@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 ritugarg.c...@gmail.com 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 cegprak...@gmail.com 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)
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