USACO Training "Humble Numbers"

-

USACO Training "Humble Numbers"

Problem Statement

In this problem we are asked to find the n-th "humble number" of a set of primes of length K <= 100.
A humble number of a set is defined as a number whose prime factors are as subset of the set.
We are asked to find the n-th humble number of a set, when considered from lowest to greatest.

Notices

  1. Possibly most importantly, we notice that any humble number of a set can be created by multiplying elements of the set together.
  2. Clearly a recursive search (i.e. recursively multiplying elements by other elements) will take both too long (O(n^n)) and use too much memory (O(n!)). We try a DP approach.

DP Approach

We calculate the humble numbers in order (i.e. from the first to the Nth). For simplicity (and correctness), we assume that 1 is the 0th humble number.
Once we have the first X humble numbers, we compute the X+1-th humble number as follows:

  1. Set best=0x7FFFFFFF (largest signed long int) and bestPrime=-1
  2. For each prime P:
  3. find the smallest previous humble number H such that H*P is greater than the last (Xth) humble number;
  4. if this is smaller than best set best to this new value and set bestPrime to P;
  5. go back to step 2 until all primes are done
  6. set the X+1th humble number to be best

Optimizations

A rather glaring optimization is that H*P for the same P will always increase as X increases. So, we can set a "start value" for each prime in an array so we won't re-check unnecessarily previous H values that were deemed too small.

The second is that the humble number array will always be sorted, so we can use  <algorithm>'s std::upper_bound. But we need to find the humble number H such that H*P is greater than the Xth humble number, not directly comparing H! We think of comparing H with (Xth)/P, but std::upper_bound automatically casts double values to const int& values. Therefore, we use <functional>'s std::less<> - then, std::upper_bound does not need to (and will not) can (Xth)/P to a int.

Code

 

Comments

Post a Comment

Popular posts from this blog

USACO Training "subset": Subset Sums

-
USACO Training "subset": Subset Sums Problem Statement Subset Sums JRM For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical. For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical: {3} and {1,2} This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions). If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum: {1,6,7} and {2,3,4,5} {2,5,7} and {1,3,4,6} {3,4,7} and {1,2,5,6} {1,2,4,7} and {3,5,6} Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways. Your program must calculate the answer, not look it up from a table. PROGRAM NAME: s
Read more

USACO 2018 Open: Talent Show

-
Image
USACO 2018 Open Talent Show In Short Given N objects with value Vi and weight Wi, a collection of elements is said to have total value-to-weight ratio as (the sum of their value) divided by (the sum of their weight). Find the best possible value-to-weight ratio of these X objects, given the restraint that the collection must have total weight at least W. Return that ratio multiplied by 1000 and floored. Analysis It would be very difficult to calculate this optimal ratio - we'd have to either try all possible collections - O(2^n), where n is at most 250, about 10 to the power of 75 calculations, OR use a greedy heuristic - get the objects with the best ratio until it satisfies the weight limit. Neither of those are too to our liking, so how about searching for it? The search space is linear and not infinite - it asks for the ratio to the closest thousandths. The search space is sorted as well - if ratio X is achievable, then ratio Y < X is also achievable, if
Read more

OpenJudge 1768: Kadane's Algorithm on 2D Arrays

-
OpenJudge 1768: Kadane's Algorithm on 2-D Arrays Simple Version Given a 2-D matrix of integers, find the sub-matrix of this whose sum is maximized, and find that sum. The size of the matrix is no larger than 100 by 100. Analysis O(N^4) - Works, but naive     That's true, an naive O(N^4) version would work. We simply need to calculate (two dimensional!) prefix sums first. A quick review to what 2-D prefix sums are:     The 2-D prefix sum of two coordinates, X and Y, is the sum of all elements that have X-coordinate less than or equal to X, and Y coordinate less than or equal to Y.     It can be calculated through tabulation by removing overlapping areas.     Let P be the prefix sum array and V be the array of values, so the formula is as follows:     P[x, y] = V[x, y] - P[x-1, y] - P[x, y-1] + P[x-1, y-1].     We need to add the P[x-1, y-1] at the end since subtracting P[x-1, y] and P[x, y-1] we had overlapped the P[x-1, y-1]. Then, all we need to
Read more