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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.
標簽： Limit following solving problem
上傳時間： 2014-01-12
上傳用戶：362279997
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
標簽： synchronization simulation algorithm function
上傳時間： 2013-12-17
上傳用戶：yulg
Euler函數: m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函數： 定義：phi(m) 表示小于等于m并且與m互質的正整數的個數。 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 在實際代碼中可以用類似素數篩法求出 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互素的數的個數 設n的素因子有p1, p2, p3, … pk 包含p1, p2…的個數為n/p1, n/p2… 包含p1*p2, p2*p3…的個數為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)
上傳時間： 2014-01-10
上傳用戶：wkchong
計算全息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)))
上傳時間： 2014-10-13
上傳用戶：wweqas
// 入口參數： // l: l = 0, 傅立葉變換 l = 1, 逆傅立葉變換 // il: il = 0,不計算傅立葉變換或逆變換模和幅角；il = 1,計算模和幅角 // n: 輸入的點數，為偶數，一般為32，64，128，...,1024等 // k: 滿足n=2^k(k>0),實質上k是n個采樣數據可以分解為偶次冪和奇次冪的次數 // pr[]: l=0時，存放N點采樣數據的實部 // l=1時, 存放傅立葉變換的N個實部 // pi[]: l=0時，存放N點采樣數據的虛部 // l=1時, 存放傅立葉變換的N個虛部 // // 出口參數： // fr[]: l=0, 返回傅立葉變換的實部 // l=1, 返回逆傅立葉變換的實部 // 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
OFDM的發射端，包括pi/4DQPSK的調制，ifft，加幀頭以及組幀
上傳時間： 2013-12-26
上傳用戶：stvnash
SMO 源程序基于libsvm，開發的支持向量機圖形界面應用程序，并提供了關于C和sigma的新的參數選擇方法，使得SVM的使用更加簡單直觀
上傳時間： 2013-12-09
上傳用戶：windwolf2000
The concept of the Altera Nios II embedded processor implementation inside Field Programmable Gate Array [FPGA] of the CCD camera for the “Pi of the Sky” experiment is presented. The digital board of the CCD camera, its most important components, current implementation of firmware [VHDL] inside the FPGA and the role of external 8051 microcontroller is briefly described. The main goal of the presented work is to get rid of the external microcontroller and to design new system with Nios II processor built inside FPGA chip. Constraints for implementing the design into the existing camera boards are discussed. New possibilities offered by a larger FPGA for next generation of cameras are considered.
標簽： implementation Programmable processor embedded
上傳時間： 2014-01-02
上傳用戶：dyctj
剛剛漏了補充這個程序，是UKF里計算sigma點的程序，不好意思
標簽： 程序
上傳時間： 2017-03-11
上傳用戶：cainaifa
本程序用于檢測音頻文件中是否具有DTMF信號，若有則將其檢出。 程序首先使用Goertzel算法求出以FRAMESIZE（默認200）為大小的一幀數據在8個DTMF頻點上的能量。 對Goertzel算法的改進，對于系數的計算不是采用2*cos[2*pi*k/N]，而是采用2*cos[2*pi*fn/fs],這樣能夠降低誤差。 確定了8個頻點的能量后運用一系列判決門限來確定有沒有DTMF信號，以及信號是什么。
上傳時間： 2017-03-15
上傳用戶：wyc199288
