for i in range(n - 1): x1, y1 = points[i] slope = defaultdict(int) # 用于统计以 points[i] 为起点的所有点的斜率 for j in range(i + 1, n): x2, y2 = points[j] dy, dx = y2 - y1, x2 - x1 g = gcd(dy, dx) # 利用公约数,化成最简形式, 3/6 == 2/4 ...
AI代码解释 classSolution:defhasGroupsSizeX(self,deck:List[int])->bool:importfunctoolsreturnfunctools.reduce(math.gcd,collections.Counter(deck).values())>1 关键点: counter实现计数 math.gcd实现求解最大公约数 reduce对列表中两两求解最大公约数 强大的内置库、简洁的python语言!
AI代码解释 #include<iostream>using namespace std;intgcd(int x,int y){return(x%y==0?y:gcd(y,x%y));}intmain(){int x,y,ans=0;cin>>x>>y;if(y%x){cout<<0;return0;}y=y/x;for(int i=1;i*i<=y;i++){if(y%i==0&&gcd(i,y/i)==1)ans+=2;}cout<<ans<<endl;}...
rt=''foriinrange(min_len): cur_s1= str1[:i+1] flag= 1forkinrange(i+1, len1, i+1):ifstr1[k :k+i+1] !=cur_s1: flag=0breakforkinrange(0, len2, i+1):ifstr2[k :k+i+1] !=cur_s1: flag=0breakifflag: rt=cur_s1returnrt#法二:classSolution:defgcdOfStrings(self, st...
(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 ...
intersection(set(total_legal_nums))) for num in legal_nums: for x in list(d): if math.gcd(num, x) == 1: d[num*x] += ct[num]*d[x] return (sum(d.values())-d[1]) % mod 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18...
Welcome to the official repository for the Generalized Category Discovery in Semantic Segmentation (GCDSS) project! Overview This repository contains the code implementation for GCDSS. We are currently in the process of organizing and refining the codebase to make it more user-friendly. Currently, ...
如果你想要关闭虚数i的支持, 可以按下Shift + Ctrl + P, 然后输入latex-sympy-calculator: Toggle Complex Number Support For Variances. 虚数功能类似这样:x = 1 + 2i,\int \cos xe^{-ikx}dx. 在VSCode 中执行 Python 表达式 你可以用快捷键Shift + Ctrl + Alt + P计算一个 Python 表达式. ...
0;};实际上求最大公约数还有更好的方式，比如辗转相除法：def GCD(a, b): if b == 0: return a return GCD(b, a % b)复杂度分析 时间复杂度：$O(log(max(a, b)))$空间复杂度：空间复杂度取决于递归的深度，因此空间复杂度为 $O(log(max(a, b)))$ 关键点分析 数论裴蜀定理 ...
Python3代码 classSolution:defgcdOfStrings(self, str1:str, str2:str) ->str: m, n =len(str1),len(str2)# solution twoifstr1 + str2 != str2 + str1:return""# 求最大公约数defgcd(a, b):returnaifb ==0elsegcd(b, a % b)returnstr1[:gcd(m, n)]if__name__ =="__main__"...