I have to check how many hundreds are there in a number and translate that number to letters. For example the number 700. I have done the following code:
DATA(lv_dmbtr) = ZDS_FG-DMBTR. //Declared local variable of type DMBTR, thus DMBTR=700.
lv_dmbtr = ZDS_FG-DMBTR MOD 100. //Finding how many times 700 is in 100 via MOD and putting the value in lv_dmbtr.
IF lv_dmbtr LE 9. //The value is less or equal than 9(if larger means that the DMBTR is larger than hundreds,
e.g. 8000)
lv_hundred = lv_dmbtr / 100. // Divide the 700 with 100, taking the number 7.
lv_hundred_check = lv_hundred MOD 1. // Then taking the value of 7 into the new variable, done in case the
lv_hundred is a decimal value, e.g. 7.32.
IF lv_hundred_check > 0.
CALL FUNCTION 'SPELL_AMOUNT'
EXPORTING
amount = lv_hundred_check
* CURRENCY = ' '
* FILLER = ' '
LANGUAGE = SY-LANGU
IMPORTING
in_words = lv_hundred_string // the value is put in the new string
EXCEPTIONS
not_found = 1
too_large = 2
OTHERS = 3.
ENDIF.
Now when I debugg the code, all the variables have the value 0. Thus, lv_dmbtr, lv_hundred, lv_hundred_check all have the value 0.
May anyone of you know where the problem may be?
Thank you in advance!
Sorry for writing a lot in the code, just wanted to clarify as much as I could what I had done.
yes so I want to display the value of a specific number 700-> seven, 1400-> four.
So the basic formula to get the hundred in a number is the following: Find out how many times 100 fits completely into your number with integer division.
99 / 100 = 0
700 / 100 = 7
701 / 100 = 7
1400 / 100 = 14
1401 / 100 = 14
Now you can simply take this number MOD 10 to get the the individual hundreds.
0 MOD 10 = 0
7 MOD 10 = 7
14 MOD 10 = 4
Keep in mind that ABAP, in contrast to many other programming languages, rounds automatically. So in code this would be:
CONSTANTS lc_hundred TYPE f VALUE '100.0'.
DATA(lv_number) = 1403.
DATA(lv_hundred_count) = CONV i( floor( ( abs( lv_number ) / lc_hundred ) ) MOD 10 ).
Related
So if I was given a sorted list/array i.e. [1,6,8,15,40], the size of the array, and the requested number..
How would you find the minimum number of values required from that list to sum to the requested number?
For example given the array [1,6,8,15,40], I requested the number 23, it would take 2 values from the list (8 and 15) to equal 23. The function would then return 2 (# of values). Furthermore, there are an unlimited number of 1s in the array (so you the function will always return a value)
Any help is appreciated
The NP-complete subset-sum problem trivially reduces to your problem: given a set S of integers and a target value s, we construct set S' having values (n+1) xk for each xk in S and set the target equal to (n+1) s. If there's a subset of the original set S summing to s, then there will be a subset of size at most n in the new set summing to (n+1) s, and such a set cannot involve extra 1s. If there is no such subset, then the subset produced as an answer must contain at least n+1 elements since it needs enough 1s to get to a multiple of n+1.
So, the problem will not admit any polynomial-time solution without a revolution in computing. With that disclaimer out of the way, you can consider some pseudopolynomial-time solutions to the problem which work well in practice if the maximum size of the set is small.
Here's a Python algorithm that will do this:
import functools
S = [1, 6, 8, 15, 40] # must contain only positive integers
#functools.lru_cache(maxsize=None) # memoizing decorator
def min_subset(k, s):
# returns the minimum size of a subset of S[:k] summing to s, including any extra 1s needed to get there
best = s # use all ones
for i, j in enumerate(S[:k]):
if j <= s:
sz = min_subset(i, s-j)+1
if sz < best: best = sz
return best
print min_subset(len(S), 23) # prints 2
This is tractable even for fairly large lists (I tested a random list of n=50 elements), provided their values are bounded. With S = [random.randint(1, 500) for _ in xrange(50)], min_subset(len(S), 8489) takes less than 10 seconds to run.
There may be a simpler solution, but if your lists are sufficiently short, you can just try every set of values, i.e.:
1 --> Not 23
6 --> Not 23
...
1 + 6 = 7 --> Not 23
1 + 8 = 9 --> Not 23
...
1 + 40 = 41 --> Not 23
6 + 8 = 14 --> Not 23
...
8 + 15 = 23 --> Oh look, it's 23, and we added 2 values
If you know your list is sorted, you can skip some tests, since if 6 + 20 > 23, then there's no need to test 6 + 40.
I have two columns that hold numbers for which I am trying to calculate the difference in % between and show the result in another column but the results seem to be wrong.
This is the code in question.
SELECT
GenPar.ParameterValue AS ClaimType,
COUNT(Submitted.ClaimNumber) AS SubmittedClaims,
COUNT(ApprovalProvision.ClaimNumber) AS ApprovedClaims,
COUNT(Declined.ClaimNumber) AS DeclinedClaims,
COUNT(Pending.ClaimNumber) AS PendingClaims,
ISNULL(SUM(SubmittedSum.SumInsured),0) AS TotalSubmittedSumInsured,
ISNULL(SUM(ApprovedSum.SumInsured),0) AS TotalApprovedSumInsured,
ISNULL(SUM(RejectedSum.SumInsured),0) AS TotalRejectedSumInsured,
ISNULL(SUM(PendingSum.SumInsured),0) AS TotalPendingSumInsured,
--This column is to show the diff in %
CASE WHEN COUNT(Submitted.ClaimNumber) <> 0 AND COUNT(ApprovalProvision.ClaimNumber) <> 0
THEN (COUNT(ApprovalProvision.ClaimNumber),0) - (COUNT(Submitted.ClaimNumber),0)
/COUNT(Submitted.ClaimNumber) * 100
ELSE 0
END
What I need is to show the difference in % between the columns SubmittedClaims and ApprovedClaims. Any column, or both may contain zeroes and it may not.
So it's: COUNT(Submitted.ClaimNumber) - COUNT(ApprovalProvision.ClaimNumber) / COUNT(Submitted.ClaimNumber) * 100 as far as I know.
I have tried this and an example of what it does is it takes 1 and 117 and returns 17 when the difference between 1 and 117 is a decrease of 99.15%. Another example is 2 and 100. This simply returns 0 whereas the difference is a decrease of 98%.
CASE WHEN COUNT(Submitted.ClaimNumber) <> 0 AND COUNT(ApprovalProvision.ClaimNumber) <> 0
THEN (COUNT(ApprovalProvision.ClaimNumber),0) - (COUNT(Submitted.ClaimNumber),0)
/COUNT(Submitted.ClaimNumber) * 100
ELSE 0
END
I've checked this link and this seems to be what I am doing.
Percentage difference between two values
I've also tried this code:
NULLIF(COUNT(Submitted.ClaimNumber),0) - NULLIF(COUNT(ApprovalProvision.ClaimNumber),0)
/ NULLIF(COUNT(Submitted.ClaimNumber),0) * 100
and this takes for example 2 and 100 and returns -4998 when the real difference is a decrease of 98%.
For completion, Submitted.ClaimNumber is this portion of code:
LEFT OUTER JOIN (SELECT * FROM Company.Schema.ClaimMain WHERE CurrentStatus=10)Submitted
ON Submitted.ClaimNumber = ClaimMain.ClaimNumber
ApprovalProvision.ClaimNumber is this:
LEFT OUTER JOIN (SELECT * FROM Company.Schema.ClaimMain WHERE CurrentStatus=15)ApprovalProvision
ON ApprovalProvision.ClaimNumber = ClaimMain.ClaimNumber
Ideally, this column would also deal with 0's. So if one value is 0 and the other is X, the result should return 0 since a percentage can't be calculated if original number is 0. If the original value is X and the new value is 0, I should show a decrease of 100%.
This will occur across all columns but there is no need to flood the page with the rest of the columns since all calculations will occur in the same manner.
Anybody see what I'm doing wrong?
I'm not familiar with why you have (x,0) as a syntax
But I see that you have
(COUNT(ApprovalProvision.ClaimNumber),0) - (COUNT(Submitted.ClaimNumber),0)
/COUNT(Submitted.ClaimNumber) * 100
shouldn't it be,
( COUNT(ApprovalProvision.ClaimNumber) - COUNT(Submitted.ClaimNumber) )
/COUNT(Submitted.ClaimNumber) * 100
It looks like it would do count of ApprovalProvision.ClaimNumber - 100 since submitted.claimnumber divided by itself is 1 times 100 is 100.
The 4900 number actually sounds right. Lets take the following example, you have 2 apples, and then you're given 98 more and got 100 apples.
An increase of 98% would have meant from 2 apples, you would have 3.96 apples.
An increase of 100% means from 2 apples you end with 4 apples. An increase of 1000% means from 2 apples you end with 22 apples. So 4000% means you end with 82 apples. 5000% means from 2 apples, you reach 102 apples.
(100-2)/2*100 = 98 / 2 = 49 * 100 = 4900, so it looks like there is a 4900% increase in number of apples if you started with 2 apples and reach 100.
Now if you had flipped the 2 and 100, say starting with 100, now you have 2,
(2-100)/100*100 = -98, so a -98% change of apples, or a 98% decrease.
Hope this solves your problem.
I'd like to calculate how many different variations of a certain amount of numbers are possible. The number of elements is variable.
Example:
I have 5 elements and each element can vary between 0 and 8. Only the first element is a bit more defined and can only vary between 1 and 8. So far I'd say I have 8*9^4 possibilities. But I have some more conditions. As soon as one of the elements gets zero the next elements should be automatically zero as well.
E.G:
6 5 4 7 8 is ok
6 3 6 8 0 is ok
3 6 7 0 5 is not possible and would turn to 3 6 7 0 0
Would somebody show me how to calculate the amount of combinations for this case and also in general, because I'd like to be able to calculate it also for 4 or 8 or 9 etc. elements. Later on I'd like to calculate this number in VBA to be able give the user a forecast how long my calculations will take.
Since once a 0 is present in the sequence, all remaining numbers in the sequence will also be 0, these are all of the possibilities: (where # below represents any digit from 1 to 8):
##### (accounts for 8^5 combinations)
####0 (accounts for 8^4 combinations)
...
#0000 (accounts for 8^1 combinations)
Therefore, the answer is (in pseudocode):
int sum = 0;
for (int x = 1; x <= 5; x++)
{
sum = sum + 8^x;
}
Or equivalently,
int prod = 0;
for (int x = 1; x <= 5; x++)
{
prod = 8*(prod+1);
}
great thank you.
Sub test()
Dim sum As Single
Dim x As Integer
For x = 1 To 6
sum = sum + 8 ^ x
Next
Debug.Print sum
End Sub
With this code I get exactly 37488. I tried also with e.g. 6 elements and it worked as well. Now I can try to estimate the calculation time
Is there a way to give while loop a condition which makes it give an an output every ten times' executions, however it continues running after this output?
I hope I made myself clear...
Thanks!
Amy
Modulo is useful for this.
As an example; In swift to do modulo you use the % symbol. Essentially modulo outputs the remainder of the given terms.
So;
Value 1 MODULO Value 2 outputs Remainder.
Furthermore;
6 % 2 = 0
6 % 5 = 1
6 % 4.5 = 1.5
Essentially you want every nth element to output a value, with n being the rate. You need to track how many loops of the while you have gone through already.
The code below will run through the while 1000 times, and print out every 10 times ( for a total of 100 prints of output. )
var execution : Int = 0
while ( execution != 1000 ) {
if ( execution % 10 == 0 ) {
print("output")
}
execution = execution + 1
}
Here is the same answer as given by adam but then in Labview.
Given a NxN matrix and a (row,column) position, what is a method to select a different position in a random (or pseudo-random) order, trying to avoid collisions as much as possible?
For example: consider a 5x5 matrix and start from (1,2)
0 0 0 0 0
0 0 X 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
I'm looking for a method like
(x,y) hash (x,y);
to jump to a different position in the matrix, avoiding collisions as much as possible
(do not care how to return two different values, it doesn't matter, just think of an array).
Of course, I can simply use
row = rand()%N;
column = rand()%N;
but it's not that good to avoid collisions.
I thought I could apply twice a simple hash method for both row and column and use the results as new coordinates, but I'm not sure this is a good solution.
Any ideas?
Can you determine the order of the walk before you start iterating? If your matrices are large, this approach isn't space-efficient, but it is straightforward and collision-free. I would do something like:
Generate an array of all of the coordinates. Remove the starting position from the list.
Shuffle the list (there's sample code for a Fisher-Yates shuffle here)
Use the shuffled list for your walk order.
Edit 2 & 3: A modular approach: Given s array elements, choose a prime p of form 2+3*n, p>s. For i=1 to p, use cells (iii)%p when that value is in range 1...s-1. (For row-length r, cell #c subscripts are c%r, c/r.)
Effectively, this method uses H(i) = (iii) mod p as a hash function. The reference shows that as i ranges from 1 to p, H(i) takes on each of the values from 0 to p-1, exactly one time each.
For example, with s=25 and p=29 or 47, this uses cells in following order:
p=29: 1 8 6 9 13 24 19 4 14 17 22 18 11 7 12 3 15 10 5 16 20 23 2 21 0
p=47: 1 8 17 14 24 13 15 18 7 4 10 2 6 21 3 22 9 12 11 23 5 19 16 20 0
according to bc code like
s=25;p=29;for(i=1;i<=p;++i){t=(i^3)%p; if(t<s){print " ",t}}
The text above shows the suggestion I made in Edit 2 of my answer. The text below shows my first answer.
Edit 0: (This is the suggestion to which Seamus's comment applied): A simple method to go through a vector in a "random appearing" way is to repeatedly add d (d>1) to an index. This will access all elements if d and s are coprime (where s=vector length). Note, my example below is in terms of a vector; you could do the same thing independently on the other axis of your matrix, with a different delta for it, except a problem mentioned below would occur. Note, "coprime" means that gcd(d,s)=1. If s is variable, you'd need gcd() code.
Example: Say s is 10. gcd(s,x) is 1 for x in {1,3,7,9} and is not 1 for x in {2,4,5,6,8,10}. Suppose we choose d=7, and start with i=0. i will take on values 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, which modulo 10 is 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 0.
Edit 1 & 3: Unfortunately this will have a problem in the two-axis case; for example, if you use d=7 for x axis, and e=3 for y-axis, while the first 21 hits will be distinct, it will then continue repeating the same 21 hits. To address this, treat the whole matrix as a vector, use d with gcd(d,s)=1, and convert cell numbers to subscripts as above.
If you just want to iterate through the matrix, what is wrong with row++; if (row == N) {row = 0; column++}?
If you iterate through the row and the column independently, and each cycles back to the beginning after N steps, then the (row, column) pair will interate through only N of the N^2 cells of the matrix.
If you want to iterate through all of the cells of the matrix in pseudo-random order, you could look at questions here on random permutations.
This is a companion answer to address a question about my previous answer: How to find an appropriate prime p >= s (where s = the number of matrix elements) to use in the hash function H(i) = (i*i*i) mod p.
We need to find a prime of form 3n+2, where n is any odd integer such that 3*n+2 >= s. Note that n odd gives 3n+2 = 3(2k+1)+2 = 6k+5 where k need not be odd. In the example code below, p = 5+6*(s/6); initializes p to be a number of form 6k+5, and p += 6; maintains p in this form.
The code below shows that half-a-dozen lines of code are enough for the calculation. Timings are shown after the code, which is reasonably fast: 12 us at s=half a million, 200 us at s=half a billion, where us denotes microseconds.
// timing how long to find primes of form 2+3*n by division
// jiw 20 Sep 2011
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
double ttime(double base) {
struct timeval tod;
gettimeofday(&tod, NULL);
return tod.tv_sec + tod.tv_usec/1e6 - base;
}
int main(int argc, char *argv[]) {
int d, s, p, par=0;
double t0=ttime(0);
++par; s=5000; if (argc > par) s = atoi(argv[par]);
p = 5+6*(s/6);
while (1) {
for (d=3; d*d<p; d+=2)
if (p%d==0) break;
if (d*d >= p) break;
p += 6;
}
printf ("p = %d after %.6f seconds\n", p, ttime(t0));
return 0;
}
Timing results on 2.5GHz Athlon 5200+:
qili ~/px > for i in 0 00 000 0000 00000 000000; do ./divide-timing 500$i; done
p = 5003 after 0.000008 seconds
p = 50021 after 0.000010 seconds
p = 500009 after 0.000012 seconds
p = 5000081 after 0.000031 seconds
p = 50000021 after 0.000072 seconds
p = 500000003 after 0.000200 seconds
qili ~/px > factor 5003 50021 500009 5000081 50000021 500000003
5003: 5003
50021: 50021
500009: 500009
5000081: 5000081
50000021: 50000021
500000003: 500000003
Update 1 Of course, timing is not determinate (ie, can vary substantially depending on the value of s, other processes on machine, etc); for example:
qili ~/px > time for i in 000 004 010 058 070 094 100 118 184; do ./divide-timing 500000$i; done
p = 500000003 after 0.000201 seconds
p = 500000009 after 0.000201 seconds
p = 500000057 after 0.000235 seconds
p = 500000069 after 0.000394 seconds
p = 500000093 after 0.000200 seconds
p = 500000099 after 0.000201 seconds
p = 500000117 after 0.000201 seconds
p = 500000183 after 0.000211 seconds
p = 500000201 after 0.000223 seconds
real 0m0.011s
user 0m0.002s
sys 0m0.004s
Consider using a double hash function to get a better distribution inside the matrix,
but given that you cannot avoid colisions, what I suggest is to use an array of sentinels
and mark the positions you visit, this way you are sure you get to visit a cell once.