学习中不段进步! 游戏思路:通过键盘控制白色小球,让它"捡起"柠黄色方块,捡起一个加1分,全部捡起游戏胜利! 游戏对象: Ground:绿色地面 player:  小球 Obstacle:围墙 PickUps:小方块 脚本 键盘控制小球移动 摄像机跟随小球移动 小球捡起方块(也就是消除方块) 得分以及胜利 小方块自己在不停的旋转 1.键盘控制小球移动 这个需要先给小球添加Rigidbody(刚体)组件,想要让小球移动就需要给小球施加一个力(下面红线是控制在这操作的代码) private

GameObject的添加就不细说了,地面,小球和碰撞小物体. 刚体组件(Rigidbody): 使物体能够模拟物理效果,比如重力,碰撞,推力等: 控制小球移动的脚本(Script,Ball的脚本): Input.GetAxis("Horizontal"):控制横向运动键,有A,D键: Input.GetAxis("Vertical"):控制纵向运动键,有W,S: AddForce(Vector3):添加力,力是矢量,有方向: 控制相机跟随物体移动(第一视角): 把

1.摄像机的跟随运动,逻辑就是保持摄像机跟主角的距离不变(Undate()函数). offset=trandform.position-player.position. Undate() { transform.position=player.position+offset; } 2.控制物体的移动(Input读取输入的用法)(其他方法等学到补充) 第一种方法: 获取物体的刚体,刚体打带有AddForce()方法:传入的是一个人向量,Input.GetAxis();通过Horizontal,Ve

using UnityEngine; using System.Collections; /// <summary> /// 相机控制 /// </summary> public class CameraController : MonoBehaviour { /// <summary> /// 玩家 /// </summary> public GameObject player; /// <summary> /// 玩家到相机的偏移 /// &

1. 游戏界面 2.代码 //FoodRotate - - 控制cube旋转 using System.Collections; using System.Collections.Generic; using UnityEngine; public class FoodRotate : MonoBehaviour { // Use this for initialization void Start () { } // Update is called once per frame void U

1-2 Unity5.2入门课程 进入Unity开发的奇幻世界 任务1:Roll A Ball项目简介 Unity官网的tutorial入门项目 方向键控制小球在平台上滚动,碰撞方块得分,消掉所有方块后胜利 任务2:初始化游戏环境 1. New Project "Roll A Ball" 2. save Main.unity into "Scenes" folder 3. create a Plane and rename it as "Ground&qu

B - Space BowlingTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87643#problem/B Description The inhabitants of planets orbiting around the pulsar PSR 2010+15 enjoy playing space bowling. A few cylin

class MtwGame { public constructor() { } private static _instance: MtwGame; public static get Instance(): MtwGame { if (this._instance == null || this._instance == undefined) { this._instance = new MtwGame(); } return this._instance; } public stage:

Magic Ball Game Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 640    Accepted Submission(s): 169 Problem Description When the magic ball game turns up, Kimi immediately falls in it. The inter

http://acm.hdu.edu.cn/showproblem.php?pid=4605 Magic Ball Game Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 309    Accepted Submission(s): 73 Problem Description When the magic ball game tu

Roll The Cube Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 345    Accepted Submission(s): 127 Problem Description This is a simple game.The goal of the game is to roll two balls to two holes

Magic Ball Game Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2489    Accepted Submission(s): 759 Problem Description When the magic ball game turns up, Kimi immediately falls in it. The int

虚拟现实 三维空间的右手笛卡尔坐标如图1所示. 图1 在航空中,pitch, yaw, roll如图2所示. pitch是围绕X轴旋转,也叫做俯仰角,如图3所示. yaw是围绕Y轴旋转,也叫偏航角,如图4所示. roll是围绕Z轴旋转,也叫翻滚角,如图5所示. (http://www.cnblogs.com/xiaobo-Linux/) 图2 图3 图4 图5

有一个筒,从A口可以放球,放进去的球可通过挡板DE使其掉进B管或C管里,现有带1-10标号的球按给定顺序从A口放入,问是否有一种控制挡板的策略可以使B管和C管中的球从下往上标号递增. 输入: 第一行输入数据组数N.接下来N行为N组具体数据,每组数据中有10个整数,代表球的放入顺序. 输出: 对于每组数据,若策略存在,输出YES:若不存在,输出NO 解法1:DFS 思路:每次判断当前小球是否大于左边容器的最上端的小球,如果可以就放,否则再去看右边的.一旦发现左右两边都不能放,那就只能判定是NO了.

function Ball(message){ alert(message); }; var ball0=new Ball("executing"); //var ball0=new Ball("executing"); //上面这行代码等价于下面的过程var ball0=new Object(); ball0.construct=Ball;//将属性construct指向代码片段Ball code ball0.construct("executing&q

#include<iostream> #include<cstring> using namespace std; int sum; ]; int n; int head; int ping(int w){ ; ;y<w;y++){ u=u*; } return u; } int main(){ freopen("ball.in","r",stdin); freopen("ball.out","w"

Color the ball 我真的该认真的复习一下以前没懂的知识了,今天看了一下线段树,以前只会用模板,现在看懂了之后,发现还有这么多巧妙的地方,好厉害啊 所以就应该尽量搞懂 弄明白每个知识点 [题目链接]Color the ball [题目类型]线段树区间更新 &题意: N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的"小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色.但是N次以后lel

Color the ball Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13645    Accepted Submission(s): 6850 Problem Description N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气

roll绕Y轴旋转 pitch绕X轴旋转 heading绕Z轴旋转 单位是弧度,可以使用osg::inDegrees(45)将45角度转换为弧度 定义一个四元数 osg::Quat q( roll,osg::Vec3d(0.0, 1.0, 0.0), pitch,osg::Vec3d(1.0, 0.0, 0.0), heading,osg::Vec3d(0.0, 0.0, 1.0)); //设置旋转 manipulator->setRotation(q);

Color the ball Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8984    Accepted Submission(s): 4594 Problem Description N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球