maya cmds pymel 选择 uv area(uv 面积) 为0 的面 cmds.selectType( pf=True ) cmds.polySelectConstraint( m=3, t=8, ta=True, tab=(0, 0.000010) ) # to get face with texture area between 0-0.000010 cmds.polySelectConstraint( m = 0, ta = False) # turn off the 2D a…
Maya cmds pymel 单位和轴向设置 import maya.cmds as cmds # 1. to make the Y-axis of the world to be the up axis: cmds.upAxis( ax='y' ) # 2. to make the Z-axis of the world to be the up axis, # and rotate the view: cmds.upAxis( ax='z', rv=True ) # 3. to query…
Find the total area covered by two rectilinear rectangles in a2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Assume that the total area is never beyond the maximum possible value of int. Cre…
Find the total area covered by two rectilinearrectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Example: Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2 Output:…
在二维平面上计算出两个由直线构成的矩形叠加覆盖后的面积. 假设面积不会超出int的范围. 详见:https://leetcode.com/problems/rectangle-area/description/ Java实现: class Solution { public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) { int sum = (C - A) * (D - B) + (H - F)…
/* 像是一道数据分析题 思路就是两个矩形面积之和减去叠加面积之和 */ public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) { //求两个面积 int a1 = (C-A)*(D-B); int a2 = (G-E)*(H-F); //求叠加面积,(低上限-高下限)*(左右线-右左线) int h1 = Math.min(D,H); int h2 = Math.max(B,F); int…