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摘要: 串行傳輸技術(shù)具有更高的傳輸速率和更低的設(shè)計(jì)成本, 已成為業(yè)界首選, 被廣泛應(yīng)用于高速通信領(lǐng)域。提出了一種新的高速串行傳輸接口的設(shè)計(jì)方案, 改進(jìn)了Aurora 協(xié)議數(shù)據(jù)幀格式定義的弊端, 并采用高速串行收發(fā)器Rocket I/O, 實(shí)現(xiàn)數(shù)據(jù)率為2.5 Gbps的高速串行傳輸。關(guān)鍵詞: 高速串行傳輸; Rocket I/O; Aurora 協(xié)議
為促使FPGA 芯片與串行傳輸技術(shù)更好地結(jié)合以滿(mǎn)足市場(chǎng)需求, Xilinx 公司適時(shí)推出了內(nèi)嵌高速串行收發(fā)器RocketI/O 的Virtex II Pro 系列FPGA 和可升級(jí)的小型鏈路層協(xié)議———Aurora 協(xié)議。Rocket I/O支持從622 Mbps 至3.125 Gbps的全雙工傳輸速率, 還具有8 B/10 B 編解碼、時(shí)鐘生成及恢復(fù)等功能, 可以理想地適用于芯片之間或背板的高速串行數(shù)據(jù)傳輸。Aurora 協(xié)議是為專(zhuān)有上層協(xié)議或行業(yè)標(biāo)準(zhǔn)的上層協(xié)議提供透明接口的第一款串行互連協(xié)議, 可用于高速線(xiàn)性通路之間的點(diǎn)到點(diǎn)串行數(shù)據(jù)傳輸, 同時(shí)其可擴(kuò)展的帶寬, 為系統(tǒng)設(shè)計(jì)人員提供了所需要的靈活性[4]。但該協(xié)議幀格式的定義存在弊端,會(huì)導(dǎo)致系統(tǒng)資源的浪費(fèi)。本文提出的設(shè)計(jì)方案可以改進(jìn)Aurora 協(xié)議的固有缺陷,提高系統(tǒng)性能, 實(shí)現(xiàn)數(shù)據(jù)率為2.5 Gbps 的高速串行傳輸, 具有良好的可行性和廣闊的應(yīng)用前景。
標(biāo)簽：
Rocket
2.5
高速串行
收發(fā)器
上傳時(shí)間：
2013-10-13
上傳用戶(hù)：lml1234lml
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題目：利用條件運(yùn)算符的嵌套來(lái)完成此題：學(xué)習(xí)成績(jī)>=90分的同學(xué)用A表示，60-89分之間的用B表示，60分以下的用C表示。 1.程序分析：(a>b)?a:b這是條件運(yùn)算符的基本例子。
標(biāo)簽：
gt
90
運(yùn)算符
嵌套
上傳時(shí)間：
2015-01-08
上傳用戶(hù)：lifangyuan12
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RSA算法 :首先, 找出三個(gè)數(shù), p, q, r, 其中 p, q 是兩個(gè)相異的質(zhì)數(shù), r 是與 (p-1)(q-1) 互質(zhì)的數(shù)...... p, q, r 這三個(gè)數(shù)便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 這個(gè) m 一定存在, 因?yàn)?r 與 (p-1)(q-1) 互質(zhì), 用輾轉(zhuǎn)相除法就可以得到了..... 再來(lái), 計(jì)算 n = pq....... m, n 這兩個(gè)數(shù)便是 public_key ，編碼過(guò)程是, 若資料為 a, 將其看成是一個(gè)大整數(shù), 假設(shè) a < n.... 如果 a >= n 的話(huà), 就將 a 表成 s 進(jìn)位 (s
標(biāo)簽：
person_key
RSA
算法
上傳時(shí)間：
2013-12-14
上傳用戶(hù)：zhuyibin
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數(shù)字運(yùn)算，判斷一個(gè)數(shù)是否接近素?cái)?shù)
A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value.
Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not.
Input
Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone.
Output
For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise.
Sample Input
10 111
2 110
10 123
6 1000
8 2314
0
Sample Output
yes
yes
no
yes
no
標(biāo)簽：
數(shù)字
運(yùn)算
上傳時(shí)間：
2015-05-21
上傳用戶(hù)：daguda
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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(hù)：siguazgb
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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(hù)：weixiao99
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上下文無(wú)關(guān)文法（Context-Free Grammar, CFG）是一個(gè)4元組G=(V, T, S, P)，其中，V和T是不相交的有限集，S∈V，P是一組有限的產(chǎn)生式規(guī)則集，形如A→α，其中A∈V，且α∈(V∪T)*。V的元素稱(chēng)為非終結(jié)符，T的元素稱(chēng)為終結(jié)符，S是一個(gè)特殊的非終結(jié)符，稱(chēng)為文法開(kāi)始符。
設(shè)G=(V, T, S, P)是一個(gè)CFG，則G產(chǎn)生的語(yǔ)言是所有可由G產(chǎn)生的字符串組成的集合，即L(G)={x∈T* | Sx}。一個(gè)語(yǔ)言L(fǎng)是上下文無(wú)關(guān)語(yǔ)言（Context-Free Language, CFL），當(dāng)且僅當(dāng)存在一個(gè)CFG G，使得L=L(G)。 *⇒
例如，設(shè)文法G：S→AB
A→aA|a
B→bB|b
則L(G)={a^nb^m | n,m>=1}
其中非終結(jié)符都是大寫(xiě)字母，開(kāi)始符都是S，終結(jié)符都是小寫(xiě)字母。
標(biāo)簽：
Context-Free
Grammar
CFG
上傳時(shí)間：
2013-12-10
上傳用戶(hù)：gaojiao1999
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一:需求分析
1. 問(wèn)題描述
魔王總是使用自己的一種非常精練而抽象的語(yǔ)言講話(huà),沒(méi)人能聽(tīng)懂,但他的語(yǔ)言是可逐步解釋成人能聽(tīng)懂的語(yǔ)言,因?yàn)樗恼Z(yǔ)言是由以下兩種形式的規(guī)則由人的語(yǔ)言逐步抽象上去的:
-----------------------------------------------------------
(1) a---> (B1)(B2)....(Bm)
(2)[(op1)(p2)...(pn)]---->[o(pn)][o(p(n-1))].....[o(p1)o]
-----------------------------------------------------------
在這兩種形式中,從左到右均表示解釋.試寫(xiě)一個(gè)魔王語(yǔ)言的解釋系統(tǒng),把
他的話(huà)解釋成人能聽(tīng)得懂的話(huà).
2. 基本要求:
用下述兩條具體規(guī)則和上述規(guī)則形式(2)實(shí)現(xiàn).設(shè)大寫(xiě)字母表示魔王語(yǔ)言的詞匯 小寫(xiě)字母表示人的語(yǔ)言的詞匯 希臘字母表示可以用大寫(xiě)字母或小寫(xiě)字母代換的變量.魔王語(yǔ)言可含人的詞匯.
(1) B --> tAdA
(2) A --> sae
3. 測(cè)試數(shù)據(jù):
B(ehnxgz)B 解釋成 tsaedsaeezegexenehetsaedsae若將小寫(xiě)字母與漢字建立下表所示的對(duì)應(yīng)關(guān)系,則魔王說(shuō)的話(huà)是:"天上一只鵝地上一只鵝鵝追鵝趕鵝下鵝蛋鵝恨鵝天上一只鵝地上一只鵝".
| t | d | s | a | e | z | g | x | n | h |
| 天 | 地 | 上 | 一只| 鵝 | 追 | 趕 | 下 | 蛋 | 恨 |
標(biāo)簽：
語(yǔ)言
抽象
分
上傳時(shí)間：
2014-12-02
上傳用戶(hù)：jkhjkh1982
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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(hù)：jeffery
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The XML Toolbox converts MATLAB data types (such as double, char, struct, complex, sparse, logical) of any level of nesting to XML format and vice versa.
For example,
>> project.name = MyProject
>> project.id = 1234
>> project.param.a = 3.1415
>> project.param.b = 42
becomes with str=xml_format(project, off )
"<project>
<name>MyProject</name>
<id>1234</id>
<param>
<a>3.1415</a>
<b>42</b>
</param>
</project>"
On the other hand, if an XML string XStr is given, this can be converted easily to a MATLAB data type or structure V with the command V=xml_parse(XStr).
標(biāo)簽：
converts
Toolbox
complex
logical
上傳時(shí)間：
2016-02-12
上傳用戶(hù)：a673761058
