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SRM 106
Thursday, August 1, 2002

The Problems

Division-II, Level 1: Insomnia (250pt)  discuss it
Submission rate - 91% (169/186)
Success rate - 93% (158/169)
Avg. Score - 216.78pt
Best Score - dre2xl, 248.09pt

I'll have to try this creative cure for insomnia sometime...

This problem can be simply solved by looping through all of the minutes and adding or subtracting sheep (+15 or -10) depending on where you are in the count. For the first hour, third hour, fifth hour, etc. add 15 sheep per minute. At all other times subtract 10 sheep per minute. Just make sure to switch from +15 to -10 and vice-versa every 60 minutes.

Division-II, Level 2 & Division-I, Level 1: Auction  discuss it

Div-II (500pt)
Submission rate - 75% (140/186)
Success rate - 59% (83/140)
Avg. Score - 313pt
Best Score - leibniz, 440.19pt

Div-I (250pt)
Submission rate - 99% (106/107)
Success rate - 79% (84/106)
Avg. Score - 196.67pt
Best Score - b0b0b0b, 241.02pt

A clever scheme for selling "counter-intuitive math problems" to math geeks was described in this problem. Given a number of email bids for a math paper, the task is to find the maximum possible profit, also given that the selling price will be equal to the maximum non-winning bid, and anyone bidding higher than the maximum non-winning bid will pay the non-winning bid price for the paper.

Probably the fastest and simplest way to solve this problem is to make a copy of the bids input, sort the copy, then iterate through the copy, looking for consecutive elements that are non-identical. In the event that consecutive elements differ, compute the profit which would ensue from using the lower of the two elements as the maximum non-winning bid. If the profit is greater than the maximum profit calculated so far, then store the maximum non-winning bid. After this loop simply generate the output by finding all of the indices with bids greater than the maximum non-winning bid.

Division-II, Level 3: Handles (1000pt)  discuss it
Submission rate - 35% (66/186)
Success rate - 23% (15/66)
Avg. Score - 481.4pt
Best Score - Gojira, 578.12pt

This problem was a straightforward parsing exercise to determine the complexity of a given string based on 11 pre-defined rules. None of these rules in themselves poses an extreme challenge, but coding all of them takes some time, and of course it only takes one bug to mess up your solution. Familiarity with easy-to-use text parsing routines in the language of your choice would be a definite asset for solving this problem in a timely fashion.

Division-I, Level 2: Dance (500pt)  discuss it
Submission rate - 49% (52/107)
Success rate - 63% (33/52)
Avg. Score - 283.76pt
Best Score - leibniz, 440.19pt

TopCoders need to have fun at dances too...

This problem asked you to find the total number of distinct combinations of dancing pairs, given a list of who everyone would be willing to dance with. The problem can be broken down by counting the number of distinct combinations with just 1 pair, the number of distinct combinations with 2 pairs, and so on. There are at most 10 elements to the input of this problem, so no more than 10 people are willing to dance with others. One good way to keep track of things for this problem (inspired by kv) would be to parse the input strings and form a bitmask for each of the dancers which describes which other dancers they were willing to dance with. Then start an array list whose elements describe all of the various dance combinations. Loop through all of the possible dancing pairs (in which person A is willing to dance with B and vice-versa) and if there is an element in the list whose bitmask does NOT indicate person A dancing with person B, then add an element to the list whose bitmask is the same as the previous bitmask, in addition to including person A dancing with person B. Also create an element in the list whose mask indicates just A dancing with B. The total number of dance combos will just be the number of elements in the list.

Division-I, Level 3: FakeCoin (900pt)  discuss it
Submission rate - 23% (25/107)
Success rate - 52% (13/25)
Avg. Score - 515pt
Best Score - SnapDragon, 679.06pt

Given that one coin out of twelve is known to be a fake, how do you figure out from a series of balance weighings which one is the fake, and whether it is heavier, lighter, etc.? dmwright came up with a rather elegant solution to this problem. He basically used two status flags for each coin to indicate whether or not that coin could be (a)"fake and heavier" or "not fake and heavier" and (b)"fake and lighter" or "not fake and lighter". All status flags for all 12 coins were initialized as being both fake and heavier and fake and lighter. Each of the balance readings were parsed, and the following steps were taken for the coins in each equation:

  1. If a coin appears on either side of an "=" sign, then it must be real (since there can only be one fake coin) so set the flags for this coin to be "not fake and heavier" and "not fake and lighter".

  2. If a coin appears to the left of a "<" sign or to the right of a ">" sign then there is no way that it can be fake and heavier, so set the "not fake and heavier" flag for this coin.

  3. If a coin appears to the right of a "<" sign or to the left of a ">" sign then there is no way that it can be fake and lighter, so set the "not fake and lighter" flag for this coin.

  4. If a coin does NOT appear to the left or right of any given "<" or ">" sign then it can not be a fake, once again for the reason that there can be only one fake coin. Set the flags for this coin to be "not fake and heavier" and "not fake and lighter".

Once these steps are completed for each of the balance equations, enough information exists to solve the problem. Count the number of fake coins by checking to see if either the "fake and heavier" or "fake and lighter" flags are set for each of the coins. If the number of fake coins is zero, then this is obviously a contradiction of the problem statement. Output "CONTRADICTION". Otherwise, output one of the following for each coin, depending on the status of its flags and the total number of fakes counted:

  1. If both flags for the coin are set to "not fake and heavier" and "not fake and lighter" then this coin is obviously not a fake, so output 'R'.

  2. Otherwise, one of the flags indicates a fake; if there is more than one fake counted (out of 12), then you have not been given enough equations to determine which coin is the fake, so output 'U' for this coin.

  3. If only one fake was counted and the "fake and heavier" flag is set, output 'H'.

  4. If only one fake was counted and the "fake and lighter" flag is set, output 'L'.

  5. If only one fake was counted and both the "fake and heavier" and "fake and lighter" flags are set, you are sure the coin is a fake, but have not been given enough equations to determine whether or not it is heavier or lighter, so output 'F'.

By Penwiper
TopCoder Member