@rashmi: there is no confusion for the third item.. so simply u buy all the third item without any offer
On Mon, Aug 15, 2011 at 12:52 AM, ritu <[email protected]> wrote: > solution is as following if problem is "buy all the n items for > minimum price if there are offers so that item j is free if customer > buys K numbers of item i" > > 1. create two parallel arrays cost[] (cost[i] = item[i] * K ) and > free[](free[i] = j ) > 2. sort cost[] > 3. now for highest priced item ,check if it freely avilable with any > lower cost item > 4. add this lower priced item with quantity K to the set > 5. else add single quantity of higher priced item to set. > 6. remove these item from array and similarly repeat for other items > > On Aug 11, 6:19 pm, cegprakash <[email protected]> wrote: > > there are n number of items available in the shop > > price[] {size n} gives the cost of each item > > and there are quantity[] {size n} means that there are quantity[i] > > number of i'th item > > > > the shop keeper provides some free items > > if you buy k nos of item i, you will get 1 item j for free (i may be > > equal to j) > > > > also there can be such offers for many items > > > > what you have to do is to buy all the items in shop with minimum > > expenditure. > > . > > > > source: own problem (i don't have the solution) > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
