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解读 Implementing data cubes efficiently
概述 论文链接：Implementing data cubes efficiently 预计算基于空间换时间实现查询性能提升，物化视图是数据立方体(data cubes)的一种实现方式。
39600编辑于 2025-02-20 - 来自专栏程序编程之旅
HDOJ 1334 Perfect Cubes(暴力)
Problem Description For hundreds of years Fermat’s Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.) It is possible, however, to find integers greater than 1 that satisfy the “perfect cube” equation a^3 = b^3 + c^3 + d^3 (e.g. a quick calculation will show that the equation 12^3 = 6^3 + 8^3 + 10^3 is indeed true). This problem requires that you write a program to find all sets of numbers {a, b, c, d} which satisfy this equation for a <= 200.
39320发布于 2021-01-21 - 来自专栏Skykguj 's Blog
CF1619B Squares and Cubes 题解
我的思路是，从 1 到 \sqrt{n} 循环筛一下 完全平方数，再从 1 到 \sqrt[3]{n} 循环筛一下 完全立方数，去掉重复的数字，然后输出 size。
35230编辑于 2022-09-09 【杭电oj】1334 - Perfect Cubes（水）
Perfect Cubes Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
27110编辑于 2025-08-26- 来自专栏饶文津的专栏
【Gym - 101164I】Cubes（dfs，剪枝）
根据n的范围，立方数最大不超过400的立方，并且个数也不会很多。 dfs，设置一个深度的上限up。从大到小枚举立方数，剪枝条件：当前层数加上至少还需要的层数>=up就return。
27810发布于 2020-06-02 - 来自专栏壹种念头
Unity基础教程系列(新)（五）——计算着色器（Rendering One Million Cubes）
这是关于学习使用Unity的基础知识的系列文章中的第五篇。这次，我们将使用计算着色器显著提高图形的分辨率。
4.7K12发布于 2021-03-10 - 来自专栏机器学习、深度学习
运动相机检测无人机-- Detecting Flying Objects using a Single Moving Camera
这里我们使用 3D descriptors 从 spatio-temporal image cubes 提取特征，对这些特征进行分类，判断其是否是感兴趣目标或威胁物体 we detect whether ， 我们将 spatio-temporal image cubes 称之为 st-cubes，看看下面的示意图吧： 下图每一行对应一个 st-cube ? 对每个 st-cubes 进行 运动补偿，建立 stabilized st-cubes 4）对每个 st-cubes进行分类，判断其是否含有感兴趣目标 5）因为每个尺度是独立处理的，在同一个位置我们在尺度空间进行非极大值抑制 这里我们介绍两个分类器用于判断每个 st-cubes 是否含有感兴趣目标 we want to train a classifier that takes as input st-cubes such Influence of the st-cubes sizes ? ?
1.2K90发布于 2018-01-03 - 来自专栏机器人课程与技术
Cozmo人工智能机器人SDK使用笔记（4）-任务部分cubes_and_objects
" % len(cubes)) lookaround.stop() if len(cubes) == 0: robot.play_anim_trigger(cozmo.anim.Triggers.MajorFail ) cubes = robot.world.wait_until_observe_num_objects(num=2, object_type=cozmo.objects.LightCube, timeout=60) lookaround.stop() if len(cubes) < 2: print("Error: need 2 Cubes but only found", len(cubes), "Cube(s)") else: # Try and pickup the 1st cube current_action ) < 3: print("Error: need 3 Cubes but only found", len(cubes), "Cube(s)") else: robot.pickup_object
79420发布于 2019-01-28 - 来自专栏python全栈教程专栏
python文档：列表
100] [1, 4, 9, 16, 25, 36, 49, 64, 81, 100] 与 immutable 的字符串不同, 列表是一个 mutable 类型，就是说，它自己的内容可以改变: >>> cubes 64 >>> cubes[3] = 64 # replace the wrong value >>> cubes [1, 8, 27, 64, 125] 你也可以在列表末尾通过 append() 方法 来添加新元素（我们将在后面介绍有关方法的详情）: >>> cubes.append(216) # add the cube of 6 >>> cubes.append(7 ** 3) # and the cube of 7 >>> cubes [1, 8, 27, 64, 125, 216, 343] 给切片赋值也是可以的，这样甚至可以改变列表大小，或者把列表整个清空: >>> letters
60420发布于 2021-10-18 - 来自专栏小樱的经验随笔
Codeforces Round #395 (Div. 2)(A.思维，B,水)
He got kit of n cubes as a birthday present from his parents. Timofey put all the cubes in a row and went to unpack other presents. Suppose the cubes are numbered from 1 to n in their order. When Timofey returned to his cubes, he understood that their order was changed. Input The first line contains single integer n (1 ≤ n ≤ 2·105) — the number of cubes.
86860发布于 2018-04-08 - 来自专栏雪萌天文台
笔记·将MineCraft3D皮肤转换为4D皮肤
0, 12, 0] }, { "name": "body", "parent": "waist", "pivot": [0, 24, 0], "cubes 16]} ] }, { "name": "head", "parent": "body", "pivot": [0, 24, 0], "cubes 0]} ] }, { "name": "hat", "parent": "head", "pivot": [0, 24, 0], "cubes 24, 3] }, { "name": "leftArm", "parent": "body", "pivot": [5, 22, 0], "cubes ] }, { "name": "leftSleeve", "parent": "leftArm", "pivot": [5, 22, 0], "cubes
65110编辑于 2024-07-29 - 来自专栏python3
python列表（Lists）
36, 49, 64, 81, 100] [1, 4, 9, 16, 25, 36, 49, 64, 81, 100] （4）不同于不能改变的字符串，列表是可以改变的，例如可以改变列表中的内容 >>> cubes = [1, 8, 27, 65, 125] # 这里的值有一些错误，因为这是一个立方表 >>> 4 ** 3 #4的立方是64，不是65 64 >>> cubes[3] = 64 # 替换错误的值 >>> cubes [1, 8, 27, 64, 125] 可以通过append（）方法，在列表结尾处添加新值。 >>> cubes.append(216) # add the cube of 6 >>> cubes.append(7 ** 3) # and the cube of 7 >>> cubes [1
89410发布于 2020-01-11 - 来自专栏叶子的开发者社区
Unity开发微信跳一跳小游戏
(cube); } if (cubes.Count > maxCubeNumber) { var cube = cubes.Dequeue if (cubes.Count > maxCubeNumber) { var cube = cubes.Dequeue(); Destroy(cube GameObject head; public GameObject cubePrefab; public Vector3 cubePlace; private Queue<GameObject> cubes void Start() { var cube = Instantiate(cubePrefab, cubePlace, Quaternion.identity); cubes.Enqueue (cube); } if (cubes.Count > maxCubeNumber) { var cube = cubes.Dequeue
63510编辑于 2023-07-13 - 来自专栏叶子的开发者社区
Unity开发微信跳一跳小游戏
Quaternion.identity); cubePlace = cube.transform.position; cubes.Enqueue (cube); } if (cubes.Count > maxCubeNumber) { var cube = cubes.Dequeue if (cubes.Count > maxCubeNumber) { var cube = cubes.Dequeue(); Destroy head; public GameObject cubePrefab; public Vector3 cubePlace; private Queue<GameObject> cubes (cube); } if (cubes.Count > maxCubeNumber) { var cube = cubes.Dequeue
78620编辑于 2023-07-30 - 来自专栏一Wa哇一天
2019-2020 ICPC Southeastern European Regional Programming Contest (SEERC 2019)-G.Projection
You are asked to compute a possible sets of cubes that must be placed inside the 3D volume such that the 3D object created with the cubes throws the shadows specified by the projection-matrices, when the If it is possible you must find exactly two sets, one with the maximum amount of cubes and one with the You can assume there is no gravitation (the cubes are located inside the 3D volume exactly where they chosen in the lexicographically smallest solution with maximum number of cubes.
1.2K20发布于 2020-04-09 - 来自专栏小鹏的专栏
用一个脚本学习 python
= [] for num in nums: cubes.append(num ** 3) # 等价 comprehension cubes = [num**3 for num in nums] # --------------------------------- cubes_of_even = [] for num in nums: if num % 2 == 0: cubes_of_even.append (num**3) # 等价 comprehension cubes_of_even = [num**3 for num in nums if num % 2 == 0] # -------------- ------------------- cubes_add_squares = [] for num in nums: if num % 2 == 0: cubes_add_squares.append (num**3) else: cubes_add_squares.append(num**2) # 等价 comprehension cubes_and_squares = [num**3 if
1.5K70发布于 2018-01-09 - 来自专栏礼拜八不工作
背包大乱斗与俄罗斯方块(代码篇)
public class TileInfo : IReset { public List<GameObject> Cubes = new List<GameObject>(); public public Color BaseColor; public void Reset() { BaseColor = Color.white; if (Cubes.Count > 0) { for (int i = 0; i < Cubes.Count; i++) PrefabPoolManager.GetInstance ().PushGameObjectByType(PrefabPoolManager.PrefabType.Cube, Cubes[i]); Cubes.Clear();
34810编辑于 2024-11-25 - 来自专栏Python基础、进阶与实战
Python基础-2 数字、字符串和列表
36,49,64,81,100] [1, 4, 9, 16, 25, 36, 49, 64, 81, 100] 与不可变（immutable）字符串不同, 列表可变（mutable）类型，其内容可以改变： >>> cubes = [1,8,27,65,125] # something's wrong here >>> cubes[3]= 64 # replace the wrong value >>> cubes [1 , 8, 27, 64, 125] append()方法可以在列表结尾添加新元素 >>> cubes.append(216) # add the cube of 6 >>> cubes.append (7 **3) # and the cube of 7 >>> cubes [1, 8, 27, 64, 125, 216, 343] 为切片赋值可以改变列表大小，甚至清空整个列表： >>> letters
2.8K20编辑于 2022-12-06 【杭电oj】1220 - Cube（数学水题）
edge length is N, it is cut by the planes that was paralleled to its side planes into N * N * N unit cubes Two unit cubes may have no common points or two common points or four common points. Your job is to calculate how many pairs of unit cubes that have no more than two common points.
25110编辑于 2025-08-26- 来自专栏神光的编程秘籍
用 Three.js 和 AudioContext 实现音乐频谱的 3D 可视化
mousedown', triggerHandler); 之后可以创建 3D 场景中的各种物体： 创建立方体： 因为频谱为 1024 个数据，我们 50个分为一组，就只需要渲染 21 个立方体： const cubes ); const cube = new THREE.Mesh( geometry, material ); cube.translateX((10 + 10) * i); cubes.add (cube); } cubes.translateX(- (10 +10) * CUBE_NUM / 2); scene.add(cubes); 立方体的物体 Mesh，分别设置几何体是 BoxGeometry ** * 花瓣分组 */ const petal = new THREE.Group(); /** * 频谱立方体 */ const cubes (cube); } cubes.translateX(- (10 +10) * CUBE_NUM / 2); var flowerTexture1 =
3.2K30编辑于 2021-12-04
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