[ad_1] You could use the given number as a seed and generate the number using the standard random number function. function getrand($num) { srand($num); return rand(1 , 30); } 0 [ad_2] solved generate unqie life time random number for each number [closed]
[ad_1] Ask yourself, in which way does i increment grow towards the final value n? How many times will the outer loop run for a given n? The same for the inner loop. I recommend you read through somthing like this or this SO post and maybe start with some examples: n = 100; i … Read more
[ad_1] It depends how many queries you can expect, and the number of identical queries you can expect. One approach is to “memoize” queries, simply to store each query and result, and look that up before doing more serious work. A more problem-specific approach – probably what your teacher is after – is to compute … Read more
[ad_1] Here’s a method using Array.forEach(): const number = 165; const candidates = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]; //looking for the lowest valid candidate, so start at infinity and work down let bestCandidate = Infinity; //only consider numbers higher than the target const higherCandidates = candidates.filter(candidate => candidate > number); … Read more
[ad_1] A loop won’t return the String you want this way, you need to create the String and, at every loop, concatenate a new one to it: public static String m(Map<String, String> mp) { Iterator<?> it = mp.entrySet().iterator(); String str = “{” + ” \”SomeAttribute\”: ” + “{“; while (it.hasNext()) { Map.Entry pair = (Map.Entry) … Read more
[ad_1] I would start small, and build up. The smallest (n=1) is simply: * that clearly doesn’t work since there are 0 neighbors (and even number). So no solution exists for n=1. Next, try n=2. Only one choice: ** This works. Now what about n=3? Doesn’t work, no solution for n=3. Now, how can you … Read more
[ad_1] You still haven’t asked any question, but I assume that your program gives wrong answer (as it really do) and you don’t know why. For example: for n=16 it prints: 1234 1234 1234 1234 The following code fragment is responsible: array[i][j]=j+1; System.out.print(array[i][j]); Firstly, you don’t print spaces, so numbers aren’t separated. Secondly, you have … Read more
[ad_1] Write an algorithm accepts a positive integer n as input, when executed, prints out a list of all self-aware arrays of length n [ad_2] solved Write an algorithm accepts a positive integer n as input, when executed, prints out a list of all self-aware arrays of length n
[ad_1] The comment you wrote in @OBu’s answer is about only a quarter right: 1*n + 2*(n-1) + 3*(n-2) + … +n*1 That equals to: Sum(i=1..n, i*(n-i+1)) = n*Sum(i) – Sum(i*i) + n = n*[n(n+1)/2] – [n(n+1)(2n+1)/6] + n If you want, feel free to compute the exact formula, but the overall complexity is O(n^3). … Read more
[ad_1] It sounds like you’re trying to verify that all items in list2 are contained within at least one item in list1. If that’s the case, this should do the trick: bool list1ContainsList2Items = list2.All(l2 => list1.Any(l1 => l1.Contains(l2))); 2 [ad_2] solved Quickest way to compare two generic lists C# [closed]
[ad_1] f(x) would be 0. But this should hardly ever occur. g(x)=0 means you had no costs to reach x (should only be the case for the starting point) h(x)=0 means the heuristics says that the costs to reach the goal from x costs not more than 0 (means that you are at the goal) … Read more
[ad_1] You problem is with floating point precision. float i = 4.2; i *= 100; printf(“%f\n”, i); Print: 419.999969 and not 4.2 as it should, in this case 419 is the value used in the coin problem resulting in 22 coins used 16 of 25, 1 of 10, 1 of 5 and 4 of 1 … Read more
[ad_1] If your a double loop that solves this problem breaks out of the loops as soon as a match is found, then both methods are similarly inefficient – they both require iterating over all of N elements M times, worst-case, where N is the length of one array and M is the length of … Read more
[ad_1] A typical fast solution to the gcd algorithm would be some iterative version like this one: def gcd(x, y): while y != 0: (x, y) = (y, x % y) return x In fact, in python you’d just use directly the gcd function provided by fractions module. And if you wanted such function to … Read more
[ad_1] It seems you are stuck because you are not familiar with the modelisation effort of going from the assignment statement to a solution design. This is a difficult task for everybody, not just beginners. But in your assignment you already got some guidelines. First you have to understand what do you get as input … Read more