xor

问题 The problem is to find all the subarrays of the given array with xor of all its elements equal to zero. For example, if array contains elements [13,8,5,3,3] , the solution should give the indices of all subarrays like 0-2 , 3-4 , 0-4 , etc. The question is similar to the one asked here The only difference is that I want the indices of all the subarrays that satisfies the equation A0 xor A1 xor...xor An = 0 回答1: This is a fairly straightforward extension of the linked question. In Python, #

问题 I read that the ^ operator is the logical XOR operator in C#, but I also thought it was the "power of" operator. What is the explanation? 回答1: It is not the power of operator of C# since there is no such operator in C#. It is just the XOR operator. For "power of", use Math.Pow. As you can see from this page on the C# Operators, ^ is listed under the "Logical (boolean and bitwise)" category, which means it can both handle boolean values, and binary values (for bitwise XOR). 来源： https:/

问题 How can the XOR operation (on two 32 bit ints) be implemented using only basic arithmetic operations? Do you have to do it bitwise after dividing by each power of 2 in turn, or is there a shortcut? I don't care about execution speed so much as about the simplest, shortest code. Edit: This is not homework, but a riddle posed on a hacker.org. The point is to implement XOR on a stack-based virtual machine with very limited operations (similar to the brainfuck language and yes - no shift or mod).

问题 How do you do the XOR bitwise operation if you only have available the AND and the OR operations? 回答1: Creating my own scripting language - ChrisScript - you just need something like: #!/bin/chrish bit XOR (bit A, bit B) { bit notA; bit notB; IF (A == 0) notA = 1 ELSE notA = 0; IF (B == 0) notB = 1 ELSE notB = 0; F = ((A && notB) || (notA && B)); RETURN F; } Even without NOT, it can be emulated like this. But this is the best solution you're going to get without having some form of inverter.

问题 I am trying to perform the exclusive or of two byte arrays and return the result as a hex string. I have converted the two byte array to their corresponding binary string. Each byte will have bits since it has 8 bytes. byte[] key = { 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17, 0x18 }; byte[] PAN = { 0x12, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x23 }; Until now I have used a method that transforms the byte array into their corresponding binary string value, e.g. "10101010101". However when I

问题 #include <stdio.h> #include <string.h> int main() { char greeting[]="\nHello World!\n"; int a; for(int i=0; i<strlen(greeting); i++) greeting[i]^=111; for(int i=0; i<strlen(greeting); i++) greeting[i]^=111; printf("%s\n",greeting); scanf("%d",&a); } Output: Hell Why does it cut everything after spotting a letter corresponding to the XOR key's number (in this case, ASCII 'w')? In mathematical logic, N^N=0 and 0^N=N , doesn't it? 回答1: Because 'o' is ASCII code 111, and XORing 111 with 111

问题 I'm stuck at one problem i.e. to find a subarray whose xor is 0. I read somewhere that this can be done using TRIE data structure but I want the starting and ending indices of the array. For example, consider an array a = [3, 6, 13, 8 15] The subarray from 0 to 3 i.e. [3, 6, 13, 8] has xor equal to 0. (3 xor 6 xor 13 xor 8 = 0) I'm in search for an algorithm than can find those indices ([0, 3] in this case). Detailed answer would be very helpful. Update I tried the brute Force approach find

问题 I would like to XOR a very big file (~50 Go). More precisely, I would like to do so by XORing each block of 32 bytes of a plaintext file (because of lack of memory) with the key 3847611839 and create (block after block) a new cipher file. Thank You for any help!! 回答1: This sounded like fun, and doesn't sound like a homework assignment. I don't have a previously xor-encrypted file to try with,but if you convert one back and forward, there's no diff. That I tried atleast. Enjoy! :) This xor's

问题 So what I know about CRC, and also did a Java implementation is this: Having an initial message as a 16-bit polynomial, for instance 0x0617 65 0000.0110.0001.0111 this one gets another 16 of 0 bits 0000.0110.0001.0111|0000.0000.0000.0000 Then, having the divisor, 0x1021 0001.0000.0010.0001 (0, 5, 12) We allign it at the start of each "1" in our initial message, and do XOR between the bits, until there are no more 1s in the initial message. In total, there will be 6 XORs in our example. The

问题 Given the XOR problem: X = xor_input = np.array([[0,0], [0,1], [1,0], [1,1]]) Y = xor_output = np.array([[0,1,1,0]]).T And a simple two layered Multi-Layered Perceptron (MLP) with sigmoid activations between them and Mean Square Error (MSE) as the loss function/optimization criterion [code]: def sigmoid(x): # Returns values that sums to one. return 1 / (1 + np.exp(-x)) def sigmoid_derivative(sx): # For backpropagation. # See https://math.stackexchange.com/a/1225116 return sx * (1 - sx) # Cost