day12 --------------------------------------------------------------- 实例019:完数 题目 一个数如果恰好等于它的因子之和,这个数就称为"完数".例如6=1+2+3.编程找出1000以内的所有完数. 分析:如果能被a整除,那么a是因子,放入一个列表然后再求和,与原被除数相等,就是完数 1 for i in range(1,int(a)): 2 list = [] 3 if i >1: 4 for j in r…
import java.util.ArrayList; import java.util.List; //一个数如果恰好等于它的因子之和,这个数就称为"完数".例如6=1+2+3.第二个完全数是28, //它有约数1.2.4.7.14.28,除去它本身28外,其余5个数相加, //编程找出1000以内的所有完数. public class Test { public static void main(String[] args) { System.out.print("完数有…
a.b只要数字a能被数字b整除,不论b是不是质数,都算是a的因子.比如:8的质因子是 2, 2, 2,但8的因子就包括 1,2,4. import math for i in range(2, 1000): factors = [] #因子列表,i 每次循环都清空 for j in range(1, math.floor(i/2)+1): if i%j == 0: factors.append(j) if sum(factor…
Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.For example:Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].Note:1.The order of…
Given an array of integers, every element appears three times except for one, which appears exactly once. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? class Solut…
Given an array of integers, every element appears twice except for one. Find that single one. class Solution { public: int singleNumber(vector<int>& nums) { int size=nums.size(); ||nums.empty()) ; ; ;i<size;++i) res^=nums[i]; return res; } };…
给定一个整数数组,除了某个元素外其余元素均出现两次.请找出这个只出现一次的元素.备注:你的算法应该是一个线性时间复杂度. 你可以不用额外空间来实现它吗? 详见:https://leetcode.com/problems/single-number/description/ Java实现: class Solution { public int singleNumber(int[] nums) { int n=nums.length; if(n==0||nums==null){ return In…
题目十:判断101-200之间有多少个素数,并输出所有素数. #!/usr/bin/env python3 # -*- coding: utf-8 -*- """ 题目十:判断101-200之间有多少个素数,并输出所有素数.""" #首先判断不是素数 lst = [] for i in range(101, 201): for x in range(2, i): if (i % x) == 0: lst.append(i) break #再找出素…
// 生成n个随机数,要求n个数的和等于100 function lessANumber(n, v) { var i, s = 0, r = [], x = v; for (i = 1; i < n; i++) { x = Math.random() * x; r.push(x.toFixed(0)); s += x; x = v - s; } r.push(x.toFixed(0)); return r; } function random(Min,Max){ var Range = Max…
第二课主要介绍第一课余下的BFPRT算法和第二课部分内容 1.BFPRT算法详解与应用 找到第K小或者第K大的数. 普通做法:先通过堆排序然后取,是n*logn的代价. // O(N*logK) public static int[] getMinKNumsByHeap(int[] arr, int k) { if (k < 1 || k > arr.length) { return arr; } int[] kHeap = new int[k];//存放第k小的数 for (int i =…
