algorithm

问题 Is it possible to solve tower of hanoi halfway? I've done extensive research to look for codes that solves the user's configuration halfway but I've yet to find one. This is for an assignment and I require the code to take over from where the user has stopped solving and continue solving it for the user, without resetting the puzzle to square one. I understand that there are recursion algorithms out there readily available but that is not what I'm searching for. I'm searching for algorithms

问题 What is the most efficient way in python for picking multiple n consecutive integers from n consecutive list, picking up one integer from each list. Here n is quite large..say in the order of 100s. L1 = [5,3,2,7,1] L2 = [3,5,6,8,9,21,2] L3 = [5,3,6,7,3,9] I'd like to print out the ranges of consecutive integers from consecutive lists, where first element is picked up from first list, second element from second list and so on and so forth: Candidate solution [5,6,7], [1,2,3], [7,8,9] 回答1: L1 =

问题 So I'm looking at different ways to compute the Cartesian product of n arrays, and I came across the rather elegant solution (here on SO) of using the following code: import itertools for array in itertools.product(*arrays): print array Looking at the python doc page (I'm using 2.7, btw) for itertools.product() , it says the code is equivalent to the following: def product(*args, **kwds): # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy # product(range(2), repeat=3) --> 000 001 010 011 100

问题 Please note that this is required for a C# .NET 2.0 project ( Linq not allowed ). I know very similar questions have been asked here and I have already produce some working code (see below) but still would like an advice as to how to make the algorithm faster given k and s conditions. This is what I've learnt so far: Dynamic programming is the most efficient way to finding ONE (not all) subsets. Please correct me if I am wrong. And is there a way of repeatedly calling the DP code to produce

问题 I'm studying Tarjan's algorithm for strongly-connected components and the way it works is clear to me. Anyway there's a line I don't understand: // Consider successors of v for each (v, w) in E do if (w.index is undefined) then // Successor w has not yet been visited; recurse on it strongconnect(w) v.lowlink := min(v.lowlink, w.lowlink) else if (w.onStack) then // Successor w is in stack S and hence in the current SCC v.lowlink := min(v.lowlink, w.index) // ************* end if end for I

问题 I'm studying Tarjan's algorithm for strongly-connected components and the way it works is clear to me. Anyway there's a line I don't understand: // Consider successors of v for each (v, w) in E do if (w.index is undefined) then // Successor w has not yet been visited; recurse on it strongconnect(w) v.lowlink := min(v.lowlink, w.lowlink) else if (w.onStack) then // Successor w is in stack S and hence in the current SCC v.lowlink := min(v.lowlink, w.index) // ************* end if end for I

问题 I am using two arrays to accomplish a task of checking if values in array1 exist in array2. If so remove the content in array1 and keep checking till array1 is empty. If they dont exist just return from the function. I am using Javascript I implemented it using classic two for loops which gives a run time of o(n*2). I would like to know if there is any other efficient way to perform this operation using any other data structure that javascript supports. Below is my current implementation for

问题 Input Integers 'n' (up to 10^14) and 'm'(up to 10^3) Output Fib(n) modulo m Sample Cases Input: 239 1000 Output: 161 Input: 2816213588 239 Output: 151 Hint given in Question As it is not possible to iterate 'n' times (because n is huge), consider using Pisano Period(repetition of remainders when every element Fibonacci series is divided by any integer) Code which I wrote (maybe wrong, but passes above-mentioned cases) n, m = map(int, input().split()) a, b = 0, 1 fib_rems = [0, 1] # remainders

问题 I encountered this problem on an online coding platform: String S contains the binary representation of value V and we need to calculate number of operations after which this value becomes zero. There are 2 operations to perform: if V is even divide by 2 if V is odd subtract by 1 For example if S = "111", the function should return 5. Explanation: String S encodes to number V = 7. It requires five operations: V = 7, which is odd: subtract 1 to obtain 6 V = 6, which is even: divide by 2 to

问题 I encountered this problem on an online coding platform: String S contains the binary representation of value V and we need to calculate number of operations after which this value becomes zero. There are 2 operations to perform: if V is even divide by 2 if V is odd subtract by 1 For example if S = "111", the function should return 5. Explanation: String S encodes to number V = 7. It requires five operations: V = 7, which is odd: subtract 1 to obtain 6 V = 6, which is even: divide by 2 to