This is a case of undefined behavior: if(!visited[j]) is undefined. visited is not initialized because the call memset(visited, sizeof(visited), false); is wrong. You are reading uninitialized variables. The declaration of memset is void *memset( void *dest, int ch, size_t count ); You are writting 0 times the value 10000 into visited. On your machine this … Read more
This is a very “basic” example: chars=”abcdefghijklmnopqrstuvwxyz” my_list = [] for c1 in chars: for c2 in chars: for c3 in chars: for c4 in chars: my_list.append(c1+c2+c3+c4) print(my_list) solved Go from aaaa to zzzz python [closed]
Let’s start with a slightly simpler problem. All keys are 8 numeric digits, and you have 10 machines. Simple enough – one machine checks 0???????, another checks 1??????? and so on. Note that this slicing doesn’t care that you’re doing multiprocessing – it’s just allocating ranges by fixing one of the digits. Your version is … Read more
If the encryption method is an accepted standard method such as AES and the key is good it is not possible to find the key given the plain and encrypted data. If the encryption method is trivial such as Caeser Cipher then it can be found. If the key is trivial such as “password” or … Read more
First of all, you need to provide a minimal reproducible example of your code. You can check here: https://stackoverflow.com/help/minimal-reproducible-example In regards to your question, using three loops can be a solution: import java.util.ArrayList; public class Main { public static void main(String[] args) { var combinations = new ArrayList<int[]>(); for (int i = 0; i < … Read more
This can be solved in Dynamic Programming (DP) by following the recursive formula: D(x) = INFINITY x < 0 D(0) = 0 D(x) = min{ D(x-1), D(x-15), D(x/2) } + 1 In bottom-up DP it will be something like: (pseudo code) D = new int[151] D[0] = 0 for (int i = 1; i <151; … Read more
Recall that a Caesar Cipher has only 25 possible shifts. Also, for text of non-trivial length, it’s highly likely that only one shift will make the input make sense. One possible approach, then, is to see if the result of the shift makes sense; if it does, then it’s probably the correct shift (e.g. compare … Read more
You could store the intermediate value in a dictionary and do a kind of a dynamic programming. numbers = {1:0} max = -1 startMax = -1 for i in range(2, 1000 000): n = i steps = 0 while n>=i: if n&2 == 0: n = n/2 else: n = 3*n + 1 steps = … Read more
Try this: with open(‘file.out’, ‘w’) as output: for n in xrange(100000000): s = “{0:08d}”.format(n) output.write(s[:2] + ‘-‘ + s[2:] + ‘\n’) … But be aware that that’s a lot of combinations, it’s possible that you’ll run out of memory, and if not anyway the resulting file will be huge and the program will take a … Read more
//… generate(apple,i); //… you have to specify the arguments in the signature of your function ‘generate’ like: void generate( string apple, int it ) { And the error: size_t i; //… for(i=0; i <= pear.length(); i++) { //logic here?! } you add an element to a string using: string.push_back( ‘a’ ); solved C++ Error: ‘no … Read more