https://cs.stackexchange.com/ is probably better for this. Also I’m pretty sure that $$ formatting only works on some StackExchange sites. But anyways, think about what this algorithm is doing at each step. We start with p = A_n. Then we take p = p*x + A_{n-1}. So what is this doing? We now have p = … Read more
If anyone will search answer, I did it! import math x = int(input(“How many cities? “)) a = math.factorial(x) print(“For”, x, “cities, there are”, a, “possible routes”) solved Traveling Salesman assignment
You caught me on a day where I don’t want to do what I’m supposed to be doing1. The code below only generates token output; it doesn’t attempt to store tokens anywhere. You can redirect the output to a file, but as others have pointed out, you’re going to need a bigger boat hard drive … Read more
The function can look as it is shown in the demonstrative program #include <iostream> #include <algorithm> #include <vector> long long int solution( const std::vector<int> &v ) { long long int max_sum = 0; for ( auto it = v.begin(); ( it = std::find_if( it, v.end(), []( int x ) { return !( x < 0 … Read more
I am so sorry , I posted this question here, as suggested by one of the users, I googled a few different things online an dhave found a ruby file which has been built for the same purpose. And , if anybody would like to access it, they may check it out online at http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/298382 … Read more
Use Base64. It is widely used and probably already supported on your target framework. Just as base 10 allows only 10 possible glyphs within a character, base 64 allows 64 glyphs per characters. To express a 32 bit number such an an IP address, you would need at a base 64 number that has at … Read more
There a number of possible measures of similarity. Ideally, you should derive one yourself that takes account of the reason why you are doing this classification, so that good similarity scores amount to something that performs well when you use it in practice. Here are a few examples. 1) Cosine similarity. Treat the two sets … Read more
Note that if r is a primitive root modulo p. Then we can reduce complexity of the sum. We have to find S = a1*1 + a1*r + a1*r^2 + … + a1*r^n. Then we write S in the closed form as S = a1*(r^n – 1) / (r – 1). Now it can be … Read more
If this is homework I guess your professor explained you how to compute Fibonnacci numbers and expect you to use the same trick. So basically remark that: |1+a+ a²+…+ a^n| |a 0 1| |1+a+ a²+…+ a^(n-1)| | a+2a²+…+na^n| = |a a 1| | a+2a²+…+(n-1)a^(n-1)| |1 | |0 0 1| |1 | |a 0 1|^n |1| … Read more
We can list out a few terms T(0) = 0 T(1) = 1 * 0 + 1 T(2) = 2 * 1 + 2 = 4 T(3) = 3 * 4 + 3 = 15 T(4) = 4 * 15 + 4 = 64 … We can note a couple of things. First, the function … Read more
The classic recursive implementation of fib is this: int fib(int n) { if (n < 2) return n; else return fib(n – 1) + fib(n – 2); } This function recurses so many times, it is used as an easy benchmark to measure function call efficiency in various languages, with a time complexity of O(φn). … Read more
If I understand you correctly then all that code is just to avoid printing numbers that have more than eight zeroes You can do that like this my $credit_card_number = “$a$b$c$d$e$Fc$g$h$i$j$k$l$m$n$o$p”; print $credit_card_number, “\n” unless $credit_card_number =~ tr/0// > 8; But I still think your enterprise is a fruitless one solved Generateing all possible 16 … Read more
A few things on readability: First codeblock: I would have the counters counting in the same direction– it will still compare each word this way. It’s not a terribly important change, but it can save the reader a step so that they don’t have to do any mental math to determine if the code is … Read more
There will be exactly n iterations of the first loop. There will be exactly n(n+1)/2 iterations of the second loop. Inside of the second loop there is roughly 1/3 of n/2 long loops, 1/3 of j long loops and 1/3 of 2 long loops. If we look only at the first of the three possible … Read more
Not sure what you were trying to do in your code, but just start at 1, check and increment. The loop will stop as soon as $i is not found in the array: function solution($A) { for($i=1; in_array($i, $A); $i++); return $i; } 1 solved Smallest integer that is not present in the array