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=
FLOOR Round towards minus infinity. FLOOR(X) rounds the elements of X to the nearest integers towards minus infinity. CEIL Round towards plus infinity. CEIL(X) rounds the elements of X to the nearest integers towards infinity. FIX Round towards zero.
pog loves szh II Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2106 Accepted Submission(s): 606 Problem Description Pog and Szh are playing games.There is a sequence with n numbers, Pog wi
一. 数据和变量: 省略号(三个英文句点):表示换行 历史指令调用:用方向键上下 数据显示格式: format short format long format rational 复数: I,j realmin:最小正实数 realmax:最大正实数 Inf:无穷大 eg:2/0 NaN:不定值 eg:0/0 二. 数组的输入