复数

是集体名词,不是整体名词啊! 这类名词在意义上是复数,而在语法形式上是单数. 这类名词作主语时,遵循“语法一致”和“意义一致”两种原则的选择. 举例常见例子： 1）police,people,cattle.通常作复数,随后的动词用复数. 2）machinery,equipment,furniture,merchandise,通常作不可数名词,随后动词用单数. 3）audiance,committee,class,crew,family,government,public,既可作单数,也可作复数用.如将各该名词所表示的集体视为一个整体,则动词用单数.如将侧重点放在组成集体的成员上,动词用复数. 比如说：那个观众很激动（谓语动词用单数）；所有观众很激动（用复数）.

一般不可数名字+S是指一类事物吧，像fish用作复数值鱼类，而不是指几条鱼

someequipment总复数。解释为，一些设备。例句：Ifyou"replanningondoingsomeseriousclimbingandyouneedsomeequipment，wedohaveanexcellentclimbingsuppliesstorejustfiveminutes"walkaway.如果你打算真正的攀岩，你需要一些设备，我们知道一个很不错的攀岩用品商店，从这里步行只需要五分钟。

这个应该就是“装备, 设备, 器材, 装置, 铁道车辆, (一企业除房地产以外的)固定资产, 才能”的意思，应该没有特殊用法

有啊，加-s 2011-10-25 6:44:38

单复同形。

equipment是不可数名词，但如果表示设备的种类时，可以在equipment后加-s,指的是多种设备。

没有。这是不可数名词!没有负数!

字典上说是不可数名词

There are some flowers in my bedroom. There are two penciles on my desk. There are some toys on my bed. There is a beautiful pillow on my bed. There is a table in this corner. There is a broom behind the door.

There are some flowers in my bedroom.There are two penciles on my desk.There are some toys on my bed.There is a beautiful pillow on my bed.There is a table in this corner.There is a broom behind the door.

玉米有三种表示法 一：corn [kC:n] n. 1〔英〕谷物； 五谷 2〔美、加、澳〕玉米 3〔苏、爱〕燕麦， [英]小麦 4老套而陈旧的笑话（指故事、音乐等） 5谷粒， (胡椒， 水果等的)子 6〔美口〕玉米威士忌酒 7〔美俚〕钱 Chinese corn 粟 Indian corn 玉蜀黍 corn flour (精磨)玉米面 corn flakes 玉米片 corn bread 玉米面包 corn chip [美]玉米煎饼 corn stalk 麦〔玉米〕杆 corn beef [美]碎咸牛肉 词性变化： corn [kC:n] vt. 1使成料状 2腌（肉类等） 3给（土地）种上谷物 4以谷物喂（牲畜） vi. (谷穗等)成熟， 结实 继承用法 corncake [`kR:nkeIk] n. 玉米饼 corncob [5kC:nkRb] n. 玉米芯， 玉米穗轴 corncrib [`kR:nkrIb] n. [美]玉米透风仓， 玉米囤 cornflower [5kC:nflaJE(r)] n. 矢车菊 cornhusker [`kR:n9hQskE(r)] n. 玉米穗剥皮人〔机〕 cornhusking [`kR:n9hQskIN] n. [美]剥玉米； 剥玉米会（一般有舞会等） cornmeal [`kR:nmi:l] n. 麦片， [美]玉米粉 机 cornmint n. 日本薄荷 corn-picker n. 玉米摘穗机 cornstock n. 玉米杆； 麦秸 cornstarch [`kR:nstB:tF] n. 玉米淀粉 cornless adj. 无籽粒的 cornlet n. 未成熟的玉米仁 习惯用语 acknowledge the corn 认错， 承认事实， (辩论中)认输 admit the corn 认错， 承认事实， (辩论中)认输 confess the corn 认错， 承认事实， (辩论中)认输 carry corn [俚]不因成功而忘乎所以； 不得意忘形（借喻马吃玉米后的状态） eat one"s corn in the blade 寅吃卯粮， (钱)未入先出， 挥霍无度 feed sb. on soft corn [美俚]奉承某人， 拍某人的马屁 measure another"s corn by one"s own bushel 以已度人， 拿自己做标准衡量别人 sell corn on the hoof 以饲养牲畜的方式出售谷物（即不直接出售谷物， 而以谷物饲养牲畜， 然后将牲畜出售） corn in Egypt 大量的食物； 充裕； 供应意想不到地多 特殊用法 barley corn 大麦粒 broom corn 高梁， 帚高梁 burnt corn 焦谷粒 canned crushed corn 玉米碎粒罐头， 罐装玉米碎粒 canned whole grain corn 整玉米粒罐头 cream-style canned corn 玉米糊罐头 dent corn 臼齿形（顶陷）玉米 dredge corn 混合谷物饲料； 脱下的玉米粒 guinnea corn (=millet) 高粱， 蜀黍 hard corn 小麦， 黑麦 kafir corn 高粱 Lent corn (grain) [英]斋谷 Maryland-style canned corn 马里兰式甜玉米罐头（盐水整粒玉米罐头） pepper corn 干胡椒， 胡椒子， 胡椒粒 rice corn 稻米； 白高粱 saracen corn (Fagopyrum) 荞麦 shelled corn 玉米粒 standing corn 未割的庄稼〔谷物〕 sugar corn 甜玉米， 餐用玉米 table corn 甜玉米， 餐用玉米 sweet corn 甜玉米， 餐用玉米 sugared popped corn 加糖玉米花 sun-red corn 日光红玉米 turkey corn 紫堇 waxy corn 蜡质种玉米 whole kernel corn 整粒玉米 二：maize [meiz] n. 玉米, 黄色 adj. 玉米色的, 黄色的 现代英汉综合大词典 maize [meiz] n. [英]【植】玉蜀黍, 玉米(美国、加拿大叫 corn) 玉米色, 黄色 the maize country [美]穷乡僻壤 词性变化 maize [meiz] adj. 玉米色的, 黄色的 特殊变化 cut maize 玉米片 true maize (=flint corn) 黄质玉米 yellow maize 黄玉米 leaf spot maize 玉米叶小斑病 white maize 白玉米 三：mealie ["mi:li] n. 玉米 mealie ["mi:li] n. 1玉米果穗, 玉蜀黍穗 2[pl. ]玉米, 玉蜀黍

There are some flowers in my bedroom. There are two penciles on my desk. There are some toys on my bed. There is a beautiful pillow on my bed. There is a table in this corner. There is a broom behind the door.

名词单复数与动词三单及动词ing一、请写出下列名词的复数形式,没有复数形式的请划出/。（25分）bus______ fox______ boy______ day______ zoo______tree______ deer______ fish______ city______ leaf______life______ milk______ foot______ horse______ mouse______tooth______ woman______ broom______ juice______ water______people______ branch______ family______ ox______ country______二、填入所给名词的正确形式。（5分）1. The ______________ are playing football now. (child)2. There are ten ___________________in our school. (woman teacher)3. Most of __________ live in __________. （German）4. There are three _________ and five _______ in the room. (Chinese, German)5. Could you please give me some __________? (milk)三、请写出下列动词的三人称单数及现在分词形式。（20分）sit_____________ _____________ swim_____________ _____________say_____________ _____________ play_____________ _____________run_____________ _____________ wash_____________ _____________cry_____________ _____________ draw_____________ _____________die_____________ _____________ make_____________ _____________see_____________ _____________ have_____________ _____________go_____________ _____________ write_____________ _____________do_____________ _____________ study_____________ _____________watch_____________ _____________ dance_____________ _____________stop_____________ ______________ sing_____________ ______________四、用do. does填空。（10分）1. _____ you ride a bike after school? Yes, I ________. 2. ______ your sister like PE? No, she ______ not. 3. What_______ the students have? They have some pens. 4. How______ Linda go to school? She goes to school on foot. 5. He ______ not speak English. He speaks Chinese. 6. ______ they watch TV on Sundays? Yes, they ______.7. My father and mother ______ not read newspapers on Saturday.8. ________ you know that girl? Yes, she is my sister.9. ________ your father have CDs? No, he _____________.10. Why ________ you ________ your homework? We don"t like it.五、用一般现在时填空。（5分）1. My mother______not_____(like) English. She_____(like) Chinese.2. ______you______(go) to school by bus? No, I_____(go) to school by car.3. Miss Wang ______ (swim) every day.4. I ___(like) English. Tom ____ ______ _____(not like) English.5. When_____ you ______ (go) to school? I _____ (go) to school at five every day.六、用进行时填空。（10分）1. I ____ _____ (read) English now.2. He ______ _______ (go) to the park now.3. We ________ ________ (have) an English class.4. What ______ they _____ (do)? They _____ _____ (sit) in the park.5. My mother ________ ________ (clean) the room now.6. The boy ________ (draw) a picture now.7. Listen. Some girls ________ (sing) in the classroom.8. My mother __________ (cook) some nice food now.9. What _____ you ______ (do) now?10. Look. They _________ (have) an English lesson.七、单项选择。(10分)（ ）1. 我在照看孩子。A. I am looking after the baby. B. I"m look aftering the baby.C. I look am aftering the baby. D. I looking after the baby.( )2. _____friend"s making______a kite.A. I, me B. My, my C. My, me D. His, his( )3. Is the woman ______ yellow your teacher?A. in B. putting on C. wearing D. having( )4. Look!The twins_____their mother do the housework.A. are wanting B. help C. are helping D. are looking( )5. _____are the birds doing? They are singing in a tree.A. Who B. What C. How D. Where( )6. Is she____something?A. eat B. eating C. eatting D. eats( )7. 你在干什么?A. What is you doing? B. What are you do?C. What are you doing? D. What do you do?( )8. What are you listening_____?A. / B. for C. at D. to( )9. 我正在听他说话。A. I listening to him. B. I"m listening to him.C. I"m listen to him. D. I"m listening him.( )10. They are_____their clothes.A. makeing B. putting C. put away D. putting on八、用现在进行时完成下列句子：（10分）1. What___________you____________(do)?2. I_________________(sing) an English song.3. What________________he__________________(mend)?4. He__________________(mend) a car.5. _________you______________(fly) a kite? Yes, ____________.6. ___________she_______________(sit) in the boat?7. __________you_____________(ask) questions?8. We_________________________(play) games now.9. Look. Three boys _______（run）.10. Listen, someone _______（sing）in the classroom.九、选择填空：（5分）（ ）1. Who ______ over there now?A. singing B. are sing C. is singing（ ）2. It"s nine ten. The students ______ a music class.A. have B. having C. are having（ ）3. Listen! The boy _______.A. crying B. is crying C. cries（ ）4. Don"t talk here. Grandparents ______.A. sleep B. is sleeping C. are sleeping（ ）5. Is the man _______ tea or milk?A. drinks B. drink C. drinking十、按要求进行句型转换：（10分）1. Look! Lily is dancing. （改为一般疑问句）________________________________________________________________2. There are three monkeys. （改为一般疑问句）________________________________________________________________3. We can see many fish in the lake. （改为一般疑问句）________________________________________________________________4. They like to ride bikes in the park. （改为一般疑问句）________________________________________________________________5. Kate is looking for her watch. （改为否定句）________________________________________________________________6. I can draw a sun in the picture. （改为否定句）________________________________________________________________7. She goes to the English classes everyday. （改为否定句）________________________________________________________________8. Mrs White is watching TV. （对划线部分提问）________________________________________________________________9. I am doing homework. （改为否定句）________________________________________________________________10. They are waiting for you at the library. （就划线部分提问）________________________________________________________________

你应该把题目贴上来呀

----- They have flowers.

并非所有的名词的复数都是不规则变化。broom，它的复数直接在后面加s。

brooms

加不加ing跟复数没关系

像词义是“物”的呢，比如你说的苹果店、花店这些一般前面的名词作形容词用，用的是单数！像词义是“人”的呢，像女医生、男医生、女老师之类的，前面的名词作形容词用复数形式！但是像运动（sport）这个单词，是比较特别的，它用于做运动（dosports）、运动用品店（Sportsshop）这些名词作形容词的用法时，它一般加复数。有些特殊的词组混在有规律的词组中，也别太钻研它，记住就好了。

Those are flower stamps.

y=e^(a+ib)xy"=(a+ib)e^(a+ib)x一般这种类型的复数首先化为e^(f(x)+i*g(x)),求导为e^(f(x)+i*g(x))*(f"(x)+i*(g"(x)))本题中f(x)=ax，g(x)=bxf"(x)=a，g"(x)=b

因为复数直接进行乘除法运算的话非常费劲，所以通常都是转化为指数形式进行计算，也就是你所说的极坐标形式；指数形式的复数进行乘除法比较简便，但计算加减法比较费劲，所以计算加减法时一般都是使用复数的代数形式。

幅值相量和有效值相量都可以写成是一般复数的极坐标形式；相量加减乘除运算与复数运算没有区别，在算加减法时用复数的一般形式比较简单，在算乘除法时用复数的极坐标形式也就是相量形式比较简单；正弦量与其相量与其相量之间可以转化。正弦量的表达式又叫时域形式，与相量形式相对应。其中，相量的模是正弦量幅值的根号2分之一倍，阻抗角是争先量的初相角。

用只有两个元素的1维数组存放复数的实部和虚部。然后根据具体的加减乘除法则进行计算。

复数加减法，一般化成直角坐标形式。

按照你要的程序 这里有个 希望对你有用 谢谢public class Complex { private double realPart; private double imaginaryPart; public Complex(double a, double b) { this.realPart = a; this.imaginaryPart = b; } public Complex add(Complex a) { Complex result = new Complex(this.realPart + a.realPart, this.imaginaryPart + a.imaginaryPart);//(why?) return result; } public Complex decrease(Complex a) { Complex result = new Complex(this.realPart - a.realPart, this.imaginaryPart - a.imaginaryPart);//(why?) return result; } public Complex multiply(Complex a) { double newReal = this.realPart*a.realPart - this.imaginaryPart * a.imaginaryPart; double newImaginary = this.realPart*a.imaginaryPart + this.imaginaryPart * a.realPart; Complex result = new Complex(newReal, newImaginary); return result; } public Complex divide(Complex a) { Complex conjugate = new Complex(this.realPart, -this.imaginaryPart); Complex multiplication = conjugate.multiply(a); multiplication.realPart /= this.realPart*this.realPart + this.imaginaryPart * this.imaginaryPart; multiplication.imaginaryPart /= this.realPart*this.realPart + this.imaginaryPart * this.imaginaryPart; return multiplication; } public String toString() { String show = this.realPart + " + " + this.imaginaryPart + "i"; return show; } public static void main(String [] args) { Complex a = new Complex (2, 3); Complex b = new Complex (1,1); System.out.println((a.add(b)).toString()); System.out.println((a.decrease(b)).toString()); System.out.println((a.multiply(b)).toString()); System.out.println((a.divide(b)).toString()); } }

相量？复数..角度幅度包裹。与正弦函数可以表示是非常相似.. 的正弦曲线的振幅的振幅值的瞬时的电压或电流，振幅的相位角的正弦.. 1的正弦函数表示...为什么更好的复数加法，减法，乘法和除法的多个.. 为什么能代表正弦函数..根据欧拉公式E ^ IX = cosx + isinx。正弦函数的虚步..复数加法是独立的每个开口的实数部分和虚数部分的加法和减法，所以更多的，而不是正弦函数相互减法.. （基尔霍夫电压和电流定律）。 3虚，乘法和除法，如U / I = JWL，U / I =-JXC这是基于正弦和排放，然后更复杂的结果添加J衍生..除了这个地方，而不是正弦相量乘法和除法..在其他地方，如计算能力，相对量不能使用。 短，减法使用相量，而不是正弦运算符计数阻抗，也可以使用在其他时间，并且不能使用！

easy~~#include<iostream>#include<math.h>using namespace std;class Complex{private: double real; double image;public: Complex(double real,double image) { this-> real=real; this->image=image; } float modulo () void display() { cout<<"("<<real<<","<<image<<")"<<endl; }};int main(){cout<<"enter the real and the image"<<endl;double A,B;cout<<"输入数的实部:";cin>>A;cout<<endl<<"输入数的虚部:";cin>>B;Complex C (A,B);C.display();cout<<"此复数的模为:"<<C.modulo();}

①先将极坐标形式化成直角坐标形式：20∠60°=20cos60°+j20sin60°=10+j17.320∠-45°=10cos-45°+j10sin-45°=7.07-j7.07②再将直角坐标式的实部与实部相加，虚部与虚部相加。这样就得到第二行。

bu ke yi

因为复数虽然是表示为a+bi的形式，但它和向量确实不是一回事啊~~ 复数终归就是一个数啊~ 2维向量的两个维是等同的，而复数的1和i可以看成是不同的单位。而且i*i=-1，这一点是向量办不到的。 另外，比如说复数可以计算几次方，向量根本没这回事。

complex complex::operator /(complex &p) {complex n;double s=p.real*p.real+p.imag*p.imag;n.real=(real*p.real+imag*p.imag)/s;n.imag=(imag*p.real-real*p.imag)/s;return n;}这是复数除号运算符重载的代码，验证了下应该没问题。

你的函数从头到尾都是结构体啊，你题目的要求是用类实现，补充你的函数是用结构体实现，你到底是要实现你题目的要求，还是补充你的函数呢？

C++的问题：定义描述复数的结构体类型变量，并实现复数之间的加减法运算和输入输出我记得在书本上直接有一个案例的，直接引用模板即可解决。

public class ComplexNumber { /** * @param args */ int shi,xu;//复数的实部和虚部 public ComplexNumber(int n,int ni){ shi = n; xu = ni; } public void ComplexShow(){ String output = ""; output+=shi; if(xu>=0){ output+="+"; } output+=xu; output+="i"; System.out.println(output); } public void ComplexShow1(){//不要换行 String output = ""; output+=shi; if(xu>=0){ output+="+"; } output+=xu; output+="i"; System.out.print(output); } public static void ComplexAdd(ComplexNumber x1,ComplexNumber x2){//实现两个复数相加 ComplexNumber cn = new ComplexNumber(0, 0);//将两个复数相加等于cn cn.shi = x1.shi + x2.shi; cn.xu = x1.xu + x2.xu; cn.ComplexShow(); } public static void ComplexMinus(ComplexNumber x1,ComplexNumber x2){//实现两个复数相减，第一个数减第二个数 ComplexNumber cn = new ComplexNumber(0, 0);//将两个复数相加等于cn cn.shi = x1.shi - x2.shi; cn.xu = x1.xu - x2.xu; cn.ComplexShow(); } public static void ComplexMultiply(ComplexNumber x1,ComplexNumber x2){//实现两个复数相乘 ComplexNumber cn = new ComplexNumber(0, 0);//将两个复数相加等于cn cn.shi = x1.shi * x2.shi - x1.xu * x2.xu; cn.xu = x1.xu * x2.shi + x2.xu * x1.shi; cn.ComplexShow(); } public static void ComplexDivide(ComplexNumber x1,ComplexNumber x2){//实现两个复数相除，第一个数除以第二个数 ComplexNumber x2_gong = new ComplexNumber(x2.shi,0-x2.xu);//求被除数的共轭复数 ComplexNumber cn = new ComplexNumber(0, 0);//将两个复数相加等于cn cn.shi = x1.shi * x2_gong.shi - x1.xu * x2_gong.xu;//x1/x2,求分子实部 cn.xu = x1.xu * x2_gong.shi + x2_gong.xu * x1.shi;//x1/x2,求分子虚部 int fenMu = x2.shi * x2.shi + x2.xu * x2.xu; if(fenMu!=0){ System.out.print("("); cn.ComplexShow1(); System.out.print(")"); System.out.println("/"+fenMu); } else System.out.println("分母为零，无法相除"); } public static void main(String[] args) { // TODO Auto-generated method stub ComplexNumber cn = new ComplexNumber(-1, -1);//初始化复数 cn.ComplexShow();//显示复数 ComplexNumber c1 = new ComplexNumber(-1, -1); ComplexNumber c2 = new ComplexNumber(1, 1); System.out.print("加："); ComplexAdd(c1, c2); System.out.print("减："); ComplexMinus(c1, c2); System.out.print("乘："); ComplexMultiply(c1, c2); System.out.print("除："); ComplexDivide(c1, c2);//自己化简 }}

先把复数算出结果，再加上常数就可以得出答案。在计算复数加常数的时候，要先把复数的短式子计算出来，再用计算出来的答案和常数相加就可以得出最终的答案。复数运算法则有加减法、乘除法，两复数之和是复数，所得和的实部数值为原两复数实部之和，所得和的虚部是原两复数虚部之和，复数的加法满足交换律和结合律，两复数之差是复数，所得差的实部是原来两个复数实部的差，所得差的虚部是原两复数虚部之差。两复数之积是复数，两复数相除计算方法与乘法相同，需要分子分母相乘时乘分母的共轭，互为共轭的两复数之积是实常数。

你不嫌麻烦，用正弦定理与余弦定理算，也可以。

1+2i+3+4i=4+6i 只能这么加

复数运算Description输入两个复数a1+b1i和a2+b2i，对此进行加（+）、减（-）、乘（*）和除（/）法运算。Input输入多组测试数据，第一行为测试数据的组数T，每组测试数据由3行组成，第1行为a1和b1，第2行为一个字符，代表a1+b1i与a2+b2i所进行的运算（可能的值有+、-、*、/），第3行为a2和b2。这里的a1、b1、a2、b2均为整数。我们的测试数据保证在进行除法运算时不会发生被0除的情况。Output每组测试数据输出一行，即计算后的结果a3+b3i的最简形式，中间不包含任何空格，其中a3和b3均保留1位小数。SampleInput311+1111-1011+-1-1#include<iostream>#include<iomanip>usingnamespacestd;classComplex{private:doublereal;doubleimag;public:Complex(){real=0;imag=0;}voidInit();Complexoperator+(Complex&c2);Complexoperator-(Complex&c2);Complexoperator*(Complex&c2);Complexoperator/(Complex&c2);doubleGet_real(){returnreal;}doubleGet_imag(){returnimag;}voiddisplay();};voidComplex::Init(){cin>>real>>imag;}ComplexComplex::operator+(Complex&c2){Complexc;c.real=real+c2.real;c.imag=imag+c2.imag;returnc;}ComplexComplex::operator-(Complex&c2){Complexc;c.real=real-c2.real;c.imag=imag-c2.imag;returnc;}ComplexComplex::operator*(Complex&c2){Complexc;c.real=real*c2.real-imag*c2.imag;c.imag=real*c2.imag+imag*c2.real;returnc;}ComplexComplex::operator/(Complex&c2){Complexc;doublet=c2.real*c2.real+c2.imag*c2.imag;c.real=(real*c2.real+imag*c2.imag)/t;c.imag=(imag*c2.real-real*c2.imag)/t;returnc;}voidComplex::display(){doublere=real,im=imag;if(re==0){if(im==0)cout<<"0.0 ";else{if(im==1)cout<<"i"<<endl;elseif(im==-1)cout<<"-i"<<endl;elsecout<<setiosflags(ios::fixed)<<setprecision(1)<<im<<"i"<<endl;}}else{if(im==0)cout<<setiosflags(ios::fixed)<<setprecision(1)<<re<<endl;elseif(im>0){if(im==1){cout<<setiosflags(ios::fixed)<<setprecision(1)<<re;cout<<"+"<<setiosflags(ios::fixed)<<setprecision(1)<<"i"<<endl;}else{cout<<setiosflags(ios::fixed)<<setprecision(1)<<re;cout<<"+"<<setiosflags(ios::fixed)<<setprecision(1)<<im<<"i"<<endl;}}else{if(im==-1){cout<<setiosflags(ios::fixed)<<setprecision(1)<<re;cout<<setiosflags(ios::fixed)<<setprecision(1)<<"-i"<<endl;}else{cout<<setiosflags(ios::fixed)<<setprecision(1)<<re;cout<<setiosflags(ios::fixed)<<setprecision(1)<<im<<"i"<<endl;}}}}intmain(){Complexc1,c2,c3;intt;cin>>t;while(t--){charch;c1.Init();cin>>ch;c2.Init();if(ch=="+")c3=c1+c2;elseif(ch=="-")c3=c1-c2;elseif(ch=="*")c3=c1*c2;elsec3=c1/c2;c3.display();}return0;}

运算过程分解，配图解

你仔细看看复变函数吧,我学过的.复变其实就相当于复数的基本运算加上微积分,里面从复数的极限、连续、导数、极数再到积分,都是有的.大体的思想还是差不多的,比如可导推出连续.不过在复数域里还是有很多与实数域相差别的地方.比如sin x在复数域里不再是有界函数,而是可以取尽复数域的所有数.复数域里面...

题中二复数之和等于-14.05-2.34i因为实部与虚部都小于0,故该复数位于复平面第三象限,它的幅角应为arctan(-2.34/-14.05)-180°,从而等于170.54°

不能

string.hpp：#ifndef __STRING_HPP__#define __STRING_HPP__#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>using namespace std;class String{ private: char* data; int len; public: String();//构造函数 String(int,char);//构造函数的存在 String(const char*); ~String();//析构函数 String(const String&);//拷贝构造函数 //赋值运算符函数 String& operator=(const String&); void show(); friend ostream& operator<< (ostream&,const String&);//插入运算符的重载 //提取运算符的重载 friend istream& operator>>(istream&,String&); const String operator +(const String&)const; String& operator+=(const String&); char& operator[](int); char* getdata(); int getlength(); operator char*(); friend bool operator==(const String&,const String&); friend bool operator!=(const String&,const String&); friend bool operator<(const String&,const String&); friend bool operator>(const String&,const String&);};#endif //__STRING_HPP__string.cpp中：#include"string.hpp"String::String()//构造函数{ data=new char[1]; len=0; *data="";}String::String(int n,char ch)//构造函数的存在{ data=new char[1+n]; len=n; char* p=data;w while(n-->0) *p++=ch;}String::String(const char* str){ /*if(str!=NULL) { len=strlen(str); data=new char[1+len]; strcpy(data,str); } else { data=new char[1]; len=0; *data=""; }*/ len=strlen(str?str:""); data=new char[1+len]; strcpy(data,str?str:"");}String::~String()//析构函数{ delete[]data; data=NULL;}String::String(const String& that)//拷贝构造函数{ //cout<<"拷贝构造函数被调用"<<endl; len=that.len; data=new char[1+len]; strcpy(data,that.data);}//赋值运算符函数String& String::operator=(const String& that){ // obj3=obj4; //初级程序员的写法 /*if(this==&that) return *this;//防止自赋值 delete[]data;//释放原内存 data=new char[1+that.len];//开辟新的内存 len=that.len; strcpy(data,that.data);//拷贝新内容 return *this;//返回自引用*/ //中级程序员的写法 //a=that; if(&that==this) return *this; char* p=new char[1+that.len]; if(!p) { cout<<"内存开辟失败"<<endl; exit(-1); } //如果开辟内存不成功的话，就抛出一个异常 delete[]data; data=strcpy(p,that.data); len=that.len; return *this; //高级程序员的写法 //a=that if(&that!=this) { String other(that); len=that.len; swap(data,other.data); } return *this;}void String::show(/*int a*/){ cout<<data<<" "<<len<<endl;}ostream& operator<<(ostream& os,const String& that){ //在插入运算符中可以把流对象os当作cout来使用 //最后还要返回流对象自身,以备连用 return os<<that.data<<" "<<that.len; //return os;}istream& operator>>(istream& is,String& that){ //return is>>that.data; char buf[256]={}; is.getline(buf,100); that.len=strlen(buf); delete[]that.data; that.data=new char[1+that.len]; strcpy(that.data,buf); return is;}const String String::operator+(const String& that) const{ String other; other.len=len+that.len; delete[]other.data; other.data=new char[1+other.len]; //strcat(strcpy(other.data,data),that.data); strcpy(other.data,data); strcat(other.data,that.data); //that.len=0; return other;}//obj1 =obj2 + obj3 //int a,b,c;//a=b+c;//ok//b+c=a;//error匿名变量只能做右值/*a=1,b=2,c=3;(a=b)=c;a=3 b=2 c=3*/String& String::operator+=(const String& that){ /*String other(*this); delete[]data; len+=that.len; data=new char[1+len]; strcat(strcpy(data,other.data),that.data);*/ *this=*this+that; return *this;}char& String::operator[](int i){ if(i<0||i>len) { cout<<"数组下标越界"<<endl; exit(-1); } return data[i];}/*obj1("abcdf");cout<<obj1[0]<<endl;//aobj1[0]="A"；cout<<obj1<<endl;//Abcdf*/char* String::getdata(){ return data;}int String::getlength(){ return len;}String::operator char*(){ static char buf[256]={}; sprintf(buf,"%s%d",data,len); return buf;}//obj1("abcdef");//cout<<(char*)obj1<<endl;//abcdef6bool operator==(const String&that,const String&other){ return strcmp(that.data,other.data)==0;}bool operator!=(const String&that,const String&other){ //return strcmp(that.data,other.data)!=0; return !(that==other);}bool operator>(const String&that,const String&other){ return strcmp(that.data,other.data)>0;}bool operator<(const String&that,const String&other){ return strcmp(that.data,other.data)<0;}至于对类的使用，你可以自己写个程序测试下吧

(600，120°)+(400，-60°)=(600，120°)+(-400，120°)=(200，120°)

因为 i^2=-1 假设上式成立 . 两边平方 计算左边 即[(i+(-i)]^2=i^2+(-i)^2+2i(-i)=-1+(-1)-2i^2=-2+2=0=右边 所以i+（-i）=0正确

#include "iostream"using namespace std;class calc{public: int a1,b1,a2,b2; char temp[32]; void add(); void cut(); void multiply(); void divide();private: void init();};void calc::init(){ cout<<"请输入第一个数的实部："; cin>>a1; cout<<"请输入第一个数的虚部："; cin>>b1; cout<<"请输入第二个数的实部："; cin>>a2; cout<<"请输入第二个数的虚部："; cin>>b2;}void calc::add(){ init(); sprintf(temp,("%d,%di + %d,%di = %d,%di"),a1,b1,a2,b2,a1+a2,b1+b2); cout<<temp<<endl;}void calc::cut(){ init(); sprintf(temp,("%d,%di - %d,%di = %d,%di"),a1,b1,a2,b2,a1-a2,b1-b2); cout<<temp<<endl;}void calc::multiply(){ init(); sprintf(temp,("%d,%di × %d,%di = %d,%di"),a1,b1,a2,b2,a1*a2-b1*b2,a1*b2+a2*b1); cout<<temp<<endl;}void calc::divide(){ init(); sprintf(temp,("%d,%di ÷ %d,%di = %d,%di"), a1,b1,a2,b2,(a1*a2+b1*b2)/(a2*a2+b2*b2),(b1*a2-a1*b2)/(a2*a2+b2*b2)); cout<<temp<<endl;}void main(){ int i; calc ll; while(1) { cout<<endl<<endl; cout<<" 1.加法 2.减法 3.乘法 4.除法 5.退出"<<endl; cin>>i; switch (i) { case 1:ll.add(); break; case 2:ll.cut(); break; case 3:ll.multiply(); break; case 4:ll.divide(); break; case 5:return; break; default: cout<<"输入错误！"<<endl; } }}

1、复数是指实数和虚数，是高等数学的基础知识，是大学一年级的第一章。我们把形如z=a+bi（a,b均为实数）的数称为复数，其中a称为实部，b称为虚部，i称为虚数单位。当z的虚部等于零时，常称z为实数；当z的虚部不等于零时，实部等于零时，常称z为纯虚数。 2、复数运算法则有：加减法、乘除法。两个复数的和依然是复数，它的实部是原来两个复数实部的和，它的虚部是原来两个虚部的和。复数的加法满足交换律和结合律。此外，复数作为幂和对数的底数、指数、真数时，其运算规则可由欧拉公式e^iθ=cos θ+i sin θ（弧度制）推导而得。

#define ARR_SIZE 20typedef struct complex{int aNum;int bNum;}Complex;void Add(Complex *A, Complex *B){ A->aNum=A->aNum+B->aNum; A->bNum=A->bNum+B->bNum;}void Minus(Complex *A, Complex *B){A->aNum=A->aNum-B->aNum;A->bNum=A->bNum-B->bNum;}char * FormatComplex(Complex *A) //格式化复数，变成a+bi{char temp[ARR_SIZE];wsprintf(temp,"%d+%d i",A->aNum,A->bNum);return temp;}

首先1 i/1-i可以化简。同乘1 i得（i方加2i加1）/2。因为i方等于-1所以这时候这项变成i。然后原来的试子就变成了i乘i方乘i的二次方乘i的三次方乘。。。。。一直乘到i的一百次方，最后的i的五千零五十次方。化简的负一

加减法需要把极坐标形式化成代数形式计算，以后把结果再化成极坐标形式。乘除需要模模相乘除，复角相加减

复数的加法按照以下规定的法则进行：设z1=a+bi,z2=c+di是任意两个复数，则它们的和是 (a+bi)+(c+di)=(a+c)+(b+d)i.两个复数的和依然是复数，它的实部是原来两个复数实部的和，它的虚部是原来两个虚部的和。复数的加法满足交换律和结合律，即对任意复数z1,z2,z3,有： z1+z2=z2+z1; (z1+z2)+z3=z1+(z2+z3). 复数的减法按照以下规定的法则进行：设z1=a+bi,z2=c+di是任意两个复数，则它们的差是 (a+bi)-(c+di)=(a-c)+(b-d)i.两个复数的差依然是复数，它的实部是原来两个复数实部的差，它的虚部是原来两个虚部的差。

这种格式只适合复数的乘除，乘方、开方计算不适合复数的加减计算，角度不能相加，角度相加是复数乘法的计算法则所以一般还是转化为a+bi（数学中的复数表示方法）或a+jb（电工学中的复数表示方法）后计算。

#include <iostream>using namespace std; class Complex{public: Complex(double r=0.0,double i=0.0)： real（r）， imag（i）{} Complex operator+(const Complex &c2) const; Complex operator-(const Complex &c2) const; Complex operator*(const Complex &c2) const; Complex operator/(const Complex &c2) const; void display() const;private: double real; double imag;};Complex Complex::operator+(const Complex &c2) const{ return Complex(real+c2.real, imag+c2.imag);} Complex Complex::operator-(const Complex &c2) const{ return Complex(real-c2.real, imag-c2.imag);}Complex Complex::operator*(const Complex &c2) const{ Complex c; c.real=real*c2.real-imag*c2.imag; c.imag=imag*c2.real+real*c2.imag; return c;}Complex Complex::operator/(const Complex &c2) const{ Complex c; double d=imag*imag+c2.imag*c2.imag; c.real=(real*c2.real+imag*c2.imag)/d; c.imag=(imag*c2.real-real*c2.imag)/d; return c;}void Complex::display() const{ cout<<"("<<real<<","<<imag<<")"<<endl;} int main(){ Complex c1(5,4),c2(2,10),c3; cout<<"c1=";c1.display(); cout<<"c2=";c2.display(); c3=c1+c2; cout<<" c3=c1+c2=";c3.display(); c3=c1-c2; cout<<" c3=c1-c2=";c3.display(); c3=c1*c2; cout<<" c3=c1*c2=";c3.display(); c3=c1/c2; cout<<" c3=c1/c2=";c3.display(); return 0;}

解析：(a+bi)+(c+di)=(a+c)+(b+d)i～～～～～～～～～(a+bi)-(c+di)=(a-c)+(b-d)i

第二步+i移到括号时应该变成-i望采纳

正解见下：#include<iostream>using namespace std;class complex//复数类声明{private: double real; double image;public: complex(double r=0.0,double i=0.0)//构造函数 { real=r; image=i; } complex operator+(complex c2);//+重载为成员函数 complex operator-(complex c2);//-重载为成员函数 complex operator*(complex c2);//*重载为成员函数 friend complex operator/(complex,complex);///重载为成员函数 bool operator==(complex c2);//==重载为成员函数 void display();};complex complex::operator +(complex c2)//重载的实现{complex c;c.real=c2.real+real;c.image=c2.image+image;return complex(c.real,c.image);}complex complex::operator -(complex c2)//重载的实现{complex c;c.real=real-c2.real;c.image=image-c2.image;return complex(c.real,c.image);}complex complex::operator *(complex c2)//重载的实现{complex c;c.real=c2.real*real;c.image=c2.image*image;return complex(c.real,c.image);}complex operator/(complex c1,complex c2)//重载的实现{return complex(c1.real/c2.real,c1.image/c2.image);}bool complex::operator ==(complex c2)//重载的实现{if((real==c2.real)||(image==c2.image))return true;else return false;}void complex::display(){cout<<"("<<real<<","<<image<<")"<<endl;}void main(){complex c1(5.0,4),c2(5.0,4),c3;cout<<"c1=";c1.display();cout<<"c2=";c2.display();c3=c1+c2;//使用重载运算符完成复数加法cout<<"c3=c1+c2=";c3.display();c3=c1-c2;//使用重载运算符完成复数减法cout<<"c3=c1-c2=";c3.display();c3=c1*c2;//使用重载运算符完成复数乘法cout<<"c3=c1*c2=";c3.display();c3=c1/c2;//使用重载运算符完成复数除法cout<<"c3=c1/c2=";c3.display();//使用重载运算符完成两个复数的比较bool result=(c1==c2);cout<<"(c1==c2)="<<result<<endl

i^2=-1原式=i-1+1+1+i=1+2i

// Lab 4: Complex.h#ifndef COMPLEX_H#define COMPLEX_H/* Write class definition for Complex */#include <iostream>//#include <double>using namespace std;#include "Complex.h"// Complex class definitionclass Complex{public: ComplexNumber(double ,double ); void setComplexNumber(double rp, double ip); // double getComplexNumber(); Complex add(const Complex & ); Complex subtract(const Complex &); void printComplex(); Complex(double ,double);private: double realPart; double imaginaryPart;};#endif// Lab 4: ComplexTest.cpp#include <iostream>using namespace std;#include "Complex.h"int main(){ Complex a( 1, 7 ), b( 9, 2 ), c(0,0); // create three Complex objects a.printComplex(); // output object a cout << " + "; b.printComplex(); // output object b cout << " = "; c = a.add( b ); // invoke add function and assign to object c c.printComplex(); // output object c cout << " "; a.setComplexNumber( 10, 1 ); // reset realPart and b.setComplexNumber( 11, 5 ); // and imaginaryPart a.printComplex(); // output object a cout << " - "; b.printComplex(); // output object b cout << " = "; c = a.subtract( b ); // invoke add function and assign to object c c.printComplex(); // output object c cout << endl;} // end main// Lab 4: Complex.cpp// Member-function definitions for class Complex.#include <iostream>using namespace std;#include "Complex.h"Complex::Complex( double real, double imaginary ){ setComplexNumber( real, imaginary );} // end Complex constructorComplex Complex::add( const Complex &right ){ /* Write a statement to return a Complex object. Add the realPart of right to the realPart of this Complex object and add the imaginaryPart of right to the imaginaryPart of this Complex object */ realPart=realPart+right.realPart; imaginaryPart=imaginaryPart+right.imaginaryPart; return Complex(realPart,imaginaryPart);} // end function addComplex Complex::subtract( const Complex &right ){ /* Write a statement to return a Complex object. Subtract the realPart of right from the realPart of this Complex object and subtract the imaginaryPart of right from the imaginaryPart of this Complex object */ realPart=realPart-right.realPart; imaginaryPart=imaginaryPart-right.imaginaryPart; return Complex(realPart,imaginaryPart);} // end function subtractvoid Complex::printComplex(){ cout << "(" << realPart << ", " << imaginaryPart << ")";} // end function printComplexvoid Complex::setComplexNumber( double rp, double ip ){ realPart = rp; imaginaryPart = ip;} // end function setComplexNumber

循环用当型循环吧 复数 你单独写个世子 去判断下

题中二复数之和等于-14.05-2.34i因为实部与虚部都小于0,故该复数位于复平面第三象限它的幅角应为arctan(-2.34/-14.05)-180°,从而等于170.54°复数分为实部和虚部，记为a+ib在直角坐标系中，横轴代表实数，纵轴代表虚数，以A（a,b）代表实数A=a+ib，在极坐标系中，以原点作为始点，A（a,b）作为终点的矢量代表该虚数，用A（r,θ）表示，其中r=(a平方+b平方）的开二次方，θ=arctg（b/a)。扩展资料：极坐标系是一个二维坐标系统。该坐标系统中的点由一个夹角和一段相对中心点——极点（相当于我们较为熟知的直角坐标系中的原点）的距离来表示。极坐标系的应用领域十分广泛，包括数学、物理、工程、航海以及机器人等领域。在两点间的关系用夹角和距离很容易表示时，极坐标系便显得尤为有用；而在平面直角坐标系中，这样的关系就只能使用三角函数来表示。对于很多类型的曲线，极坐标方程是最简单的表达形式，甚至对于某些曲线来说，只有极坐标方程能够表示。参考资料来源：百度百科-极坐标

那个i，你把它当成字母输出

你必须有一个输入格式约定，无论简单或复杂。例如：必须有实部和虚部，格式为3-2i1-i1+i等。然后才好处理。这里放入charstr[80]；了，如果是键盘输入流，也用类似格式。#include<stdio.h>main(){charstr[80]="1-i";//1+i2-2i2+3i0-iintr,m;inti,flag,s;flag=sscanf(str,"%d%di",&r,&m);if(flag==1){s=1;for(i=0;i<strlen(str);i++)if(str[i]=="-")s=-1;elseif(str[i]=="+")s=1;m=s;}printf("real=%dimag=%d ",r,m);return0;}

U1uff0bU2uff1dj30uff0b100uff1d104.4U000200cb16.70u20e3ufe0fV

因为复数虽然是表示为a+bi的形式，但它和向量确实不是一回事啊~~ 复数终归就是一个数啊~ 2维向量的两个维是等同的，而复数的1和i可以看成是不同的单位。而且i*i=-1，这一点是向量办不到的。 另外，比如说复数可以计算几次方，向量根本没这回事。

#include <iostream>using namespace std;class Complex{public: Complex(double pr=0, double pi=0) { real=pr; imag=pi; } Complex operator+(Complex c); Complex operator-(Complex c); Complex operator*(Complex c); Complex operator/(Complex c); void display();private: double real; double imag;};Complex Complex::operator+(Complex c){ return Complex(real+c.real, imag+c.imag);}Complex Complex::operator-(Complex c){ return Complex(real-c.real, imag-c.imag);}Complex Complex::operator*(Complex c){ double r,i; r=real*c.real+imag*c.imag; i=imag*c.real+real*c.imag; return Complex(r,i);}Complex Complex::operator/(Complex c){ double r,i,d; r=real*c.real+imag*c.imag; i=imag*c.real+real*c.imag; d=c.real*c.real+c.imag*c.imag; if(d>0) return Complex(r/d,i/d); else{ cout <<"Error!"; exit(1); }} void Complex::display(){ cout <<real<<"+"<<imag<<"i"<<endl;}int main(){ Complex c1(3,4),c2(-5,6),c3,c4,c5,c6; c3=c1+c2; c4=c1-c2; c5=c1*c2; c6=c1/c2; c3.display(); c4.display(); c5.display(); c6.display(); return 0;}

去括号的时候要把括号前的“系数”乘以括号内的每一项，也就是说如果括号前是加就直接去掉，如果是减就要改变括号内的每一项符号，再去掉

不知道是不是 ～//复数类。public class Complex{ private double real,im; //实部，虚部 public Complex(double real, double im) //构造方法 { this.real = real; this.im = im; } public Complex(double real) //构造方法重载 { this(real,0); } public Complex() { this(0,0); } public Complex(Complex c) //拷贝构造方法 { this(c.real,c.im); } public boolean equals(Complex c) //比较两个对象是否相等 { return this.real==c.real && this.im==c.im; } public String toString() { return "("+this.real+"+"+this.im+"i)"; } public void add(Complex c) //两个对象相加 { //改变当前对象，没有返回新对象 this.real += c.real; this.im += c.im; } public Complex plus(Complex c) //两个对象相加，与add()方法参数一样不能重载 { //返回新创建对象，没有改变当前对象 return new Complex(this.real+c.real, this.im+c.im); } public void subtract(Complex c) //两个对象相减 { //改变当前对象，没有返回新对象 this.real -= c.real; this.im -= c.im; } public Complex minus(Complex c) //两个对象相减，与subtract()方法参数一样不能重载 { //返回新创建的对象，没有改变当前对象 return new Complex(this.real-c.real, this.im-c.im); }}class Complex__ex{ public static void main(String args[]) { Complex a = new Complex(1,2); Complex b = new Complex(3,5); Complex c = a.plus(b); //返回新创建对象 System.out.println(a+" + "+b+" = "+c); } }/*程序运行结果如下：(1.0+2.0i) + (3.0+5.0i) = (40.0+7.0i)*/

in cmplx.h#ifndef CMPLX1_H_#define CMPLX1_H_#ifdef __cplusplusextern "C"{#endiftypedef struct _CMPLX{ float c_real; float c_vir;}CMPLX, *PCMPLX;CMPLX CMPLX_(float real, float vir);//operations between CMPLXCMPLX C_add(CMPLX c_l, CMPLX c_r);CMPLX C_sub(CMPLX c_l, CMPLX c_r);CMPLX C_mul(CMPLX c_l, CMPLX c_r);CMPLX C_div(CMPLX c_l, CMPLX c_r);void C_out(CMPLX data);#ifdef __cplusplus}#endif#endif /* CMPLX1_H_ */in cmplx.c#include <stdio.h>#include <string.h>#include "cmplx.h"CMPLX CMPLX_(float real, float vir){ CMPLX d; d.c_real = real; d.c_vir = vir; return d;}CMPLX C_add(CMPLX c_l, CMPLX c_r){ c_l.c_real += c_r.c_real; c_l.c_vir += c_r.c_vir; return c_l;}CMPLX C_sub(CMPLX c_l, CMPLX c_r){ c_l.c_real -= c_r.c_real; c_l.c_vir -= c_r.c_vir; return c_l;}CMPLX C_mul(CMPLX c_l, CMPLX c_r){ int rl = c_l.c_real; int vl = c_l.c_vir; int rr = c_r.c_real; int vr = c_r.c_vir; c_l.c_real = rl*rr - vl*vr; c_l.c_vir = rl*vr + rr*vl; return c_l;}CMPLX C_div(CMPLX c_l, CMPLX c_r){ CMPLX c_r_r = CMPLX_(c_r.c_real, -c_r.c_vir);/*a-b*i*/ float c_d = (c_r.c_real*c_r.c_real + c_r.c_vir*c_r.c_vir); if(c_d == 0) { memset(&c_l, 0x00, sizeof(CMPLX)); return c_l; } c_l = C_mul(c_l, c_r_r); c_l.c_real /= c_d; c_l.c_vir /= c_d; return c_l;}void C_out(CMPLX data){ if(data.c_real!=(float)0.0 && data.c_vir!=(float)0.0) printf("%f%c%fi ", data.c_real,data.c_vir<0?("-"):("+"), data.c_vir<0?(-data.c_vir):(data.c_vir)); else if(data.c_real == (float)0) printf("%fi ", data.c_vir); else if(data.c_vir == (float)0) printf("%f ", data.c_real);}in P4.c#include <stdio.h>#include "cmplx.h"int test(){ CMPLX dat1 = CMPLX_(2, 1); CMPLX dat2 = CMPLX_(1, 1); CMPLX dat_sum = C_add(dat1, dat2); CMPLX dat_sub = C_sub(dat1, dat2); CMPLX dat_mul = C_mul(dat1, dat2); CMPLX dat_div = C_div(dat1, dat2); C_out(dat_sum); C_out(dat_sub); C_out(dat_mul); C_out(dat_div); return 1;}int main(){ test(); return 0;}

D 对于复数的加减法运算法则判断出①对；对于②向量a的性质 ，但 是实数，但 不一定是实数，如z=i，就不成立，故②错；对于③复数加法的几何意义判断出③对，故选D

不知道是不是 ～//复数类。public class Complex{ private double real,im; //实部，虚部 public Complex(double real, double im) //构造方法 { this.real = real; this.im = im; } public Complex(double real) //构造方法重载 { this(real,0); } public Complex() { this(0,0); } public Complex(Complex c) //拷贝构造方法 { this(c.real,c.im); } public boolean equals(Complex c) //比较两个对象是否相等 { return this.real==c.real && this.im==c.im; } public String toString() { return "("+this.real+"+"+this.im+"i)"; } public void add(Complex c) //两个对象相加 { //改变当前对象，没有返回新对象 this.real += c.real; this.im += c.im; } public Complex plus(Complex c) //两个对象相加，与add()方法参数一样不能重载 { //返回新创建对象，没有改变当前对象 return new Complex(this.real+c.real, this.im+c.im); } public void subtract(Complex c) //两个对象相减 { //改变当前对象，没有返回新对象 this.real -= c.real; this.im -= c.im; } public Complex minus(Complex c) //两个对象相减，与subtract()方法参数一样不能重载 { //返回新创建的对象，没有改变当前对象 return new Complex(this.real-c.real, this.im-c.im); }}class Complex__ex{ public static void main(String args[]) { Complex a = new Complex(1,2); Complex b = new Complex(3,5); Complex c = a.plus(b); //返回新创建对象 System.out.println(a+" + "+b+" = "+c); } }/*程序运行结果如下：(1.0+2.0i) + (3.0+5.0i) = (40.0+7.0i)*/

首先1i/1-i可以化简.同乘1i得（i方加2i加1）/2.因为i方等于-1所以这时候这项变成i.然后原来的试子就变成了i乘i方乘i的二次方乘i的三次方乘.一直乘到i的一百次方,最后的i的五千零五十次方.化简的负一

#include "iostream.h"class Complex{ private: double real,imag; public: Complex(double r,double i) { real=r; imag=i; } /*friend Complex operator +(Complex& a,Complex& b)//友元运算符重载 { double r=a.real+b.real; double i=a.imag+b.imag; return Complex(r,i); }*/ Complex operator +(Complex&b)//类运算符重载 { double r=real+b.real; double i=imag+b.imag; return Complex(r,i); } show() { cout<<real<<"+"<<imag<<"i"; } };void main(){ Complex c1(6,3),c2(3,6),c=c1+c2; c1.show(); cout<<"+"; c2.show(); cout<<"="; c.show(); cout<<endl;}以上重载"+",重载"-"以此类似,仅供参考

给你弄了几行~你照着写下去吧~好久没有动手了~#includeclassComplex{public:Complex(doubler=0,doublev=0):real(r),virt(v){}friendComplexoperator+(Complexa,Complexb);friendostream&operator<<(ostream&out,Complex&a);private:doublereal;doublevirt;};ostream&operator<<(ostream&out,Complex&a){returnout<评论00加载更多

part of the students study用复数，因为part　of　the　students就是多于一人。　除非是one　of the students, 那后面就用单数studies

sheep的复数形式为sheep,单复数同形；lampsbabies

desk/table lamps.世界上最著名的那盏台灯就在这儿。Home to the world"s most famous desk lamp.

如下：1、father父亲；爸爸、dad爸爸（口语）、mother母亲；妈妈、mom妈妈（口语）、man男人、woman女人、grandmother（外）祖母、grandma（口语）（外）祖母、grandfather（外）祖父、grandpa（口语）（外）祖父、sister姐妹、brother兄弟。2、peach桃、pear梨、orange橙子、watermelon西瓜、apple苹果、banana香蕉、strawberry草莓、grape葡萄。3、bus公共汽车、bike自行车、taxi出租车、jeep吉普车、desk课桌、chair椅子、walkman随身听、lamp台灯、zoo动物园。

mechanics [mi04k03niks]knives [naivz]neighbours [04neib05z]housewives [04hauswaivz]houses [04hausiz]garage attendants [04g03rɑ:01 0504tend05nts]shop assistants [040600p 0504sist05ntz]lamp-posts [04l03mp p05usts]cars [kɑ:z]days [deiz]garages [04g03rɑ:01iz]crashes [04kr0306iz]cinemas [04sinim05z]cities [04sitiz]films [filmz]suitcases [04su:tkeisiz]women [04wimin]vegetables [04ved01it05blz]children [04t06ildr05n]