binary

问题 I am trying to solve a binary puzzle, my strategy is to transform a grid in zeros and ones, and what I want to make sure is that every row has the same amount of 0 and 1. Is there a any way to count how many 1s and 0s a number has without iterating through the number? What I am currently doing is: def binary(num, length=4): return format(num, '#0{}b'.format(length + 2)).replace('0b', '') n = binary(112, 8) // '01110000' and then n.count('0') n.count('1') Is there any more efficient

问题 I am trying to solve a binary puzzle, my strategy is to transform a grid in zeros and ones, and what I want to make sure is that every row has the same amount of 0 and 1. Is there a any way to count how many 1s and 0s a number has without iterating through the number? What I am currently doing is: def binary(num, length=4): return format(num, '#0{}b'.format(length + 2)).replace('0b', '') n = binary(112, 8) // '01110000' and then n.count('0') n.count('1') Is there any more efficient

问题 I am trying to solve a binary puzzle, my strategy is to transform a grid in zeros and ones, and what I want to make sure is that every row has the same amount of 0 and 1. Is there a any way to count how many 1s and 0s a number has without iterating through the number? What I am currently doing is: def binary(num, length=4): return format(num, '#0{}b'.format(length + 2)).replace('0b', '') n = binary(112, 8) // '01110000' and then n.count('0') n.count('1') Is there any more efficient

问题 I am trying to solve a binary puzzle, my strategy is to transform a grid in zeros and ones, and what I want to make sure is that every row has the same amount of 0 and 1. Is there a any way to count how many 1s and 0s a number has without iterating through the number? What I am currently doing is: def binary(num, length=4): return format(num, '#0{}b'.format(length + 2)).replace('0b', '') n = binary(112, 8) // '01110000' and then n.count('0') n.count('1') Is there any more efficient

问题 I am struggling to pixelate an image which is made up of RGB values stored in a binary (P6) PPM file. The steps to pixelate the image are as follows: Read in the binary data and store it in a 1-dimensional array Iterate through the data stored in this array, in cells of size 'c' (row x columns). This variable 'c' can be changed by the user, but for this program it's currently set to int c = 4; meaning it iterates through pixel data in block dimensions of 4 x 4 Now find the average colour

问题 I am struggling to pixelate an image which is made up of RGB values stored in a binary (P6) PPM file. The steps to pixelate the image are as follows: Read in the binary data and store it in a 1-dimensional array Iterate through the data stored in this array, in cells of size 'c' (row x columns). This variable 'c' can be changed by the user, but for this program it's currently set to int c = 4; meaning it iterates through pixel data in block dimensions of 4 x 4 Now find the average colour

问题 A = 110000000 - 384 Blue+Red B = 011000010 - 194 Green+Black+Red A & B = C = 010000000 - 128 Red How can I check if B contains all the bits in A and perhaps others? In the case above I would like to get "false". I'm using XCode & objective-c but that shouldn't matter as far as I know 回答1: B contains A if A & B (ie, the intersection) is equal to A: (a & b) == a Which is analogous to a ⊆ b ↔ (a ∩ b) = a from set theory. 回答2: If you mean exactly the same bits, the test is A == B . If you mean B

问题 A = 110000000 - 384 Blue+Red B = 011000010 - 194 Green+Black+Red A & B = C = 010000000 - 128 Red How can I check if B contains all the bits in A and perhaps others? In the case above I would like to get "false". I'm using XCode & objective-c but that shouldn't matter as far as I know 回答1: B contains A if A & B (ie, the intersection) is equal to A: (a & b) == a Which is analogous to a ⊆ b ↔ (a ∩ b) = a from set theory. 回答2: If you mean exactly the same bits, the test is A == B . If you mean B

问题 Example, N = 3 The first N integers for value 3 is 1 , 2 , 3 Binary of 1 is 1 2 is 10 3 is 11 Concatenations of N=3 of binary values will be 11011 And the decimal value returned for the binary value 11011 is 27 The code I am using below only works for first integers N<=15 String input = ""; for(int i = 1;i<=n;i++) { input += (Integer.toBinaryString(i)); } return Integer.parseInt(input,2); For larger N numbers, any ideas on solving using modulo 10^9 + 7 (since concatenation is large) 回答1: Note

问题 I am communicating with an API using HTTP.client in Python 3.6.2. In order to upload a file it requires a three stage process. I have managed to talk successfully using POST methods and the server returns data as I expect. However, the stage that requires the actual file to be uploaded is a PUT method - and I cannot figure out how to syntax the code to include a pointer to the actual file on my storage - the file is an mp4 video file. Here is a snippet of the code with my noob annotations :)