math

问题 I'm creating a little game in Javascript, and I have trees that can hold a certain amount of max experience, and each tree have a varied amount of branches that also needs to be leveled up with experience, and they should total to the tree max experience. If it was that easy, I would just divide them equally, like max / branches , but since it's for a game, I need to make a formula where the first branch needs no experience, and then each branch experience needed steadily increase (while

问题 # import packages we need later import matplotlib.pyplot as plt import numpy as np What I am doing Inspired by this question & answer, I am fitting a series of Legendre polynomials to a time series: curve1 = \ np.asarray([942.153,353.081,53.088,125.110,140.851,188.170,70.536,-122.473,-369.061,-407.945,88.734,484.334,267.762,65.831,74.010,-55.781,-260.024,-466.830,-524.511,-76.833,-36.779,-117.366,218.578,175.662,185.653,299.285,215.276,546.048,1210.132,3087.326,7052.849,13867.824,27156.939

问题 # import packages we need later import matplotlib.pyplot as plt import numpy as np What I am doing Inspired by this question & answer, I am fitting a series of Legendre polynomials to a time series: curve1 = \ np.asarray([942.153,353.081,53.088,125.110,140.851,188.170,70.536,-122.473,-369.061,-407.945,88.734,484.334,267.762,65.831,74.010,-55.781,-260.024,-466.830,-524.511,-76.833,-36.779,-117.366,218.578,175.662,185.653,299.285,215.276,546.048,1210.132,3087.326,7052.849,13867.824,27156.939

问题 From What’s New In Python 3.7 we can see that there is new math.remainder. It says Return the IEEE 754-style remainder of x with respect to y. For finite x and finite nonzero y, this is the difference x - n*y , where n is the closest integer to the exact value of the quotient x / y . If x / y is exactly halfway between two consecutive integers, the nearest even integer is used for n . The remainder r = remainder(x, y) thus always satisfies abs(r) <= 0.5 * abs(y) . Special cases follow IEEE

问题 From What’s New In Python 3.7 we can see that there is new math.remainder. It says Return the IEEE 754-style remainder of x with respect to y. For finite x and finite nonzero y, this is the difference x - n*y , where n is the closest integer to the exact value of the quotient x / y . If x / y is exactly halfway between two consecutive integers, the nearest even integer is used for n . The remainder r = remainder(x, y) thus always satisfies abs(r) <= 0.5 * abs(y) . Special cases follow IEEE

问题 I'm currently working on a modified version of the Shunting Yard Algorithm that would work with variables, but I cant figure out how to get it to work. For example, I would want the algorithm to re-write 2 * (2x + 5) - 5 to 4x + 5. Any ideas / links to already implemented algorithms that does this already? 回答1: Take the expression: 2 * (2x + 5) - 5 Add the * symbol to make it more understandable for the computer: 2 * (2*x + 5) - 5 Parse it using the Shunting Yard Algorithm, it becomes: 2 2 x

问题 So, I want to write a function in code using some sort of algorithm to calculate any number to any power, including decimals. I use JavaScript and it already has an inbuilt pow function: Math.pow(2, 0.413) // 2^0.413 = 1.331451613236371, took under 1 second. Now I want to write my own like this: function pow(x, y) { // Algorithm } This is a function that calculates the square root of any number (x^0.5), and it's very accurate with only 10 loops: function sqrt(x, p) { // p = precision

问题 So, I want to write a function in code using some sort of algorithm to calculate any number to any power, including decimals. I use JavaScript and it already has an inbuilt pow function: Math.pow(2, 0.413) // 2^0.413 = 1.331451613236371, took under 1 second. Now I want to write my own like this: function pow(x, y) { // Algorithm } This is a function that calculates the square root of any number (x^0.5), and it's very accurate with only 10 loops: function sqrt(x, p) { // p = precision

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