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  • Prolog: find all numbers of unique digits that can be formed from a list of digits

    - by animo
    The best thing I could come up with so far is this function: numberFromList([X], X) :- digit(X), !. numberFromList(List, N) :- member(X, List), delete(List, X, LX), numberFromList(LX, NX), N is NX * 10 + X. where digit/1 is a function verifying if an atom is a decimal digit. The numberFromList(List, N) finds all the numbers that can be formed with all digits from List. E.g. [2, 3] -> 23, 32. but I want to get this result: [2, 3] -> 2, 3, 23, 32 I spent a lot of hours thinking about this and I suspect you might use something like append(L, _, List) at some point to get lists of lesser length. I would appreciate any contribution.

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  • The subsets-sum problem and the solvability of NP-complete problems

    - by G.E.M.
    I was reading about the subset-sums problem when I came up with what appears to be a general-purpose algorithm for solving it: (defun subset-contains-sum (set sum) (let ((subsets) (new-subset) (new-sum)) (dolist (element set) (dolist (subset-sum subsets) (setf new-subset (cons element (car subset-sum))) (setf new-sum (+ element (cdr subset-sum))) (if (= new-sum sum) (return-from subset-contains-sum new-subset)) (setf subsets (cons (cons new-subset new-sum) subsets))) (setf subsets (cons (cons element element) subsets))))) "set" is a list not containing duplicates and "sum" is the sum to search subsets for. "subsets" is a list of cons cells where the "car" is a subset list and the "cdr" is the sum of that subset. New subsets are created from old ones in O(1) time by just cons'ing the element to the front. I am not sure what the runtime complexity of it is, but appears that with each element "sum" grows by, the size of "subsets" doubles, plus one, so it appears to me to at least be quadratic. I am posting this because my impression before was that NP-complete problems tend to be intractable and that the best one can usually hope for is a heuristic, but this appears to be a general-purpose solution that will, assuming you have the CPU cycles, always give you the correct answer. How many other NP-complete problems can be solved like this one?

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  • a simple program that counts and sums digits. How can I make it work?

    - by user1710386
    So I've to write a simple program(that loops) where you can enter an int and it spews out the number count and the sum of the numbers. Since I am such a tard when it comes to programming, I just scavanged the code online and tried to piece it together. I guess the sum block screws with n, but I am not really sure. Anyway, I would really appreciate it if somebody could point out mistakes and show me how can I make it work. #include <iostream> using namespace std; int main() { while(1) { int i,p,n,sum=0; //sum block cout<<"enter an int: "; cin>>n; { while(n!=0) { p=n % 10; sum+=p; n=n/10; } cout<<"int digit sum: "<<sum <<endl; } { int count = 0; while(n) { n /= 10; ++count; } cout <<"number of digits: " << count << '\n';} } }

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  • [SQL] Getting the sum of several columns from two tables

    - by romaintaz
    I want to get the sum of several columns from 2 different tables (these tables share the same structure). If I only consider one table, I would write this kind of query: SELECT MONTH_REF, SUM(amount1), SUM(amount2) FROM T_FOO WHERE seller = XXX GROUP BY MONTH_REF; However, I would like to also work with the data from the table T_BAR, and then have a select query that return the following columns: MONTH_REF SUM(T_FOO.amount1) + SUM(T_BAR.amount1) SUM(T_FOO.amount2) + SUM(T_BAR.amount2) everything grouped by the value of MONTH_REF. Note that a record for a given MONTH_REF can be found in one table but not in the other table. In this case, I would like to get the sum of T_FOO.amount1 + 0 (or 0 + T_BAR.amount1). How can I write my SQL query to get this information? For information, my database is Oracle 10g.

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  • MySQL how to sum subgroup first then sum total

    - by Sunry
    My data in table_1 and table_2. table_1 id_1 num ids_2 3 33 666,777,888 4 44 111,222,333 table_2 id_2 num 111 1 222 2 333 3 666 6 777 7 888 8 I only know how to do what I want with two steps: First LEFT JOIN to get: SELECT 1.id_1, sum(2.num) FROM table_1 AS 1 LEFT JOIN table_2 AS 2 on FIND_IN_SET(2.id_2, 1.ids_2) GROUP BY 1.id_1; id_1 sum(2.num) 3 6+7+8 4 1+2+3 Then LEFT JOIN with table_1 again to sum(table_1.num+sum(2.num)): id_1 sum(table_1.num+sum(table_2.num)) 3 6+7+8+33 4 1+2+3+44 Can I do it in only one SQL?

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  • C++ Loop - Need variable to accumulate sum

    - by user1780064
    I'm writing a program to ask the user to enter a value between 5 and 21 (inclusive). If the number entered is not in this range, it prints, "Please try again". If the number is within the range, I need to take that number, and print the sum of all the numbers from 1 to the value entered. So if the user entered "7", the sum would be "28". I successfully wrote the first loop, in the case of the number not being within the range, but cannot figure out how to run the second loop- whether to use a while, do-while, or for loop. Please advise. #include <iostream> int main () { int uservalue; int count; int sum; //Prompt user for input do { cout << "Enter a value from 5 to 21: "; cin >> uservalue; if (uservalue < 5 || uservalue > 21) cout << "Value out of range. Try again..." << endl; } while (uservalue < 5 || uservalue > 21); cout << endl; //Loop to accumulate sum for (count = 1, count < uservalue, count++;) { sum = uservalue + count; if (uservalue <= 5 || uservalue <= 21) cout << the sum is " << sum << endl; } return 0; }

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  • linq get sum of two columns in one query

    - by Axarydax
    Hi, let's say that I have a table called Items (ID int, Done int, Total int) I can do it by two queries: int total = m.Items.Sum(p=>p.Total) int done = m.Items.Sum(p=>p.Done) But I'd like to do it in one query, something like this: var x = from p in m.Items select new { Sum(p.Total), Sum(p.Done)}; Surely there is a way to call aggregate functions from LINQ syntax...?

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  • Get sum of two columns in one LINQ query

    - by Axarydax
    Hi, let's say that I have a table called Items (ID int, Done int, Total int) I can do it by two queries: int total = m.Items.Sum(p=>p.Total) int done = m.Items.Sum(p=>p.Done) But I'd like to do it in one query, something like this: var x = from p in m.Items select new { Sum(p.Total), Sum(p.Done)}; Surely there is a way to call aggregate functions from LINQ syntax...?

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  • MSSQL Sum query

    - by ldb
    today my problem is this i have 2 column and i wish check if the sum of that columns isn't Higher then a value(485 for example) and if is do a query...i though to do SELECT * FROM table WHERE ColumnA+ColumnB<485 But isn't working... i've already tried with SELECT Sum(ColumnA)+Sum(ColumnB) AS Total FROM table but it gives me 1 column with the sum of all rows, i instead want a row for every sum. so how can i do..? xD i hope you understood if not just ask that i try to explain it better! and thanks in advice for who want to help me! EDIT: I Found out XD the problem was that the columns was Smallint and the result of 1 or more rows was more than 32k so it wasn't working! Thanks At all!!

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  • Python - Number of Significant Digits in results of division

    - by russ
    Newbie here. I have the following code: myADC = 128 maxVoltage = 5.0 maxADC = 255.0 VoltsPerADC = maxVoltage/maxADC myVolts = myADC * VoltsPerADC print "myADC = {0: >3}".format(myADC) print "VoltsPerADC = {0: >7}".format(VoltsPerADC) print VoltsPerADC print "myVolts = {0: >7}".format(myVolts) print myVolts This outputs the following: myADC = 128 VoltsPerADC = 0.0196078 0.0196078431373 myVolts = 2.5098 2.50980392157 I have been searching for an explanation of how the number of significant digits is determined by default, but have had trouble locating an explanation that makes sense to me. This link link text suggests that by default the "print" statement prints numbers to 10 significant figures, but that does not seem to be the case in my results. How are the number of significant digits/precision determined? Can someone shed some light on this for me. Thanks in advance for your time and patience.

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  • select SUM in Cake

    - by SpawnCxy
    When I tried using follow code to get the sum of the column named Serviceinfo.services $conditions = array('fields'=>array('SUM(Serviceinfo.services) as servicecount'),'conditions'=>array('time_id BETWEEN ? AND ?'=>array($startid,$endtid))); $services = $this->Serviceinfo->find('all',$conditions); I had to fetch the sum data with services[0][0]['servicecount'] which seems a little weird.And What I expect is services['Serviceinfo']['servicecount'],or more simple one.Then how can I fix this?Thanks in advance!

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  • Create Sum of calculated rows in Microsoft Reporting Services

    - by kd7iwp
    This seems like it should be simple but I can't find anything yet. In Reporting Services I have a table with up to 6 rows that all have calculated values and dynamic visibility. I would like to sum these rows. Basically I have a number of invoice items and want to make a total. I can't change anything on the DB side since my stored procedures are used elsewhere in the system. Each row pulls data from a different dataset as well, so I can't do a sum of the dataset. Can I sum all the rows with a table footer? Similarly to totaling a number of rows in Excel? It seems very redundant to put my visibility expression from each row into my footer row to calculate the sum.

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  • How to sum properties of the objects within an array in Ruby

    - by Ernst Fitschen
    I understand that in order to sum array elements in Ruby one can use the inject method, i.e. array = [1,2,3,4,5]; puts array.inject(0, &:+) But how do I sum the properties of objects within an object array e.g. There's an array of objects and each object has a property "cash" for example. So I want to sum their cash balances into one total. Something like... array.cash.inject(0, &:+) (but this doesn't work) I realise I could probably make a new array composed only of the property cash and sum this, but I'm looking for a cleaner method if possible!

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  • Calculate the SUM of the Column which has Time DataType:

    - by thevan
    I want to calculate the Sum of the Field which has Time DataType. My Table is Below: TableA: TotalTime ------------- 12:18:00 12:18:00 Here I want to sum the two time fields. I tried the below Query SELECT CAST( DATEADD(MS, SUM(DATEDIFF(MS, '00:00:00.000', CONVERT(TIME, TotalTime))), '00:00:00.000' ) AS TOTALTIME) FROM [TableA] But it gives the Output as TOTALTIME ----------------- 00:36:00.0000000 But My Desired Output would be like below: TOTALTIME ----------------- 24:36:00 How to get this Output?

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  • making a combined sum of two columns

    - by bsandrabr
    I have a table (apples) containing: cid date_am date_pm ---------------------- 1 1 1 2 2 1 3 1 3 1 1 2 I asked a question earlier (badly) about how I would rank the customers in order of the number of ones(1) they had. The solution was (based on one column): SELECT cid, sum( date_pm ) AS No_of_ones FROM apples WHERE date_am =1 GROUP BY cid ORDER BY no_of_ones DESC This works great for one column but how would I do the same for the sum of the two columns. ie. SELECT cid, sum( date_pm ) AS No_of_ones FROM apples WHERE date_am =1 add to SELECT cid, sum( date_am ) AS No_of_ones FROM apples WHERE date_pm =1 GROUP by cid ORDER by no_of_ones(added) hope I've managed to make that clear enough for you to help -thanks

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  • Finding the sum of 2D Arrays in Ruby

    - by Bragaadeesh
    Hi, I have an array of two dimensional Arrays. I want to create a new two dimensional array which finds the sum of these values in the 2D arrays. Sum at x,y of new array = Sum at x,y of arr1 + Sum at x,y of arr2 + .... |1,2,4| |1,1,1| |1,1,1| |2,4,6| |1,1,1| |1,1,1| |2,4,6| |1,1,1| |1,1,1| |2,4,6| |1,1,1| |1,1,1| Now adding the above two dimensional arrays will result in, |3,4,6| |4,6,8| |4,6,8| |4,6,8| How to achieve this in Ruby (not in any other languages). I have written a method, but it looks very long and ugly.

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  • Sum of distinc rows after a 1-many table join

    - by Lock
    I have 2 tables that I am joining. Table 1 has 1-many relationship with table 2. That is, table 2 can return multiple rows for a single row of table 1. Because of this, the records of table 1 is duplicated for as many rows as are on table 2.. which is expected. Now, I have a sum on one of the columns from table 1, but because of the multiple rows that get returned on the join, the sum is obviously multiplying. Is there a way to get this number back to its original number? I tried dividing by the count of rows from table 2 but this didnt quite give me the expected result. Are there any analytical functions that could do this? I almost want something like "if this row has not yet been counted in the sum, add it to the sum"

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  • Django custom SQL returning single row of results when query returns 2?

    - by Alvin
    I have a custom SQL call that is returning different results to the template than I get when I run the same query against the database directly, 1 row vs 2 Query - copied from Django Debug Toolbar: SELECT ((Sum(new_recruit_interviews) / Sum(opportunities_offered)) * 100) as avg_recruit, ((Sum(inspections) / Sum(presentations)) * 100) as avg_inspect, ((Sum(contracts_signed) / Sum(roof_approvals)) * 100) as avg_contracts, ((Sum(adjusters) / Sum(contracts_signed)) * 100) as avg_adjusters, ((Sum(roof_approvals) / Sum(adjusters)) *100) as roof_approval_avg, ((Sum(roof_turned_in) / Sum(adjusters)) * 100) as roof_jobs_avg, Sum(roof_turned_in) as roof_jobs_total, ((Sum(siding_approvals) / Sum(adjusters)) *100) as siding_approval_avg, ((Sum(siding_turned_in) / Sum(adjusters)) * 100) as siding_jobs_avg, Sum(siding_turned_in) as siding_jobs_total, ((Sum(gutter_approvals) / Sum(adjusters)) *100) as gutter_approval_avg, ((Sum(gutter_turned_in) / Sum(adjusters)) * 100) as gutter_jobs_avg, Sum(gutter_turned_in) as gutter_jobs_total, ((Sum(window_approvals) / Sum(adjusters)) *100) as window_approval_avg, ((Sum(window_turned_in) / Sum(adjusters)) * 100) as window_jobs_avg, Sum(window_turned_in) as window_jobs_total, (Sum(roof_turned_in) + Sum(siding_turned_in) + Sum(gutter_turned_in) + Sum(window_turned_in)) as total_jobs, (((Sum(collections_jobs_new) + Sum(collections_jobs_previous)) / (Sum(roof_turned_in) + Sum(siding_turned_in) + Sum(gutter_turned_in) + Sum(window_turned_in))) * 100) as total_collections, sales_report_salesmen.location_id as detail_id, business_unit_location.title as title FROM sales_report_salesmen Inner Join auth_user ON sales_report_salesmen.user_id = auth_user.id Inner Join business_unit_location ON sales_report_salesmen.location_id = business_unit_location.id GROUP BY location_id Results from direct query running the above query: INSERT INTO `` (`avg_recruit`, `avg_inspect`, `avg_contracts`, `avg_adjusters`, `roof_approval_avg`, `roof_jobs_avg`, `roof_jobs_total`, `siding_approval_avg`, `siding_jobs_avg`, `siding_jobs_total`, `gutter_approval_avg`, `gutter_jobs_avg`, `gutter_jobs_total`, `window_approval_avg`, `window_jobs_avg`, `window_jobs_total`, `total_jobs`, `total_collections`, `detail_id`, `title`) VALUES (95.3968, 92.8178, 106.9622, 90.2928, 103.5420, 103.5670, 4152, 100.2494, 106.8845, 4285, 120.1297, 86.2559, 3458, 92.9658, 106.1611, 4256, 16151, 4.281469, 12, 'St Paul, MN'); VALUES (90.2982, 73.3723, 97.8474, 104.5433, 97.7585, 86.1848, 1884, 109.9268, 109.3321, 2390, 81.0156, 96.4318, 2108, 91.7200, 123.8792, 2708, 9090, 4.531573, 13, 'Denver, CO'); Results from template: {'roof_jobs_total': Decimal('4152'), 'gutter_jobs_total': Decimal('3458'), 'avg_adjusters': Decimal('90.2928'), 'title': u'St Paul, MN', 'window_approval_avg': Decimal('92.9658'), 'total_collections': Decimal('4.281469'), 'gutter_approval_avg': Decimal('120.1297'), 'avg_recruit': Decimal('95.3968'), 'siding_approval_avg': Decimal('100.2494'), 'window_jobs_total': Decimal('4256'), 'detail_id': 12L, 'siding_jobs_avg': Decimal('106.8845'), 'avg_inspect': Decimal('92.8178'), 'roof_approval_avg': Decimal('103.5420'), 'roof_jobs_avg': Decimal('103.5670'), 'total_jobs': Decimal('16151'), 'window_jobs_avg': Decimal('106.1611'), 'avg_contracts': Decimal('106.9622'), 'gutter_jobs_avg': Decimal('86.2559'), 'siding_jobs_total': Decimal('4285')} Tried tweaking it a few ways and running the results through various for loops, keep getting the same result where my results are a single row through the Django template and the expected results (through console) have 2 rows The row that is coming back is the same as the first row returned through the console query so I'm thinking that it is running correctly just a matter of passing the results through... for good measure this is the code I'm using to generate the query (yes it's a bit ugly, been playing with it) def sql_grouped(table, fields, group_by=False, where=False): from django.db import connection query = 'SELECT %s FROM %s' % (fields, table) if where: query = query + ' WHERE %s' % (where) if group_by: query = query + ' GROUP BY %s' % (group_by) cursor = connection.cursor() cursor.execute(query) desc = cursor.description data = [dict(zip([col[0] for col in desc], row)) for row in cursor.fetchall()] return data[0] any feedback is greatly appreciated - been tinkering with this since I realized I could skip a few steps by generating my averages directly within the SQL rather than post-process

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  • Zend Framework how to echo value of SUM query

    - by Rick de Graaf
    Hello, I created a query for the zend framework, in which I try to retrieve the sum of a column, in this case the column named 'time'. This is the query I use: $this->timequery = $this->db_tasks->fetchAll($this->db_tasks->select()->from('tasks', 'SUM(time)')->where('projectnumber =' . $this->value_project)); $this->view->sumtime = $this->timequery; Echoing the query tells me this is right. But I can't echo the result properly. Currently I'm using: echo $this->sumtime['SUM(time)']; Returning the following error: Catchable fatal error: Object of class Zend_Db_Table_Row could not be converted to string in C:\xampp\htdocs\BManagement\application\views\scripts\tasks\index.phtml on line 46 Line 46 being the line with the echo in my view. I've been searching now for two days on how to figure this out, or achieve the same result in a different way. Tried to serialize the value, but that didn't work either. Is there somebody who knows how to achieve the total sum of a database column? Any help is greatly appriciated! note: Pretty new to zend framework...

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  • Reporting Services: Two Tables One Sum

    - by Neomoon
    My report is as follows: One table provides financial information with sums at the group footer (Grouping is called "StockTable_Shipped"). The group is controlled by a boolean value (1=shows shipped data, 0 = shows received data) The second table is a variance report for data that has been shipped (boolean value of 1) and has a sum at the bottom of the table. My ultimate goal is to take the sum from table1 where shipped=1 and subtract it from the variance sum from table2. This will be placed in a textbox at the bottom of the report. I understand if this sounds confusing but I would be more then happy to provide more information.

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  • SQL..Sum two rows

    - by Cha
    Sample data: LOCATION NAME LABEL1 LABEL2 SERVICE TIME NY Andrew A B HOUSE 2555 NY Andrew A B CAR 35 NJ Copley C A HOUSE 1025 NY Copley A B HOUSE 650 VA Dalton D C PET 25 What I want to do ias add another column where in it shows sum(Time) of rows with same data except for the Service.Also, the services that I need are only the sum of car and house.Is this possible? If not can you help me with the right query Sample output I need: LOCATION NAME LABEL1 LABEL2 SERVICE TIME SUM NY Andrew A B HOUSE 2555 **2590** NY Andrew A B CAR 35 NJ Copley C A HOUSE 1025 1025 NY Copley A B HOUSE 650 650

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  • I can't make this query work with SUM function

    - by Mehper C. Palavuzlar
    This query gives an error: select ep, case when ob is null and b2b_ob is null then 'a' when ob is not null or b2b_ob is not null then 'b' else null end as type, sum(b2b_d + b2b_t - b2b_i) as sales from table where ... group by ep, type Error: ORA-00904: "TYPE": invalid identifier When I run it with group by ep, the error message becomes: ORA-00979: not a GROUP BY expression The whole query works OK if I remove the lines sum(b2b_d+b2b_t-b2b_i) as sales and group by ..., so the problem should be related to SUM and GROUP BY functions. How can I make this work? Thanks in advance for your help.

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  • SELECT SUM returns a row when there are no records

    - by Jose L Martinez-Avial
    Hi, I'm finding some problems with a query that returns the sum of a field from a table for all the records that meet certain conditions. I expected to receive a "No records found' when there were no records, but instead I'm receiving a null result. SQL> SELECT * FROM DUAL WHERE 1=2; no rows selected SQL> SELECT SUM(dummy) FROM DUAL WHERE 1=2; SUM(DUMMY) ---------- SQL> Is there any way to not receive any record in that case?

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  • Changing a SUM returned NULL to zero

    - by Lee_McIntosh
    I have a Stored Procedure as follows: CREATE PROC [dbo].[Incidents] (@SiteName varchar(200)) AS SELECT ( SELECT SUM(i.Logged) FROM tbl_Sites s INNER JOIN tbl_Incidents i ON s.Location = i.Location WHERE s.Sites = @SiteName AND i.[month] = DATEADD(mm, DATEDIFF(mm, 0, GetDate()) -1,0) GROUP BY s.Sites ) AS LoggedIncidents 'tbl_Sites contains a list of reported on sites. 'tbl_Incidents containts a generated list of total incidents by site/date (monthly) 'If a site doesnt have any incidents that month it wont be listed. The problem I'm having is that a site doesnt have any Incidents this month and as such i get a NULL value returned for that site when i run this sproc, but i need to have a zero/0 returned to be used within a chart in SSRS. I've tried the using coalesce and isnull to no avail. SELECT COALESCE(SUM(c.Logged,0)) SELECT SUM(ISNULL(c.Logged,0)) Is there a way to get this formatted correctly? Cheers, Lee

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  • Complex Join - involving date ranges and sum...

    - by calumbrodie
    I have two tables that I need to join... I want to join table1 and table2 on 'id' - however in table two id is not unique. I only want one value returned for table two, and this value represents the sum of a column called 'total_sold' - within a specified date range (say one month).. SELECT ta.id, tb.total_sold as total_sold_this_week FROM table_a as ta LEFT JOIN table_b as tb ON ta.id=tb.id AND tb.date_sold BETWEEN ADDDATE(NOW(),INTERVAL -3 WEEK) AND NOW() this works but does not SUM the rows - only returning one row for each id... how do I get the sum from table b instead of only one row??? Please criticise if format of question could use more work - I can rewrite and provide sample data if required - this is a trivialised version of a much larger problem. -Thanks

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