BRINGING NLP TO JAVA CODING - CHATGPT CODE DEVELOPERTBUC, AlexandruLUCHICI, AndreiBOTEZATU, Mihai AlexandruJournal of Information Systems & Operations Management
public class ArrayModeInPlaceSnippet { /** * Returns the mode of the array. * * @param arr array to find mode in it * @return mode of array */ public static int modeArrayInPlace(int[] arr) { if (arr.length == 0) { return 0; } Arrays.sort(arr); int mode = arr[0]; int...
returnm * (n / gcd(m, n)); // parentheses importantto avoid overflow } //returna * b, stavingoff overflow as much as possible by cross-cancellation publicRational times(Rational b) { Rationala = this; // reduce p1/q2 andp2/q1, thenmultiply, wherea = p1/q1...
(python) morris inorder traversal W56 retainAll Qualcomm-Atheros 0 Tree 生成器模式 人工智能 大数据 硬件 云通讯 通讯 dongt Game Theory industrial wireless AP 量化交易 通信 2.4G&5G GCD Overload Exception Handling Game DP 离散化 Vue min yelp Oriented (python) 后端 802.11r 大数据 Euler Path ...
21-40 – Might need to refactor. 41-60 – You could possibly justify the code. (i.e. cost of refactor prevents action) 61-? – Danger, investigate when possible. Alright, with that let’s use it. Given the following Ruby: # euclids_equation.rb def gcd(x, y) while y != 0 r ...
int lcm(int a, int b){ return a * b / gcd(a, b); } 对于最大公约数问题,因为需要计算 a % b ,而这个操作是比较耗时的,可以使用 编程之美:2.7 的方法,利用减法和移位操作来替换它。对于a 和 b 的最大公约数 f(a, b),有:\1. 如果 a 和 b 均为偶数,f(a, b) = 2*f(a/2, b/...
HDU 1695 GCD(莫比乌斯反演) GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15050 Accepted Submission(s): 5762 Proble...HDU - 1695 GCD —— 莫比乌斯反演 GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 ...
foriinrange(31, -1, -1):# 从右向左找最多xor=1的儿子 bit = (n>>i) &1 ifbit ^1incur.children:# 找到就向儿子走(第i位的二进制可以满足异或结果为 1) cur = cur.children[bit ^1] res += (1<<i) else:# 没找到就向bit走,因为有可能之后会找到xor=1的儿子 ...
Solve nonlinear equations ti-83, worksheets for adding and subtracting integers 8th grade, dividing polynomials calculator, how to find lowest common multiple for a fraction in vba, online maths for dummies, pdf math problems with answers.
b[i] must fulfill that gcd(a[i], b[i+1]) = a[i+1] <=> b[i+1] € {x s.t. gcd(x/a[i+1], a[i]/a[i+1]) = 1 and a[i+1]|x and x <= m} Those are the number of co-primes numbers of a[i]/a[i+1] in range [0, (m/a[i+1])] To get the number of...