dsp 223

Http://www.cnblogs.com/scncart/articles/1805553.htmlComparison between fixed-point DSP and floating-point DSP It may be helpful for entry-level DSP developers. This article focuses on the comparison between a fixed-point DSP and a floating-point DSP, mainly from three aspec

Original articleReprint please register source HTTP://BLOG.CSDN.NET/TOSTQTutorial directory: http://blog.csdn.net/tostq/article/details/51245979In this section we will run the first multi-core DSP program, familiar with the CCS development environment, and learn to use the CCS debugging tool, the main content is as follows:(1) New CCS project(2) Import target simulation module(3) Using the Debug toolFirst, the new CCS projectSelect File/new/ccs Projec

"223-rectangle Area (rectangular region)" " leetcode-interview algorithm classic-java Implementation" "All topic Directory Index" code Download "Https://github.com/Wang-Jun-Chao" Original title Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined to bottom left corner and top right corner as shown in the figure.Rectangle AreaAssume that "total" is never beyond the maximum possible value of int.The mai

AbstractIf the DSP Builder is installed in MATLAB, after the DSP Builder is removed in a day, as long as the MATLAB is moved together, there will be negative information. How can this problem be solved? IntroductionEnvironment: MATLAB r200b Previously, I installed DSP Builder 7.2 and 8.0, but after Quartus II 8.1, I didn't install

EMCV: OpenCV that can be run on the DSP EMCV Project Home: HTTP://SF.NET/PROJECTS/EMCVEMCV all called embedded computer Vision Library, is aComputer Vision Library running on the DM64X series DSP. EMCV provides a fully consistent function interface with OPENCV, and with EMCV, you can easily port your OPENCV algorithm to a DSP without even changing one line of co

/** 223. Rectangle areas * 12.12 by Mingyang * Large area minus small area * Here we test instructions do not understand wrong oh, here is to find all the place is cover, not only by the two overlap cover place * so we need To find two of the sum of the area minus the middle repeat two times the small area*/ Public intComputearea (intAintBintCintDintEintFintG,intH) {intMinu = 0; if(E >= C | | G B) Minu= 0; ElseMinu= (Math.min (G, C)-Math.max (A, E

no.223 Rectangle AreaFind the total area covered by and rectilinear rectangles in a 2D plane. Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.Find the area where two rectangles intersectVertical rectangles, each of which is represented by a coordinate of two points in the lower left and upper right cornersIn fact, the intersecti

Tags: Post get nbsp path pre character false input git commandWhen adding a file to be submitted using git Add, if the file name is in Chinese, it will display garbled characters such as 274\232\350\256\256\346\200\273\347\273\223.Solution: At the bash prompt, enter:FalseIf the Core.quotepath is set to False, the characters above 0x80 will not be quote. Chinese display is normal.Reference:http://blog.csdn.net/tyro_java/article/details/53439537Linux gi

Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.Credits:Special thanks to @mithmatt for adding this problem, creating the above image and all test cases.Problem Solving Ideas:The Java implementation is as follows: public int Computearea (int A, int B, int C, int D, int E, int F, int G, int H) { i

Title:Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.Tips:This question to determine whether there are overlapping parts.Code:classSolution { Public: intComputearea (intAintBintCintDintEintFintGintH) {if(C B)return(c-a) * (d-b) + (G-E) * (H-F); return(c-a) * (d-b) + (G-E) * (h-f)-(min (c,g)-Max (A,e))

-E) + (d-b) * (C-A); We could also see this, since A is the left edge of C, E was the left edge of g,the left coordinate forThe overlaied rectangles is:---------------------------------------------------intleft =Math.max (A, E); the right coordinate forThe overlaied rectangles is:-------------------------------------------------------intright =math.min (C, G); The same analysis could apply to the vertical direction. Skill:to compute the overall area, we could use:total area-Overlaid area.Solutio

{2 Public:3 intComputearea (intAintBintCintDintEintFintGintH) {4 intoverlap=0;5 //Case 1:6 if(C>=ecD) {7Overlap = (C-E) * (H-B);8 }9 //Case 2:Ten if(C>=ecd) { OneOverlap = (C-E) * (dB); A } - //Case 3: - if(C>=ecD) { theOverlap = (C-E) * (dF); - } - //Case 4: - if(E>=aeF) { +Overlap = (C-E) * (H-F); - } + //Case 5: A if(A>=eaH) { atOverlap = (c-a) * (H-B); - } -

Topic:Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.Answer:There's nothing to say, two rectangles added, minus the area of the coincident part if there is a coincident part1 classSolution {2 Public:3 intComputearea (intAintBintCintDintEintFintGintH) {4 intsum;5sum= (c-a) * (d-b) + (G-E) * (H

Topic:Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.First determine whether two rectangles intersect, do not intersect directly return two rectangular area, the intersection needs to calculate the intersection of the lower left and right points of the coordinates, with two rectangular area minus the inte

Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.Idea: Need to calculate the area of the square cover, the key is to calculate the area of the coincident area, using the following calculation method;On the left side of the coincident area (bottom) (right, top)left = Math.max (A, E);Bottom = Math.max (B, F);

leftmost part of the overlapping section is the leftmost side of the Rect1 (A)The rightmost part of the overlap may be the rightmost (C) of the rect1, or it may be the rightmost (G) of the Rect22, calculate the length(1) When D The topmost part of the overlapping section is the top of the Rect1 (D)The bottom of the overlapping section may be the bottom (B) of the rect1, or it may be the bottom of the Rect2 (F)(2) When D > H,The topmost part of the overlapping section is the top of the Rect2 (H)

Topic linksTitle Requirements:Find the total area covered by and rectilinear rectangles in a 2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.　　Assume the total area is never beyond the maximum possible value of int.Credits:Special thanks to @mithmatt for adding this problem, creating the above image and all test cases.To better illustrate the problem, here are a few diagrams specifically drawn:　　The area we require = two rectangular area

1. Description of the problemFind the area of two rectangles.　　2. Methods and IdeasThe intersection area of two rectangles can be calculated first, and then the area is rectangular with two rectangular area and minus intersection area.Note: Although the area does not exceed the maximum value of int, the middle edge length operation may be more than, pay attention to the processing details, otherwise easily overflow.　　Class Solution {public:intComputearea (intAintBintCintDintEintFintGintH) {intUn

Rectangle AreaTotal Accepted: 2205 Total Submissions: 8138 Find the total area covered by and rectilinear rectangles in a2D plane.Each rectangle was defined by its bottom left corner and top right corner as shown in the figure.Assume the total area is never beyond the maximum possible value of int.IdeasThe area of two regions is calculated, and then the area of the overlapping is subtracted, which is the desired one.[CODE]public class Solution {public int computearea (int A, int B, int C, in

\) /c1> with \ (j-i+1-1\), they is the same!So our tasks now are to calculate \ (fac_{i,1}^{k}\), which is \ (i + k-2 \choose i-1\). But we cannot pre process the factorial of \ (k\), as \ (k\) can is up to \ (10^9\). But \ (fac_{1,1}\) are always 1!Think the transfer between \ (fac_{i-1,1}\) and \ (fac_{i,1}\), that is:\[\frac{(i-1+k-2)!} {(I-1-1)! (K-1)!} \rightarrow \frac{(i+k-2)!} {(I-1)! (K-1)!} \]We just need to multiply the first expression by \ (i+k-2\) and \ ((i-1) ^{-1}\, (mod\,10^9+7)