// 移頻選頻原理 //Fvco=[(P*B)+A]*Frefin/R //P=32 //loop filter 100k----prescribe //R=12.8M/100K=128---Parameter1 //Fvco=頻點(diǎn)*2+170280 -1400 //B=Fvco/32-----------Parameter2 //A=Fvco-32*B
標(biāo)簽： 100 prescribe Frefin filter
上傳時(shí)間： 2013-12-26
上傳用戶：dancnc
(1) 、用下述兩條具體規(guī)則和規(guī)則形式實(shí)現(xiàn).設(shè)大寫字母表示魔王語(yǔ)言的詞匯 小寫字母表示人的語(yǔ)言詞匯 希臘字母表示可以用大寫字母或小寫字母代換的變量.魔王語(yǔ)言可含人的詞匯. (2) 、B→tAdA A→sae (3) 、將魔王語(yǔ)言B(ehnxgz)B解釋成人的語(yǔ)言.每個(gè)字母對(duì)應(yīng)下列的語(yǔ)言.
上傳時(shí)間： 2013-12-30
上傳用戶：ayfeixiao
1.有三根桿子A,B,C。A桿上有若干碟子 2.每次移動(dòng)一塊碟子,小的只能疊在大的上面 3.把所有碟子從A桿全部移到C桿上 經(jīng)過(guò)研究發(fā)現(xiàn)，漢諾塔的破解很簡(jiǎn)單，就是按照移動(dòng)規(guī)則向一個(gè)方向移動(dòng)金片： 如3階漢諾塔的移動(dòng)：A→C,A→B,C→B,A→C,B→A,B→C,A→C 此外，漢諾塔問(wèn)題也是程序設(shè)計(jì)中的經(jīng)典遞歸問(wèn)題
標(biāo)簽： 移動(dòng) 發(fā)現(xiàn)
上傳時(shí)間： 2016-07-25
上傳用戶：gxrui1991
1. 下列說(shuō)法正確的是 （ ） A. Java語(yǔ)言不區(qū)分大小寫 B. Java程序以類為基本單位 C. JVM為Java虛擬機(jī)JVM的英文縮寫 D. 運(yùn)行Java程序需要先安裝JDK 2. 下列說(shuō)法中錯(cuò)誤的是 ( ) A. Java語(yǔ)言是編譯執(zhí)行的 B. Java中使用了多進(jìn)程技術(shù) C. Java的單行注視以//開頭 D. Java語(yǔ)言具有很高的安全性 3. 下面不屬于Java語(yǔ)言特點(diǎn)的一項(xiàng)是( ) A. 安全性 B. 分布式 C. 移植性 D. 編譯執(zhí)行 4. 下列語(yǔ)句中，正確的項(xiàng)是 ( ) A . int $e,a,b=10 B. char c,d=’a’ C. float e=0.0d D. double c=0.0f
上傳時(shí)間： 2017-01-04
上傳用戶：netwolf
4． 假設(shè)有兩個(gè)按數(shù)據(jù)元素值非遞減有序排列的線性表A和B，均以單鏈表作為存儲(chǔ)結(jié)構(gòu)。編寫算法將A表和B表歸并成一個(gè)按元素值遞減有序（即非遞增有序，允許值相同）排列的線性表C。
標(biāo)簽： 數(shù)據(jù) 元素 減 線性
上傳時(shí)間： 2013-12-20
上傳用戶：笨小孩
剖析Intel IA32 架構(gòu)下C 語(yǔ)言及CPU 浮點(diǎn)數(shù)機(jī)制 Version 0.01 哈爾濱工業(yè)大學(xué) 謝煜波 （email: xieyubo@126.com 網(wǎng)址:http://purec.binghua.com） （QQ:13916830 哈工大紫丁香BBSID:iamxiaohan） 前言 這兩天翻看一本C 語(yǔ)言書的時(shí)候，發(fā)現(xiàn)上面有一段這樣寫到 例：將同一實(shí)型數(shù)分別賦值給單精度實(shí)型和雙精度實(shí)型，然后打印輸出。 #include <stdio.h> main() { float a double b a = 123456.789e4 b = 123456.789e4 printf(“％f\n％f\n”,a,b) } 運(yùn)行結(jié)果如下：
標(biāo)簽： Version xieyubo Intel email
上傳時(shí)間： 2013-12-25
上傳用戶：徐孺
源代碼\用動(dòng)態(tài)規(guī)劃算法計(jì)算序列關(guān)系個(gè)數(shù) 用關(guān)系"<"和"="將3個(gè)數(shù)a，b，c依次序排列時(shí)，有13種不同的序列關(guān)系： a=b=c,a=b<c,a<b=v,a<b<c,a<c<b a=c<b,b<a=c,b<a<c,b<c<a,b=c<a c<a=b,c<a<b,c<b<a 若要將n個(gè)數(shù)依序列，設(shè)計(jì)一個(gè)動(dòng)態(tài)規(guī)劃算法，計(jì)算出有多少種不同的序列關(guān)系， 要求算法只占用O(n),只耗時(shí)O(n*n).
標(biāo)簽： lt 源代碼 動(dòng)態(tài)規(guī)劃 序列
上傳時(shí)間： 2013-12-26
上傳用戶：siguazgb
The government of a small but important country has decided that the alphabet needs to be streamlined and reordered. Uppercase letters will be eliminated. They will issue a royal decree in the form of a String of B and A characters. The first character in the decree specifies whether a must come ( B )Before b in the new alphabet or ( A )After b . The second character determines the relative placement of b and c , etc. So, for example, "BAA" means that a must come Before b , b must come After c , and c must come After d . Any letters beyond these requirements are to be excluded, so if the decree specifies k comparisons then the new alphabet will contain the first k+1 lowercase letters of the current alphabet. Create a class Alphabet that contains the method choices that takes the decree as input and returns the number of possible new alphabets that conform to the decree. If more than 1,000,000,000 are possible, return -1. Definition
標(biāo)簽： government streamline important alphabet
上傳時(shí)間： 2015-06-09
上傳用戶：weixiao99
BNB20 Finds the constrained minimum of a function of several possibly integer variables. ％ Usage: [errmsg,Z,X,t,c,fail] = ％ BNB20(fun,x0,xstatus,xlb,xub,A,B,Aeq,Beq,nonlcon,settings,options,P1,P2,...) ％ ％ BNB solves problems of the form: ％ Minimize F(x) subject to: xlb <= x0 <=xub ％ A*x <= B Aeq*x=Beq ％ C(x)<=0 Ceq(x)=0 ％ x(i) is continuous for xstatus(i)=0 ％ x(i) integer for xstatus(i)= 1 ％ x(i) fixed for xstatus(i)=2 ％
標(biāo)簽： constrained variables function possibly
上傳時(shí)間： 2014-01-13
上傳用戶：youth25
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
標(biāo)簽： represented integers group items
上傳時(shí)間： 2016-01-17
上傳用戶：jeffery
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