ICL 2010 Advantage Round Solutions

BITS ACM Advantage Round ICL Solutions 2010 Mayank Abhishek APOGEE 2010

Addition of Big Numbers Use Windows/Linux calculator or solve manually. Coding in C/C++ is the last resort. If you use Linux then this Shell Script solves all input files in one go: for i in `seq 0 9`; do cat $i.in | sed 's/ /+/' | bc | tail -`head -1 $i.in` >$i.out ; done ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 1

Powered Fractions Sort on the basis of . Any sort algorithm will do. Since most input files contained only 2 fractions you could also solve them manually. For finding the GCD to reduce the fractions you could use, function gcd(a, b) if a = 0 return b while b ≠ 0 if a > b a := a − b else b := b − a return a ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 2

Playing With Fibonacci Numbers Do you really need the solution for this? The only catch was to not use recursion which could blow up the runtime. ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 3

Playing With Tiles The following recursive formula solves the problem, F(1) = 1 F(2) = 2 F(N) = F(N-1) + F(N-2) How to arrive on this formula? For each tiling arrangement of size N-1 you can add a vertical tile to get an arrangement of N tiles. For each arrangement of size N-2 you can add 2 horizontal tiles to get an arrangement of N tiles. Solution for N=1 and N=2 are trivial. We do not count adding two vertical tiles in front of all arrangements of size N-2. Why? Again in this problem, do not use recursion. Iterate wisely. Take care of the limits of the data type you used for the variables as the numbers generated will be huge. Always keep F(N) as F(N) mod 10000. ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 4

Three Way Partitions Create an array SUM[N] where SUM[i] stores the sum of all the numbers upto i in the list. Now you can calculate the sum of numbers between positions k and m (both inclusive) in one step it will be, SUM[k]-SUM[m-1] Now we have to place two splitters. Try all combinations i.e NC2. Take MIN over the MAX from each combination. ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 5

More Players In The Game Create an array SUM[N] where, SUM[i] = SUM[i-1] + VALUE[i], if VALUE[i] is non-zero SUM[i] = 0, otherwise Take the MAX over the array SUM[N] ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 6

Chocolate Cookies The array WAYS[M][N] will contain the number of ways of reaching i, j from 1,1 for each i, j. WAYS[1][1] = 1, WAYS[i][j] = WAYS[i-1][j]+WAYS[i][j-1], if i, j is not blocked WAYS[i][j] = 0, otherwise Handle the cases for first row and first column. Also, be careful to keep only the last 4 digits as values grow very large. ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 7

The Winner Takes It All For each pair of buildings (both inclusive) calculate two things,  The minimum height of buildings between them.  The sum of widths of all buildings between them. Multiply these two to get the largest rectangular area between the two buildings which includes both. Take the MAX over all areas so found. This can be done in O(N2), which was expected. An O(NlogN) solution is also possible, but was not required. ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 8

Poof Go Clouds & Pyramids - Pyramids Poof Go Clouds can be solved by using a simple DFS (Depth First Search) [http://en.wikipedia.org/wiki/Depth-first_search] or BFS (Breadth First Search) [http://en.wikipedia.org/wiki/Breadth-first_search]. Pyramids – Pyramids was Longest increasing subsequence [http://en.wikipedia.org/wiki/Longest_increasing_subsequence]. ICL 2010 ADVANTAGE ROUND SOLUTIONS Page 9

ICL 2010 Advantage Round Solutions

by Mayank Abhishek on Mar 22, 2010

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ICL 2010 Advantage Round Solutions — Document Transcript