10980  Lowest Price in Town
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10980  Lowest Price in Town
Ok, I have no idea what I'm missing here, it sounds simple.
Just divide k with number of items in each combo (including single item "combo"), if not even, add one, multiply that number with the price of the combo (or "combo")... print out the minimum of those?
I thought that one was the easiest in the set, and I see a lot of people got it, but I didn't.
Any tricky input that someone can think of?
Darko
P.S. I was doing it in Java, so I suspect something with I/O, but doesn't have to be
Just divide k with number of items in each combo (including single item "combo"), if not even, add one, multiply that number with the price of the combo (or "combo")... print out the minimum of those?
I thought that one was the easiest in the set, and I see a lot of people got it, but I didn't.
Any tricky input that someone can think of?
Darko
P.S. I was doing it in Java, so I suspect something with I/O, but doesn't have to be
Once again I'm sorry if the word "buy" in output confuses you... The last sample test case is added to make things clearer.
Clearly "Get 3 for $40.00" looks clearer than "Buy 3 for $40.00", but I can do little to change now...
Clearly "Get 3 for $40.00" looks clearer than "Buy 3 for $40.00", but I can do little to change now...
7th Contest of Newbies
Date: December 31st, 2011 (Saturday)
Time: 12:00  16:00 (UTC)
URL: http://uva.onlinejudge.org
Date: December 31st, 2011 (Saturday)
Time: 12:00  16:00 (UTC)
URL: http://uva.onlinejudge.org
I ran out of ideas...
Darko
Code: Select all
22.00 2
2 22.00
4 60.00
2 4
25.00 2
2 48.00
2 46.00
2
22.00 2
2 22.00
4 40.00
1 2 3
10.00 3
2 15.00
3 20.00
50 334.00
1 2 3 4 5 17 49 50 51 53 99 100
100.00 3
2 0.01
100 0.00
17 10.00
1 100
12.00 0
1 2 100
999.99 1
51 999.99
1 50 51 52 99 100
Code: Select all
Case 1:
Buy 2 for $22.00
Buy 4 for $44.00
Case 2:
Buy 2 for $46.00
Case 3:
Buy 1 for $22.00
Buy 2 for $22.00
Buy 3 for $40.00
Case 4:
Buy 1 for $10.00
Buy 2 for $15.00
Buy 3 for $20.00
Buy 4 for $30.00
Buy 5 for $35.00
Buy 17 for $115.00
Buy 49 for $330.00
Buy 50 for $334.00
Buy 51 for $340.00
Buy 53 for $354.00
Buy 99 for $660.00
Buy 100 for $668.00
Case 5:
Buy 1 for $0.00
Buy 100 for $0.00
Case 6:
Buy 1 for $12.00
Buy 2 for $24.00
Buy 100 for $1200.00
Case 7:
Buy 1 for $999.99
Buy 50 for $999.99
Buy 51 for $999.99
Buy 52 for $1999.98
Buy 99 for $1999.98
Buy 100 for $1999.98

 Guru
 Posts: 1080
 Joined: Thu Dec 19, 2002 7:37 pm
Its a spoiler:mamun wrote:Would anybody give an idea how to form the dynamic relation here?
Calculate the array cheapest[], where cheapest[n] is the best price to buy n items, and calculate this array for n=1, 2, 3, etc. (increasing n).
Now there are many ways to buy n items:
 buy them for n times the unit price;
 buy k times an offer (m for the price of p), where k*m >= n, for k*p;
 buy a items and then b items for cheapest[a]+cheapest, where a+b = n, 0 < a,b <n;
but one of these ways is the cheapest, so store that in cheapest[n].
Can someone tell me (or give me a hint) WHEN this doesn't work ("comboSize" is N (1 for the single item) and "price" is P from the input). I really can't figure it out.
Darko
Code: Select all
cost = new long[110];
for (int i = 1; i < 110; i++) {
if (i == 2 && cost[1] == 0)
break;
long min = Long.MAX_VALUE;
for (int j = 0; j < prices.length; j++) {
if (prices[j].price == 0) {
min = 0;
break;
}
int q = i / prices[j].comboSize;
int r = i % prices[j].comboSize;
if(r == i  i == 1){
q = 1;
r = 0;
}
long c = q * prices[j].price + cost[r];
if (c < min)
min = c;
}
cost[i] = min;
}
What's prices.length?
What are you doing in
You mat try this I/O
Input
Output
What are you doing in
Code: Select all
int q = i / prices[j].comboSize;
int r = i % prices[j].comboSize;
Input
Code: Select all
10.00 2
3 25.00
4 32.00
3 4 5 6 7
Code: Select all
Case 1:
Buy 3 for $25.00
Buy 4 for $32.00
Buy 5 for $42.00
Buy 6 for $50.00
Buy 7 for $57.00

 Guru
 Posts: 647
 Joined: Wed Jun 26, 2002 10:12 pm
 Location: Hong Kong and New York City
 Contact:
Looking up to 110 isn't enough. Also, there's a simplier recurrence..Sometimes you need to buy more than 100 items for cheaper..
Check out http://www.algorithmist.com !
Thanks for the responses, guys.
prices.length is the length of the array holding the (comboSize,price) pairs (M+1 of them).
I increased 101 to 110 just in desperation, those are K's (that are never larger than 100) (I just tried 200 in another desperate attempt...)
I know that that method is slow, I keep dividing the same things and use only already calculated values for the remainders. But it runs in one second, so it's fine. I don't understand why I'm getting WA.
Yes, q is quotient and r is reminder. As in:
K = 10
N = 3, P = 10.00
I calculate the cost as:
q = 10/3 = 3
r = 1
c = q*price+cost[r] = 3*10.00 + cost[1]
Then I get minimum of those c's for all (N,P)'s (and I consider the initial one as (1, unitPrice)).
Even if you buy more than 100 items, it doesn't matter, I consider only the cost of the deal, not if there are more than 100 (if K<comboSize, I set q to 1 and r to 0).
If you think that 100 is the key, can you please provide some input, as I said, I ran out of ideas.
Darko
prices.length is the length of the array holding the (comboSize,price) pairs (M+1 of them).
I increased 101 to 110 just in desperation, those are K's (that are never larger than 100) (I just tried 200 in another desperate attempt...)
I know that that method is slow, I keep dividing the same things and use only already calculated values for the remainders. But it runs in one second, so it's fine. I don't understand why I'm getting WA.
Yes, q is quotient and r is reminder. As in:
K = 10
N = 3, P = 10.00
I calculate the cost as:
q = 10/3 = 3
r = 1
c = q*price+cost[r] = 3*10.00 + cost[1]
Then I get minimum of those c's for all (N,P)'s (and I consider the initial one as (1, unitPrice)).
Even if you buy more than 100 items, it doesn't matter, I consider only the cost of the deal, not if there are more than 100 (if K<comboSize, I set q to 1 and r to 0).
If you think that 100 is the key, can you please provide some input, as I said, I ran out of ideas.
Darko