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From: Otniel Michael Aaron Bono pgsql-sql(at)postgresql(dot)org Re: About Div 2006-07-26 01:43:28 20060726014328.38920.qmail@web53202.mail.yahoo.com (view raw or whole thread) 2006-07-25 07:57:26 from Otniel Michael  2006-07-25 15:00:35 from "Aaron Bono"   2006-07-26 01:43:28 from Otniel Michael    2006-07-26 02:34:31 from Ross Johnson    2006-07-26 14:49:25 from "Aaron Bono" pgsql-sql
```Mr. Aaron. I am sorry, your solution didn't match in my case.
A = 1
B = 1
C = 1
D = 1
E = 1
F = 1
G = 4

G have 4 candy. Its too much for G.

In my case, the solution is :
A = 1
B = 1
C = 1
D = 1
E = 2
F = 2
G = 2

The extra candy is given to three child.

Do you have the other solution? I need function in postgresql for my case.
Because my loop is too slow.

Aaron Bono <postgresql(at)aranya(dot)com> wrote: On 7/25/06, Otniel Michael <otmic_ie(at)yahoo(dot)com> wrote: Dear All,

I have a problem with this case :

I have 10 candy for 7 child (A, B, C, D, E, F, G).

Table X :
code   value
-------  --------
A        0
B        0
C        0
D        0
E        0
F        0
G       0

And I want divide it with this algorithm :
A = 10 / 7 = 1
B = (10-1) / (7-1)  = 9 / 6 = 1
C = (10-2) / (7-2) = 8 / 5 = 1
D = (10-3) / (7-3) = 7 / 4 = 1
E = (10-4) / (7-4) = 6 / 3 = 2
F = (10-6) / (7-5) = 4 / 2 = 2
G = (10-8) / (7-6) = 2 / 1 = 2

In my first solution i use loop - for each record in my function.
But it is too slow in a lot of data.
Did postgresql have a function for my case?

No loop necessary.  This is a simple math problem:

dividedamount := candy / childcount;
extra = candy % childcount;

So the first (childcount - extra) get (dividedamount) pieces of candy and the last (extra) get (dividedamount + 1) pieces of candy.

==================================================================
Aaron Bono
Aranya Software Technologies, Inc.
http://www.aranya.com
==================================================================

--
"He who is quick to become angry will commit folly, and a crafty man is hated"

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