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=
In the math class, the evil teacher gave you one unprecedented problem! Here f(n) is the n-th fibonacci number (n >= 0)! Where f(0) = f(1) = 1 and for any n > 1, f(n) = f(n - 1) + f(n - 2). For example, f(2) = 2, f(3) = 3, f(4) = 5 ... The teacher u
问题是这样,如果我们知道两个向量v1和v2,计算从v1转到v2的旋转矩阵和四元数,由于旋转矩阵和四元数可以互转,所以我们先计算四元数. 我们可以认为v1绕着向量u旋转θ角度到v2,u垂直于v1-v2平面. 四元数q可以表示为cos(θ/2)+sin(θ/2)u,即:q0=cos(θ/2),q1=sin(θ/2)u.x,q2=sin(θ/2)u.y,q3=sin(θ/2)u.z 所以我们求出u和θ/2即可,u等于v1与v2的叉积,不要忘了单位化:θ/2用向量夹角公式就能求. ma
#pragma strict public var m_pA : Vector3 = new Vector3(2.0f, 4.0f, 0.0f); public var m_pB : Vector3 = new Vector3(-4.0f, 2.0f,0.0f); private var m_pTemp : Vector3 = new Vector3(0.0f,0.0f,0.0f); private var m_fTemp : float = 0.0f; private var m_fAngle
这个月月初我们一行三人去湖南参加了ccpc湖南程序设计比赛,虽然路途遥远,六月的湘潭天气燥热,不过在一起的努力之下,拿到了一块铜牌,也算没空手而归啦.不过通过比赛,还是发现我们的差距,希望这几个月自己努力思考,积极刷题,为九月份acm网络赛做准备! 言归正传说说这道题目,这也是这次比赛想到AC比较高的题目,不过我们还是没能完成,下面我就来总结一下此题的一些思路和方法. Magic Triangle Problem Description: Huangriq is a respectful acm