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 {
標(biāo)簽： AOrigin APoint Point PointToAngle
上傳時間： 2016-10-31
上傳用戶：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 {
標(biāo)簽： AOrigin APoint Point PointToAngle
上傳時間： 2016-10-31
上傳用戶：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 {
標(biāo)簽： AOrigin APoint Point PointToAngle
上傳時間： 2013-12-18
上傳用戶：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.
標(biāo)簽： Limit following solving problem
上傳時間： 2014-01-12
上傳用戶：362279997
一個模糊控制器的仿真設(shè)計(jì)，被控對象為某離心機(jī)，本例采用Fuzzy，Simulink進(jìn)行仿真設(shè)計(jì)，最后給出被控對象的輸出響應(yīng)曲線。
標(biāo)簽： 模糊控制器 仿真設(shè)計(jì)
上傳時間： 2014-01-06
上傳用戶：腳趾頭
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.
標(biāo)簽： multimedia Semantic analysis research
上傳時間： 2016-11-24
上傳用戶：蟲蟲蟲蟲蟲蟲
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
標(biāo)簽： synchronization simulation algorithm function
上傳時間： 2013-12-17
上傳用戶：yulg
Euler函數(shù): m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函數(shù)： 定義：phi(m) 表示小于等于m并且與m互質(zhì)的正整數(shù)的個數(shù)。 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 則有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m 在實(shí)際代碼中可以用類似素?cái)?shù)篩法求出 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：定義phi(p) 為比p小的與p互素的數(shù)的個數(shù) 設(shè)n的素因子有p1, p2, p3, … pk 包含p1, p2…的個數(shù)為n/p1, n/p2… 包含p1*p2, p2*p3…的個數(shù)為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)
標(biāo)簽： Euler lt phi 函數(shù)
上傳時間： 2014-01-10
上傳用戶：wkchong
計(jì)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 ％ 這表示邊長嗎？ cgh=zeros(64*s) th=max(max(abs(Fs)))
標(biāo)簽： 20 close clear zeros
上傳時間： 2014-10-13
上傳用戶：wweqas
// 入口參數(shù)： // l: l = 0, 傅立葉變換 l = 1, 逆傅立葉變換 // il: il = 0,不計(jì)算傅立葉變換或逆變換模和幅角；il = 1,計(jì)算模和幅角 // n: 輸入的點(diǎn)數(shù)，為偶數(shù)，一般為32，64，128，...,1024等 // k: 滿足n=2^k(k>0),實(shí)質(zhì)上k是n個采樣數(shù)據(jù)可以分解為偶次冪和奇次冪的次數(shù) // pr[]: l=0時，存放N點(diǎn)采樣數(shù)據(jù)的實(shí)部 // l=1時, 存放傅立葉變換的N個實(shí)部 // pi[]: l=0時，存放N點(diǎn)采樣數(shù)據(jù)的虛部 // l=1時, 存放傅立葉變換的N個虛部 // // 出口參數(shù)： // fr[]: l=0, 返回傅立葉變換的實(shí)部 // l=1, 返回逆傅立葉變換的實(shí)部 // fi[]: l=0, 返回傅立葉變換的虛部 // l=1, 返回逆傅立葉變換的虛部 // pr[]: il = 1,i = 0 時，返回傅立葉變換的模 // il = 1,i = 1 時，返回逆傅立葉變換的模 // pi[]: il = 1,i = 0 時，返回傅立葉變換的輻角 // il = 1,i = 1 時，返回逆傅立葉變換的輻角
上傳時間： 2017-01-03
上傳用戶：ynsnjs
蟲蟲下載站版權(quán)所有 京ICP備2021023401號-1
