Usaco Mar09 Bronze

Part of USACO Mar09

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BRONZE PROBLEMS
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Four problems numbered 11 through 14
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Problem 11: Payback [Jarrah Lacko, 2009]
"Never a borrower nor a lender be." O how Bessie wishes she had taken that
advice! She has borrowed from or lent money to each of N (1 <= N <=
100,000) friends, conveniently labeled 1..N.
Payback day has finally come. She knows she is owed more money than
she owes to the other cows. They have all lined up in a straight
line, cow i standing i meters from the barn. Bessie is going to
traverse the line collecting money from those who owe her and
reimbursing money to those she owes.
As she moves down the line, she can request any cow who owes her
money to give her the money. When she has enough money to pay off
any or all of her debts, she can pay the (recently collected) money
to those she owes. Cow i owes Bessie D_i money (-1,000 <= D_i <=
1,000; D_i != 0). A negative debt means that Bessie owes money to
the cow instead of vice-versa.
Bessie starts at the barn, location 0. What is the minimum distance
she must travel to collect her money and pay all those she owes? She
must end her travels at the end of the line.
PROBLEM NAME: iou
INPUT FORMAT:
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single integer: D_i
SAMPLE INPUT (file iou.in):
5
100
-200
250
-200
200
INPUT DETAILS:
Three cows owe Bessie; she owes two. When she's done, Bessie will have 150
money.
OUTPUT FORMAT:
* Line 1: A single integer that is the total metric distance Bessie
must travel in order to collect or pay each cow.
SAMPLE OUTPUT (file iou.out):
9
OUTPUT DETAILS:
Barn 100 -200 250 -200 200
| | | | | |
***>**+**>*****>**+
* < Bessie has 350
-**<***
* < Bessie has 150
***>****>****>**+
* < Bessie has 350
-**<***
*
***>*** < Bessie ends her travels with 150*
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Problem 12: Plumbing the Pond [Rob Kolstad, 2009]
Bessie's drinks water from a pond in the northwest part of the farm.
It has an interesting bottom in that it is full of little hills and
valleys. She wonders how deep it is.
She trolls across the pond in her little boat with a very old radar
set that tends to have spurious readings. She knows the deepest
part is relatively flat and has decided that she'll believe the
largest depth number only if it is verified by the fact that the
same depth appears in an adjacent reading.
The pond is modeled as an R x C (1 <= R <= 50; 1 <= C <= 50) grid
of (positive integer) depth readings D_rc (0 <= D_rc <= 1,000,000);
some readings might be 0 -- those are not part of the pond. A depth
reading of 10 means "depth of 10".
Find the greatest depth that appears in at least two 'adjacent'
readings (where 'adjacent' means in any of the potentially eight
squares that border a square on each of its sides and its diagonals).
She knows the pond has at least one pair of positive, adjacent
readings.
PROBLEM NAME: plumb
INPUT FORMAT:
* Line 1: Two space-separated integers: R and C
* Lines 2..R+1: Line i+1 contains C space-separated integers that
represent the depth of the pond across row i: D_rc
SAMPLE INPUT (file plumb.in):
4 3
0 1 0
1 2 0
1 5 1
2 3 4
INPUT DETAILS:
The pond has 4 rows, 3 columns.
OUTPUT FORMAT:
* Line 1: A single integer that is the depth of the pond determined by
following Bessie's rules.
SAMPLE OUTPUT (file plumb.out):
1
OUTPUT DETAILS:
Even though 5 is the deepest reading Bessie gets, and the number 2 occurs
twice, 1 is the largest number that occurs in two adjacent cells.
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Problem 13: The Perfect Cow [Rob Kolstad, 2009]
For the 39th year in a row, Farmer John was named "Dairy Farmer of
the Year". The Dairy Association wants him to exhibit his most
perfect cow at the upcoming Cow Convention in Herbster, Wisconsin
on the frigid shores of Lake Superior.
FJ keeps up with scientific literature and knows that beauty is
actually a trend toward the average rather than the existence of
some superior trait. Thus, he wants to find his most average cow
and share her beauty with the other Dairy Farmers during the weekend
of revelry and partying at the convention.
Happily, each of the N*N cows (2 <= N <= 99; N odd) has her beauty
rating (1 <= R_ij <= 1,000) inscribed on a tag in her ear, just
like in this picture.
Cows aren't so good at math, so FJ lines them up into an N x N
square. He asks them to find the median cow in each row (the median
cow is the one whose beauty number is bigger than or equal to half
the cows in her group and also smaller than or equal to half the
cows in her group -- the middle number of the group). From those N
medians, FJ then finds the median of those numbers and brings that
cow to the convention.
Given a set of N x N cows, find the beauty index of the most perfect
(average) cow.
PROBLEM NAME: perfect
INPUT FORMAT:
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains N space-separated integers that are
the N beauty indices for row i of the cow square
SAMPLE INPUT (file perfect.in):
5
1 5 3 9 5
2 5 3 8 1
6 3 5 9 2
8 8 3 3 2
5 4 4 4 4
INPUT DETAILS:
N=5, so there are 25 cows. They line up in a 5 x 5 square as shown.
OUTPUT FORMAT:
* Line 1: A single integer that is the index of the most perfect cow
as described in the task.
SAMPLE OUTPUT (file perfect.out):
4
OUTPUT DETAILS:
The medians of the rows are, respectively, 5, 3, 5, 3 and 4. The
median of those five values is 4.
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Problem 14: Dairy Queen [Traditional, 2009]
Bessie, always in need of an income, has decided to leverage her
dairy skills by taking a part-time job at the local Dairy Queen
restaurant. She is running a special cash register (since she has
hooves instead of fingers and thus requires special accommodation).
Her job is making change for customers as they make a purchase.
As she was counting out 83 cents in change, she wondered: "How many
ways can I count out 83 cents? I can use three quarters and eight
pennies, seven dimes and three pennies, 83 pennies... there must
be a zillion ways!"
How many different ways can one make change for N (1 <= N <= 300)
cents using coins from a set of C (1 <= C <= 8) coins of supplied
values C_i (1 <= C_i <= 200)? "Different" means differing counts
of coins.
Thus, 8 cents can be made, in American currency, with 1 five-cent
piece + 3 one-cent pieces and also with 8 one-cent pieces. Using 3
one-cent pieces + 1 five-cent piece is the same as 1 five-cent piece
+ 3 one-cent pieces, so one can create eight cents in just two
different ways. Note that some coin systems are poor at making
change and result in an answer of '0'.
Coin values in the input file are listed in descending order from
largest to smallest. All coin values will be distinct.
HINT: Consider recursion as a solution technique.
PROBLEM NAME: dq
INPUT FORMAT:
* Line 1: Two space-separated integers: N and C
* Lines 2..C+1: Line i+1 contains a single integer: C_i
SAMPLE INPUT (file dq.in):
83 5
50
25
10
5
1
INPUT DETAILS:
Eight-three cents; standard American coins with values 50, 25, 10, 5, and 1
OUTPUT FORMAT:
* Line 1: A single line with a single integer that is the number of
ways to create N cents of change using the supplied coins. The
answer is guaranteed to fit into a signed 32-bit integer.
SAMPLE OUTPUT (file dq.out):
159
OUTPUT DETAILS:
Here are 15 of the 159 ways of making 83 cents:
0 x 50 0 x 25 0 x 10 0 x 5 83 x 1
0 x 50 0 x 25 0 x 10 1 x 5 78 x 1
0 x 50 0 x 25 0 x 10 2 x 5 73 x 1
0 x 50 0 x 25 0 x 10 3 x 5 68 x 1
0 x 50 0 x 25 0 x 10 4 x 5 63 x 1
0 x 50 0 x 25 0 x 10 5 x 5 58 x 1
0 x 50 0 x 25 0 x 10 6 x 5 53 x 1
0 x 50 0 x 25 0 x 10 7 x 5 48 x 1
0 x 50 0 x 25 0 x 10 8 x 5 43 x 1
0 x 50 0 x 25 0 x 10 9 x 5 38 x 1
0 x 50 0 x 25 0 x 10 10 x 5 33 x 1
0 x 50 0 x 25 0 x 10 11 x 5 28 x 1
0 x 50 0 x 25 0 x 10 12 x 5 23 x 1
0 x 50 0 x 25 0 x 10 13 x 5 18 x 1
0 x 50 0 x 25 0 x 10 14 x 5 13 x 1
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page revision: 0, last edited: 06 Jun 2009 21:41