原地址:http://www.cnblogs.com/yk250/p/5663907.html 文中表述仅为本人理解,若有偏差和错误请指正! 接着 ZAM 3D 制作简单的3D字幕 流程(一) .本篇将是ZAM 3D制作动画的一些关键点. 我们可以将各个字体工具自己的喜好填充好Materials,当然,也可以采用默认,一般默认灯光下是偏金黄色的,这个可以在左下角进行调节.左下第一个是当前选中的灯光设置,第二个则是全局的灯光设置.(如果你拖动了布局那就另当别论了.)比如我这里将字体染成了葫芦娃的红…

原地址-> http://www.cnblogs.com/yk250/p/5662788.html 介绍:对经常使用Blend做动画的人来说,ZAM 3D 也很好上手,专业制作3D素材的XAML导出动画和布局. 效果图如下:用ZAM 3D制作 然后导出XAML.详细步骤,以后放出. 生成的布局XAML如下: <Grid> <Viewbox Stretch="None"> <Viewport3D x:Name="ZAM3DViewport3D…

前言,小程序本身是带有地区选着器的(网站:https://mp.weixin.qq.com/debug/wxadoc/dev/component/picker.html),由于自己开发的程序的数据是很久以前mysql数据,与小程序自带的地区名称并不一样,所以只能自己写了. (详细思路写在了最底部) 第一步:把mysql地区数据变成小程序数组格式 1.用php代码,导出数据mysql的地区数据 public function area(){ $sql = "SELECT * FROM c_regi…

1>------ 已启动全部重新生成: 项目: ZERO_CHECK, 配置: Debug x64 ------1> Checking Build System1> CMake does not need to re-run because O:/OpenSceneGraph/OpenSceneGraph_build/CMakeFiles/generate.stamp is up-to-date.1> CMake does not need to re-run because O:…

Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know, uncle Ve…

Stealing Harry Potter's Precious Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in…

Stealing a Cake Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1164    Accepted Submission(s): 320 Problem Description There is a big round cake on the ground. A small ant plans to steal a sma…

题目链接:hdu 4771 Stealing Harry Potter's Precious 题目大意:在一个N*M的银行里,贼的位置在'@',如今给出n个宝物的位置.如今贼要将全部的宝物拿到手.问最短的路径,不须要考虑离开. 解题思路:由于宝物最多才4个,加上贼的位置,枚举两两位置,用bfs求两点距离,假设存在两点间不能到达,那么肯定是不能取全然部的宝物. 然后枚举取宝物的顺序.维护ans最小. #include <cstdio> #include <cstring> #incl…

注意--你可能会爆内存-- 假设一个直接爆搜索词-- 队列存储器元件被减少到-- #include<iostream> #include<map> #include<string> #include<cstring> #include<cstdio> #include<cstdlib> #include<cmath> #include<queue> #include<vector> #include…

pid=4454" target="_blank" style="">题目链接:hdu 4454 Stealing a Cake 题目大意:给定一个起始点s,一个圆形.一个矩形.如今从起点開始,移动到圆形再移动到矩形.求最短距离. 解题思路:在圆周上三分就可以.即对角度[0,2*pi]三分.计算点和矩形的距离能够选点和矩形四条边的距离最短值. #include <cstdio> #include <cstring> #incl…