private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：zhyiroy
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：sunjet
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2013-12-18
上传用户：rocketrevenge
For solving the following problem: "There is No Free Lunch" Time Limit: 1 Second Memory Limit: 32768 KB One day, CYJJ found an interesting piece of commercial from newspaper: the Cyber-restaurant was offering a kind of "Lunch Special" which was said that one could "buy one get two for free". That is, if you buy one of the dishes on their menu, denoted by di with price pi , you may get the two neighboring dishes di-1 and di+1 for free! If you pick up d1, then you may get d2 and the last one dn for free, and if you choose the last one dn, you may get dn-1 and d1 for free. However, after investigation CYJJ realized that there was no free lunch at all. The price pi of the i-th dish was actually calculated by adding up twice the cost ci of the dish and half of the costs of the two "free" dishes. Now given all the prices on the menu, you are asked to help CYJJ find the cost of each of the dishes.
标签： Limit following solving problem
上传时间： 2014-01-12
上传用户：362279997
一个模糊控制器的仿真设计，被控对象为某离心机，本例采用Fuzzy，Simulink进行仿真设计，最后给出被控对象的输出响应曲线。
上传时间： 2014-01-06
上传用户：脚趾头
Semantic analysis of multimedia content is an on going research area that has gained a lot of attention over the last few years. Additionally, machine learning techniques are widely used for multimedia analysis with great success. This work presents a combined approach to semantic adaptation of neural network classifiers in multimedia framework. It is based on a fuzzy reasoning engine which is able to evaluate the outputs and the confidence levels of the neural network classifier, using a knowledge base. Improved image segmentation results are obtained, which are used for adaptation of the network classifier, further increasing its ability to provide accurate classification of the specific content.
标签： multimedia Semantic analysis research
上传时间： 2016-11-24
上传用户：虫虫虫虫虫虫
it is a simulation about ML synchronization algorithm in OFDM systems,you can slao see a function picture in its output,that s useful for a beginner
标签： synchronization simulation algorithm function
上传时间： 2013-12-17
上传用户：yulg
Euler函数: m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函数： 定义：phi(m) 表示小于等于m并且与m互质的正整数的个数。 phi(m) = p1^(r1-1)*(p1-1) * p2^(r2-1)*(p2-1) * …… * pn^(rn-1)*(pn-1) = m*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pn) = p1^(r1-1)*p2^(r2-1)* …… * pn^(rn-1)*phi(p1*p2*……*pn) 定理：若(a , m) = 1 则有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m 在实际代码中可以用类似素数筛法求出 for (i = 1 i < MAXN i++) phi[i] = i for (i = 2 i < MAXN i++) if (phi[i] == i) { for (j = i j < MAXN j += i) { phi[j] /= i phi[j] *= i - 1 } } 容斥原理：定义phi(p) 为比p小的与p互素的数的个数 设n的素因子有p1, p2, p3, … pk 包含p1, p2…的个数为n/p1, n/p2… 包含p1*p2, p2*p3…的个数为n/(p1*p2)… phi(n) = n - sigm_[i = 1](n/pi) + sigm_[i!=j](n/(pi*pj)) - …… +- n/(p1*p2……pk) = n*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pk)
上传时间： 2014-01-10
上传用户：wkchong
计算全息close all clc clear A=zeros(64) A(15:20,20:40)=1 A(15:50,20:25)=1 A(45:50,20:40)=1 A(30:34,20:35)=1 ％ ppp=exp(rand(64)*pi*2*i) A=A.*ppp ％ Author s email: zjliu2001@163.com figure imshow(abs(A),[]) Fa=fft2(fftshift(A)) Fs=fftshift(Fa) Am=abs(Fs) ％ amplitude Ph=angle(Fs) ％ phase s=11 ％ 这表示边长吗？ cgh=zeros(64*s) th=max(max(abs(Fs)))
上传时间： 2014-10-13
上传用户：wweqas
// 入口参数： // l: l = 0, 傅立叶变换 l = 1, 逆傅立叶变换 // il: il = 0,不计算傅立叶变换或逆变换模和幅角；il = 1,计算模和幅角 // n: 输入的点数，为偶数，一般为32，64，128，...,1024等 // k: 满足n=2^k(k>0),实质上k是n个采样数据可以分解为偶次幂和奇次幂的次数 // pr[]: l=0时，存放N点采样数据的实部 // l=1时, 存放傅立叶变换的N个实部 // pi[]: l=0时，存放N点采样数据的虚部 // l=1时, 存放傅立叶变换的N个虚部 // // 出口参数： // fr[]: l=0, 返回傅立叶变换的实部 // l=1, 返回逆傅立叶变换的实部 // fi[]: l=0, 返回傅立叶变换的虚部 // l=1, 返回逆傅立叶变换的虚部 // pr[]: il = 1,i = 0 时，返回傅立叶变换的模 // il = 1,i = 1 时，返回逆傅立叶变换的模 // pi[]: il = 1,i = 0 时，返回傅立叶变换的辐角 // il = 1,i = 1 时，返回逆傅立叶变换的辐角
上传时间： 2017-01-03
上传用户：ynsnjs
