基础知识 复数表示 C = R + jI 极坐标:C = |C|(cosθ + jsinθ) 欧拉公式:C = |C|ejθ 有关更多的时域与复频域的知识可以学习复变函数与积分变换,本篇文章只给出DFT公式,性质,以及实现方法 二维离散傅里叶变换(DFT) 其中f(x,y)为原图像,F(u,v)为傅里叶变换以后的结果,根据欧拉公式可得,每个F(u,v)值都为复数,由实部和虚部组成 代码示例 void dft(short** in_array, double** re_array, double*

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14391   Accepted: 6685 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 17880   Accepted: 6709 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1

clc,clear all,close all; load woman; [c,s]=wavedec2(X,2,'haar');%进行2尺度二维离散小波分解.分解小波函数haar %多尺度二维离散小波重构(逆变换) Y=waverec2(c,s,'haar'); figure; subplot(1,2,1),imshow(X,map),title('原始图像'); subplot(1,2,2),imshow(Y,map),title('重构图像'); a1=wrcoef2('a',c,s,'ha

clc,clear all,close all; load woman; %单尺度二维离散小波分解.分解小波函数haar [cA,cH,cV,cD]=dwt2(X,'haar'); %单尺度二维离散小波重构(逆变换) Y=idwt2(cA,cH,cV,cD,'haar'); figure; subplot(1,2,1),imshow(X,map),title('原始图像'); subplot(1,2,2),imshow(Y,map),title('重构图像'); clear all;close

对X进行N尺度小波分解 [C,S]=wavedec2(X,N,'wname'); clc,clear all,close all; load woman; [c,s]=wavedec2(X,2,'db1');%进行2尺度二维离散小波分解.分解小波函数-db1 [cH1,cV1,cD1]=detcoef2('all',c,s,1);%尺度1的所有方向的高频系数 [cH2,cV2,cD2]=detcoef2('all',c,s,2);%尺度2的所有方向的高频系数 cA1=appcoef2(c,s,'

clc,clear all,close all; load woman; [cA,cH,cV,cD]=dwt2(X,'haar');%单尺度二维离散小波分解.分解小波函数haar figure,imshow(X,map),axis image; figure; subplot(2,2,1),imshow(uint8(cA));axis off;title('低频系数图像'); subplot(2,2,2),imshow(uint8(cH));axis off;title('水平高频系数图像');

相关软件参见:混沌数学之离散点集图形DEMO 相关代码: // http://wenku.baidu.com/link?url=yg_gE7LUXCg2mXRp-ZZdfRXXIkcNj8YOhvN7dKLJxzWIu6M0g33-W3y3culjalCYfNc5VQefVJEiEwi_woXP69H8W4x-bF22nIRnD2lsn53 // 二维离散电路混沌系统 class CircuitChaotic : public DiscreteEquation { public: Circuit

二维区域和检索 - 矩阵不可变 给定一个二维矩阵,计算其子矩形范围内元素的总和,该子矩阵的左上角为 (row1, col1) ,右下角为 (row2, col2). 上图子矩阵左上角 (row1, col1) = (2, 1) ,右下角(row2, col2) = (4, 3),该子矩形内元素的总和为 8. 示例: 给定 matrix = [ [3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3,

304. 二维区域和检索 - 矩阵不可变 给定一个二维矩阵,计算其子矩形范围内元素的总和,该子矩阵的左上角为 (row1, col1) ,右下角为 (row2, col2). Range Sum Query 2D 上图子矩阵左上角 (row1, col1) = (2, 1) ,右下角(row2, col2) = (4, 3),该子矩形内元素的总和为 8. 示例: 给定 matrix = [ [3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4

题目: Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2

//商家的等级信息是一个二维数组,求出最小折扣和最大折扣array(0=>array('levelname'=>'银','dis'=>7.5), 1=>array('levelname'=>'金卡','dis'=>8.0), 2=>array('levelname'=>'白金卡','dis'=>9.0)); function getMinAndMaxInArray($arr) { if(empty($arr)) { return array(10,10

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (with the red border) is defined by (row1, col1) = (2, 1)and (row2, col

[题目链接] http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0531 [题目大意] 给出一张图,和一些矩形障碍物,求该图没被障碍物覆盖的部分被划分为几个连通块 [题解] 首先对图中的点进行离散化,对于一个障碍物来说, 我们将其看做左闭右开上闭下开的图形,所以在离散的时候只要离散障碍物的点即可. 之后我们利用imos积累法得出哪些部分是障碍物,就可以统计连通块了. [代码] #include <cstdio> #include

题:https://codeforces.com/contest/1262/problem/E 分析:预处理出阵列中的矩阵,然后二分答案还原题目的烧火过程,判断是否满足要求 #include<bits/stdc++.h> using namespace std; #define pb push_back typedef long long ll; ; int main(){ int n,m; ios::sync_with_stdio(),cout.tie(); cin>>n>

clc,clear all,close all; load woman; [cA,cH,cV,cD]=swt2(X,2,'haar');%用haar小波基进行2尺度平稳小波分解 Y=iswt2(cA,cH,cV,cD,'haar'); figure; subplot(1,2,1),imshow(uint8(X)),title('原图'); subplot(1,2,2),imshow(uint8(Y)),title('重构');

对信号X进行N尺度平稳小波分解 [A,H,V,D]=swt2(X,N,'wname'); clc,clear all,close all; load woman; [cA,cH,cV,cD]=swt2(X,2,'haar');%用haar小波基进行2尺度平稳小波分解 cA1=cA(:,:,1);cH1=cH(:,:,1);cV1=cV(:,:,1);cD1=cD(:,:,1);%尺度1低.高频系数 cA2=cA(:,:,2);cH2=cH(:,:,2);cV2=cV(:,:,2);cD2=cD(

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 33682   Accepted: 12194 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14489   Accepted: 6735 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The

图像处理中常用的正交变换除了傅里叶变换以外,还有一些其它常用的正交变换,其中离散余弦变换DCT就是一种,这是JPEG图像压缩算法里的核心算法,这里我们也主要讲解JPEG压缩算法里所使用8*8矩阵的二维离散余弦正变换. 一维离散余弦变换 一般表达式 要弄懂二维离散余弦变换,首先我们需要先了解它在一维下的情况,具体表达式如下: 式中F(u)是第u个余弦变换值,u是广义频率变量,u=1,2,….,N-1;f(x)是时域N点序列.x= 1,2,….,N-1; 矩阵表示法 更为简洁的定义离散余弦变换是采用