod 10
sum = sum +d
n = n\10 'using \ for integer division
wend
response.write("Sum of digits of : "&request.querystring("n")&" is = "&sum)
%>
For those who dont understand what just happened see below:
Operator / : Performs a floating division.
Operator \ : Performs a Integer division.
Thats why I was not getting correct results.
Hope this helps!
I have this textbox named txtnum in which I have to enter a 15 digit number and allocate it to variable num. I want to split the number into individual characters so that j can carry out calculations on each. Something like: product= arraynum[2]*2 . how do I split the string in the text box into array characters?
Nothing is built-in (as far as I know), but it I easy enough to write a function which takes a string and returns an array of characters:
Function ToArray(s As String) As Variant
Dim A As Variant
Dim i As Long, n As Long
n = Len(s)
ReDim A(0 To n - 1)
For i = 0 To n - 1
A(i) = Mid(s, i + 1, 1)
Next i
ToArray = A
End Function
Having done this, there is little actual gain from using a function like this as opposed to simply using Mid().
Here is another option:
Dim s As Variant
s = "012345678901234"
s = StrConv(s, vbUnicode)
s = Split(s, vbNullChar)
s will contain an array of characters.
I am going to use the following formula on every cell of a 4 × 6 (m×n in the code) matrix to get the normalized matrix:
The matrix in Calc is:
I use the following Basic code in LibreOffice:
REM ***** BASIC *****
Sub Main
Normalize(5,3)
End Sub
Sub Normalize (ByVal n As Integer,ByVal m As Integer)
Dim Doc As Object
Dim Sheet As Object
Dim SrcCell 'Cell in the source matrix
Dim TargetCell 'Cell in the target matrix where normalized values are saved
Dim TempCell As Object
Dim I 'index
Dim J 'index
Dim JJ 'inner index
Dim Sum 'Sigma x_ij^2 (j=0 to m)
Dim m 'maximum row index
Dim n 'maximum column index
Doc = ThisComponent
Sheet = Doc.Sheets(0)
For I = 0 to n 'traverse columns
For J=0 to m 'traverse rows
SrcCell = Sheet.getCellByPosition(I,J)
'Now apply the normalization formula for this cell
'Run a new loop to run formula on this cell
Sum = 0 'Reset Sum to 0
For JJ=0 to m
TempCell = Sheet.getCellByPosition(I,JJ)
Sum = Sum + (TempCell.Value^2)
Next
TargetCell = Sheet.getCellByPosition(I+n+1,J) 'Place the normalized cells in a new matrix cell, n+1 cells away
'Put the sum in the formula
TargetCell.Value = SrcCell.Value/Sqr(Sum)
Next
Next
End Sub
I want am going to have the normalized matrix appear on the right side of the original one. But nothing appears. What am I doing wrong?
The code uses n and m as parameters but then declares them, destroying the values. To fix, remove the following two lines. For readability, move these comments near Sub Normalize.
Dim m 'maximum row index
Dim n 'maximum column index
To find such problems with the Basic IDE debugger, press Breakpoint On/Off in the toolbar to set a couple of breakpoints, and also Enable Watch.
I'm working with vb6 and I want to generate multiple randum numbers (the range from to is detirmend by user and also the number of generated answers) and send them to a listbox
But I don't want to duplicate generated numbers So..
I want before sending the generated number to the listbox to check if it already exists in the lisbox. if it already exists then generate another number if it does't then send it the the listbox
here is what I have till now
max and min are the range to chose numbers between
answers is the number of generated numbers
Randomize
For i = 1 To answers Step 1
generated = CInt(Int((max - min + 1) * Rnd() + min))
For n = 0 To List1.ListCount
If List1.List(n) <> gen Then
List1.AddItem (gen)
Else
If List1.List = gen Then
'I don't know what to do from here
'(how to go back to generate another number)
Next n
Next i
Thank you in advance
keep in minde I need to keep things simple
Thank you soo much
Use a boolean value to keep the result if same value generated is in list.
Private Sub AddRandomNumbers()
Dim blnIfFound As Boolean
Dim max As Integer
Dim min As Integer
Dim answers As Integer
max = 10
min = 1
answers = 5
Randomize
Do While List1.ListCount < answers
generated = CInt(Int((max - min + 1) * Rnd() + min))
blnIfFound = False
For n = 0 To List1.ListCount
If List1.List(n) = generated Then
blnIfFound = True
Exit For
End If
Next n
If blnIfFound = False Then List1.AddItem (generated)
Loop
End Sub
Not sure how best to explain it, other than using an example...
Imagine having a client with 10 outstanding invoices, and one day they provide you with a cheque, but do not tell you which invoices it's for.
What would be the best way to return all the possible combination of values which can produce the required total?
My current thinking is a kind of brute force method, which involves using a self-calling function that runs though all the possibilities (see current version).
For example, with 3 numbers, there are 15 ways to add them together:
A
A + B
A + B + C
A + C
A + C + B
B
B + A
B + A + C
B + C
B + C + A
C
C + A
C + A + B
C + B
C + B + A
Which, if you remove the duplicates, give you 7 unique ways to add them together:
A
A + B
A + B + C
A + C
B
B + C
C
However, this kind of falls apart after you have:
15 numbers (32,767 possibilities / ~2 seconds to calculate)
16 numbers (65,535 possibilities / ~6 seconds to calculate)
17 numbers (131,071 possibilities / ~9 seconds to calculate)
18 numbers (262,143 possibilities / ~20 seconds to calculate)
Where, I would like this function to handle at least 100 numbers.
So, any ideas on how to improve it? (in any language)
This is a pretty common variation of the subset sum problem, and it is indeed quite hard. The section on the Pseudo-polynomial time dynamic programming solution on the page linked is what you're after.
This is strictly for the number of possibilities and does not consider overlap. I am unsure what you want.
Consider the states that any single value could be at one time - it could either be included or excluded. That is two different states so the number of different states for all n items will be 2^n. However there is one state that is not wanted; that state is when none of the numbers are included.
And thus, for any n numbers, the number of combinations is equal to 2^n-1.
def setNumbers(n): return 2**n-1
print(setNumbers(15))
These findings are very closely related to combinations and permutations.
Instead, though, I think you may be after telling whether given a set of values any combination of them sum to a value k. For this Bill the Lizard pointed you in the right direction.
Following from that, and bearing in mind I haven't read the whole Wikipedia article, I propose this algorithm in Python:
def combs(arr):
r = set()
for i in range(len(arr)):
v = arr[i]
new = set()
new.add(v)
for a in r: new.add(a+v)
r |= new
return r
def subsetSum(arr, val):
middle = len(arr)//2
seta = combs(arr[:middle])
setb = combs(arr[middle:])
for a in seta:
if (val-a) in setb:
return True
return False
print(subsetSum([2, 3, 5, 8, 9], 8))
Basically the algorithm works as this:
Splits the list into 2 lists of approximately half the length. [O(n)]
Finds the set of subset sums. [O(2n/2 n)]
Loops through the first set of up to 2floor(n/2)-1 values seeing if the another value in the second set would total to k. [O(2n/2 n)]
So I think overall it runs in O(2n/2 n) - still pretty slow but much better.
Sounds like a bin packing problem. Those are NP-complete, i.e. it's nearly impossible to find a perfect solution for large problem sets. But you can get pretty close using heuristics, which are probably applicable to your problem even if it's not strictly a bin packing problem.
This is a variant on a similar problem.
But you can solve this by creating a counter with n bits. Where n is the amount of numbers. Then you count from 000 to 111 (n 1's) and for each number a 1 is equivalent to an available number:
001 = A
010 = B
011 = A+B
100 = C
101 = A+C
110 = B+C
111 = A+B+C
(But that was not the question, ah well I leave it as a target).
It's not strictly a bin packing problem. It's a what combination of values could have produced another value.
It's more like the change making problem, which has a bunch of papers detailing how to solve it. Google pointed me here: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.57.3243
I don't know how often it would work in practice as there are many exceptions to this oversimplified case, but here's a thought:
In a perfect world, the invoices are going to be paid up to a certain point. People will pay A, or A+B, or A+B+C, but not A+C - if they've received invoice C then they've received invoice B already. In the perfect world the problem is not to find to a combination, it's to find a point along a line.
Rather than brute forcing every combination of invoice totals, you could iterate through the outstanding invoices in order of date issued, and simply add each invoice amount to a running total which you compare with the target figure.
Back in the real world, it's a trivially quick check you can do before launching into the heavy number-crunching, or chasing them up. Any hits it gets are a bonus :)
Here is an optimized Object-Oriented version of the exact integer solution to the Subset Sums problem(Horowitz, Sahni 1974). On my laptop (which is nothing special) this vb.net Class solves 1900 subset sums a second (for 20 items):
Option Explicit On
Public Class SubsetSum
'Class to solve exact integer Subset Sum problems'
''
' 06-sep-09 RBarryYoung Created.'
Dim Power2() As Integer = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32764}
Public ForceMatch As Boolean
Public watch As New Stopwatch
Public w0 As Integer, w1 As Integer, w1a As Integer, w2 As Integer, w3 As Integer, w4 As Integer
Public Function SolveMany(ByVal ItemCount As Integer, ByVal Range As Integer, ByVal Iterations As Integer) As Integer
' Solve many subset sum problems in sequence.'
''
' 06-sep-09 RBarryYoung Created.'
Dim TotalFound As Integer
Dim Items() As Integer
ReDim Items(ItemCount - 1)
'First create our list of selectable items:'
Randomize()
For item As Integer = 0 To Items.GetUpperBound(0)
Items(item) = Rnd() * Range
Next
For iteration As Integer = 1 To Iterations
Dim TargetSum As Integer
If ForceMatch Then
'Use a random value but make sure that it can be matched:'
' First, make a random bitmask to use:'
Dim bits As Integer = Rnd() * (2 ^ (Items.GetUpperBound(0) + 1) - 1)
' Now enumerate the bits and match them to the Items:'
Dim sum As Integer = 0
For b As Integer = 0 To Items.GetUpperBound(0)
'build the sum from the corresponding items:'
If b < 16 Then
If Power2(b) = (bits And Power2(b)) Then
sum = sum + Items(b)
End If
Else
If Power2(b - 15) * Power2(15) = (bits And (Power2(b - 15) * Power2(15))) Then
sum = sum + Items(b)
End If
End If
Next
TargetSum = sum
Else
'Use a completely random Target Sum (low chance of matching): (Range / 2^ItemCount)'
TargetSum = ((Rnd() * Range / 4) + Range * (3.0 / 8.0)) * ItemCount
End If
'Now see if there is a match'
If SolveOne(TargetSum, ItemCount, Range, Items) Then TotalFound += 1
Next
Return TotalFound
End Function
Public Function SolveOne(ByVal TargetSum As Integer, ByVal ItemCount As Integer _
, ByVal Range As Integer, ByRef Items() As Integer) As Boolean
' Solve a single Subset Sum problem: determine if the TargetSum can be made from'
'the integer items.'
'first split the items into two half-lists: [O(n)]'
Dim H1() As Integer, H2() As Integer
Dim hu1 As Integer, hu2 As Integer
If ItemCount Mod 2 = 0 Then
'even is easy:'
hu1 = (ItemCount / 2) - 1 : hu2 = (ItemCount / 2) - 1
ReDim H1((ItemCount / 2) - 1), H2((ItemCount / 2) - 1)
Else
'odd is a little harder, give the first half the extra item:'
hu1 = ((ItemCount + 1) / 2) - 1 : hu2 = ((ItemCount - 1) / 2) - 1
ReDim H1(hu1), H2(hu2)
End If
For i As Integer = 0 To ItemCount - 1 Step 2
H1(i / 2) = Items(i)
'make sure that H2 doesnt run over on the last item of an odd-numbered list:'
If (i + 1) <= ItemCount - 1 Then
H2(i / 2) = Items(i + 1)
End If
Next
'Now generate all of the sums for each half-list: [O( 2^(n/2) * n )] **(this is the slowest step)'
Dim S1() As Integer, S2() As Integer
Dim sum1 As Integer, sum2 As Integer
Dim su1 As Integer = 2 ^ (hu1 + 1) - 1, su2 As Integer = 2 ^ (hu2 + 1) - 1
ReDim S1(su1), S2(su2)
For i As Integer = 0 To su1
' Use the binary bitmask of our enumerator(i) to select items to use in our candidate sums:'
sum1 = 0 : sum2 = 0
For b As Integer = 0 To hu1
If 0 < (i And Power2(b)) Then
sum1 += H1(b)
If i <= su2 Then sum2 += H2(b)
End If
Next
S1(i) = sum1
If i <= su2 Then S2(i) = sum2
Next
'Sort both lists: [O( 2^(n/2) * n )] **(this is the 2nd slowest step)'
Array.Sort(S1)
Array.Sort(S2)
' Start the first half-sums from lowest to highest,'
'and the second half sums from highest to lowest.'
Dim i1 As Integer = 0, i2 As Integer = su2
' Now do a merge-match on the lists (but reversing S2) and looking to '
'match their sum to the target sum: [O( 2^(n/2) )]'
Dim sum As Integer
Do While i1 <= su1 And i2 >= 0
sum = S1(i1) + S2(i2)
If sum < TargetSum Then
'if the Sum is too low, then we need to increase the ascending side (S1):'
i1 += 1
ElseIf sum > TargetSum Then
'if the Sum is too high, then we need to decrease the descending side (S2):'
i2 -= 1
Else
'Sums match:'
Return True
End If
Loop
'if we got here, then there are no matches to the TargetSum'
Return False
End Function
End Class
Here is the Forms code to go along with it:
Public Class frmSubsetSum
Dim ssm As New SubsetSum
Private Sub btnGo_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnGo.Click
Dim Total As Integer
Dim datStart As Date, datEnd As Date
Dim Iterations As Integer, Range As Integer, NumberCount As Integer
Iterations = CInt(txtIterations.Text)
Range = CInt(txtRange.Text)
NumberCount = CInt(txtNumberCount.Text)
ssm.ForceMatch = chkForceMatch.Checked
datStart = Now
Total = ssm.SolveMany(NumberCount, Range, Iterations)
datEnd = Now()
lblStart.Text = datStart.TimeOfDay.ToString
lblEnd.Text = datEnd.TimeOfDay.ToString
lblRate.Text = Format(Iterations / (datEnd - datStart).TotalMilliseconds * 1000, "####0.0")
ListBox1.Items.Insert(0, "Found " & Total.ToString & " Matches out of " & Iterations.ToString & " tries.")
ListBox1.Items.Insert(1, "Tics 0:" & ssm.w0 _
& " 1:" & Format(ssm.w1 - ssm.w0, "###,###,##0") _
& " 1a:" & Format(ssm.w1a - ssm.w1, "###,###,##0") _
& " 2:" & Format(ssm.w2 - ssm.w1a, "###,###,##0") _
& " 3:" & Format(ssm.w3 - ssm.w2, "###,###,##0") _
& " 4:" & Format(ssm.w4 - ssm.w3, "###,###,##0") _
& ", tics/sec = " & Stopwatch.Frequency)
End Sub
End Class
Let me know if you have any questions.
For the record, here is some fairly simple Java code that uses recursion to solve this problem. It is optimised for simplicity rather than performance, although with 100 elements it seems to be quite fast. With 1000 elements it takes dramatically longer, so if you are processing larger amounts of data you could better use a more sophisticated algorithm.
public static List<Double> getMatchingAmounts(Double goal, List<Double> amounts) {
List<Double> remaining = new ArrayList<Double>(amounts);
for (final Double amount : amounts) {
if (amount > goal) {
continue;
} else if (amount.equals(goal)) {
return new ArrayList<Double>(){{ add(amount); }};
}
remaining.remove(amount);
List<Double> matchingAmounts = getMatchingAmounts(goal - amount, remaining);
if (matchingAmounts != null) {
matchingAmounts.add(amount);
return matchingAmounts;
}
}
return null;
}