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1. 下列說(shuō)法正確的是 （ ） A. Java語(yǔ)言不區(qū)分大小寫(xiě) B. Java程序以類(lèi)為基本單位 C. JVM為Java虛擬機(jī)JVM的英文縮寫(xiě) 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(kāi)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(hù)：netwolf
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(shí)間： 2017-02-11
上傳用戶(hù)：waizhang
特點(diǎn)(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(lèi)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(jì)(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(shí)間： 2013-11-24
上傳用戶(hù)：541657925
TLC2543是TI公司的12位串行模數(shù)轉(zhuǎn)換器，使用開(kāi)關(guān)電容逐次逼近技術(shù)完成A/D轉(zhuǎn)換過(guò)程。由于是串行輸入結(jié)構(gòu)，能夠節(jié)省51系列單片機(jī)I/O資源；且價(jià)格適中，分辨率較高，因此在儀器儀表中有較為廣泛的應(yīng)用。 TLC2543的特點(diǎn) （1）12位分辯率A/D轉(zhuǎn)換器； （2）在工作溫度范圍內(nèi)10μs轉(zhuǎn)換時(shí)間； （3）11個(gè)模擬輸入通道； （4）3路內(nèi)置自測(cè)試方式； （5）采樣率為66kbps； （6）線性誤差±1LSBmax； （7）有轉(zhuǎn)換結(jié)束輸出EOC； （8）具有單、雙極性輸出； （9）可編程的MSB或LSB前導(dǎo)； （10）可編程輸出數(shù)據(jù)長(zhǎng)度。 TLC2543的引腳排列及說(shuō)明 TLC2543有兩種封裝形式：DB、DW或N封裝以及FN封裝，這兩種封裝的引腳排列如圖1，引腳說(shuō)明見(jiàn)表1 TLC2543電路圖和程序欣賞 #include<reg52.h> #include<intrins.h> #define uchar unsigned char #define uint unsigned int sbit clock=P1^0; sbit d_in=P1^1; sbit d_out=P1^2; sbit _cs=P1^3; uchar a1,b1,c1,d1; float sum,sum1; double sum_final1; double sum_final; uchar duan[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f}; uchar wei[]={0xf7,0xfb,0xfd,0xfe}; void delay(unsigned char b) //50us { unsigned char a; for(;b>0;b--) for(a=22;a>0;a--); } void display(uchar a,uchar b,uchar c,uchar d) { P0=duan[a]|0x80; P2=wei[0]; delay(5); P2=0xff; P0=duan[b]; P2=wei[1]; delay(5); P2=0xff; P0=duan[c]; P2=wei[2]; delay(5); P2=0xff; P0=duan[d]; P2=wei[3]; delay(5); P2=0xff; } uint read(uchar port) { uchar i,al=0,ah=0; unsigned long ad; clock=0; _cs=0; port<<=4; for(i=0;i<4;i++) { d_in=port&0x80; clock=1; clock=0; port<<=1; } d_in=0; for(i=0;i<8;i++) { clock=1; clock=0; } _cs=1; delay(5); _cs=0; for(i=0;i<4;i++) { clock=1; ah<<=1; if(d_out)ah|=0x01; clock=0; } for(i=0;i<8;i++) { clock=1; al<<=1; if(d_out) al|=0x01; clock=0; } _cs=1; ad=(uint)ah; ad<<=8; ad|=al; return(ad); } void main() { uchar j; sum=0;sum1=0; sum_final=0; sum_final1=0; while(1) { for(j=0;j<128;j++) { sum1+=read(1); display(a1,b1,c1,d1); } sum=sum1/128; sum1=0; sum_final1=(sum/4095)*5; sum_final=sum_final1*1000; a1=(int)sum_final/1000; b1=(int)sum_final%1000/100; c1=(int)sum_final%1000%100/10; d1=(int)sum_final%10; display(a1,b1,c1,d1); } }
上傳時(shí)間： 2013-11-19
上傳用戶(hù)：shen1230
#include<iom16v.h> #include<macros.h> #define uint unsigned int #define uchar unsigned char uint a,b,c,d=0; void delay(c) { for for(a=0;a<c;a++) for(b=0;b<12;b++); }; uchar tab[]={ 0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90,
標(biāo)簽： AVR 單片機(jī) 數(shù)碼管
上傳時(shí)間： 2013-10-21
上傳用戶(hù)：13788529953
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
上傳時(shí)間： 2015-05-21
上傳用戶(hù)：daguda
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
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
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
Generate 100 samples of a zero-mean white noise sequence with variance , by using a uniform random number generator. a Compute the autocorrelation of for . b Compute the periodogram estimate and plot it. c Generate 10 different realizations of , and compute the corresponding sample autocorrelation sequences , and . Compute the average autocorrelation sequence as and the corresponding periodogram for . d Compute and plot the average periodogram using the Bartlett method. e Comment on the results in parts (a) through (d).
標(biāo)簽： zero-mean Generate sequence variance
上傳時(shí)間： 2016-03-04
上傳用戶(hù)：朗朗乾坤
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