MATLAB中求矩阵非零元的坐标: 方法1: index=find(a); [i,j]=ind2sub(size(a),index); disp([i,j]) 方法2: [i,j]=find(a>0|a<0) %列出所有非零元的坐标 [i,j]=find(a==k) %找出等于k值的矩阵元素的坐标 所用函数简介: IND2SUB Multiple subscripts from linear index. IND2SUB is used to determine the equivalent
求矩阵的模: function count = juZhenDeMo(a,b) [r,c] = size(a);%求a的行列 [r1,c1] = size(b);%求b的行列 count = 0; for j=1:r-r1+1%所求的行数中取 for i=1:c-c1+1%所有的列数中取 d = a(j:j+r1-1,i:i+c1-1); e = double(d==b); if(sum(e(:))==r1*c1) count = count + 1; end end end<pre name=
29 [程序 29 求矩阵对角线之和] 题目:求一个 3*3 矩阵对角线元素之和 程序分析:利用双重 for 循环控制输入二维数组,再将 a[i][i]累加后输出. package cskaoyan; public class cskaoyan29 { @org.junit.Test public void diagonal() { java.util.Scanner in = new java.util.Scanner(System.in); int[][] arr = new int[3][
Matrix Swapping II Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1543 Accepted Submission(s): 1036 Problem Description Given an N * M matrix with each entry equal to 0 or 1. We can find so
Matrix Power Series Time Limit: 3000MS Memory Limit: 131072K Total Submissions: 15417 Accepted: 6602 Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + - + Ak. Input The input contains exactly one test cas