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特點(FEATURES) 精確度0.1%滿刻度 (Accuracy 0.1%F.S.) 可作各式數(shù)學(xué)演算式功能如:A+B/A-B/AxB/A/B/A&B(Hi or Lo)/|A| (Math functioA+B/A-B/AxB/A/B/A&B(Hi&Lo)/|A|/etc.....) 16 BIT 類比輸出功能(16 bit DAC isolating analog output function) 輸入/輸出1/輸出2絕緣耐壓2仟伏特/1分鐘(Dielectric strength 2KVac/1min. (input/output1/output2/power)) 寬范圍交直流兩用電源設(shè)計(Wide input range for auxiliary power) 尺寸小,穩(wěn)定性高(Dimension small and High stability)
標(biāo)簽：
微電腦
數(shù)學(xué)演算
輸出
隔離傳送器
上傳時間：
2013-11-24
上傳用戶：541657925
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/*--------- 8051內(nèi)核特殊功能寄存器 -------------*/
sfr ACC = 0xE0; //累加器
sfr B = 0xF0; //B 寄存器
sfr PSW = 0xD0; //程序狀態(tài)字寄存器
sbit CY = PSW^7; //進位標(biāo)志位
sbit AC = PSW^6; //輔助進位標(biāo)志位
sbit F0 = PSW^5; //用戶標(biāo)志位0
sbit RS1 = PSW^4; //工作寄存器組選擇控制位
sbit RS0 = PSW^3; //工作寄存器組選擇控制位
sbit OV = PSW^2; //溢出標(biāo)志位
sbit F1 = PSW^1; //用戶標(biāo)志位1
sbit P = PSW^0; //奇偶標(biāo)志位
sfr SP = 0x81; //堆棧指針寄存器
sfr DPL = 0x82; //數(shù)據(jù)指針0低字節(jié)
sfr DPH = 0x83; //數(shù)據(jù)指針0高字節(jié)
/*------------ 系統(tǒng)管理特殊功能寄存器 -------------*/
sfr PCON = 0x87; //電源控制寄存器
sfr AUXR = 0x8E; //輔助寄存器
sfr AUXR1 = 0xA2; //輔助寄存器1
sfr WAKE_CLKO = 0x8F; //時鐘輸出和喚醒控制寄存器
sfr CLK_DIV = 0x97; //時鐘分頻控制寄存器
sfr BUS_SPEED = 0xA1; //總線速度控制寄存器
/*----------- 中斷控制特殊功能寄存器 --------------*/
sfr IE = 0xA8; //中斷允許寄存器
sbit EA = IE^7; //總中斷允許位
sbit ELVD = IE^6; //低電壓檢測中斷控制位
8051
標(biāo)簽：
80C51
特殊功能寄存器
地址
上傳時間：
2013-10-30
上傳用戶：yxgi5
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關(guān)鍵詞 I/O 口擴展芯片、I2C、SMBus摘要CAT9554A 是一款將I2C/SMBus 接口擴展成8 位并行輸入/輸出I/O 口的器件
標(biāo)簽：
9554A
9554
CAT
IO
上傳時間：
2013-12-27
上傳用戶：txfyddz
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題目：利用條件運算符的嵌套來完成此題：學(xué)習(xí)成績>=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這是條件運算符的基本例子。
標(biāo)簽：
gt
90
運算符
嵌套
上傳時間：
2015-01-08
上傳用戶：lifangyuan12
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RSA算法 :首先, 找出三個數(shù), p, q, r, 其中 p, q 是兩個相異的質(zhì)數(shù), r 是與 (p-1)(q-1) 互質(zhì)的數(shù)...... p, q, r 這三個數(shù)便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 這個 m 一定存在, 因為 r 與 (p-1)(q-1) 互質(zhì), 用輾轉(zhuǎn)相除法就可以得到了..... 再來, 計算 n = pq....... m, n 這兩個數(shù)便是 public_key ，編碼過程是, 若資料為 a, 將其看成是一個大整數(shù), 假設(shè) a < n.... 如果 a >= n 的話, 就將 a 表成 s 進位 (s
標(biāo)簽：
person_key
RSA
算法
上傳時間：
2013-12-14
上傳用戶：zhuyibin
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數(shù)字運算，判斷一個數(shù)是否接近素數(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
運算
上傳時間：
2015-05-21
上傳用戶：daguda
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上下文無關(guān)文法（Context-Free Grammar, CFG）是一個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的元素稱為非終結(jié)符，T的元素稱為終結(jié)符，S是一個特殊的非終結(jié)符，稱為文法開始符。
設(shè)G=(V, T, S, P)是一個CFG，則G產(chǎn)生的語言是所有可由G產(chǎn)生的字符串組成的集合，即L(G)={x∈T* | Sx}。一個語言L是上下文無關(guān)語言（Context-Free Language, CFL），當(dāng)且僅當(dāng)存在一個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)符都是大寫字母，開始符都是S，終結(jié)符都是小寫字母。
標(biāo)簽：
Context-Free
Grammar
CFG
上傳時間：
2013-12-10
上傳用戶：gaojiao1999
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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
上傳時間：
2016-01-17
上傳用戶：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
上傳時間：
2016-02-12
上傳用戶：a673761058
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漢諾塔?。?！
Simulate the movement of the Towers of Hanoi puzzle Bonus is possible for using animation
eg. if n = 2 A→B A→C B→C
if n = 3 A→C A→B C→B A→C B→A B→C A→C
標(biāo)簽：
the
animation
Simulate
movement
上傳時間：
2017-02-11
上傳用戶：waizhang
