2020年整合高级语言程序设计C++平时作业答案要点名师精品资料

1、一、分析程序，写输出结果1 #include<iostream.h> 201709 #include<math.h> void main() int m, k, i ; for( m=1; m<=10; m+=2 ) k = m/3;for( i=2; i<=k; i+ )if( m%i ) cout << m << " 运行结果： 7 92 #include<iostream.h> void fun();void main()int i;for(i=1;i<5;i+) fun(); cout<<

2、;endl;void fun() static int a;int b=2;a += 2 ;cout<< a+b <<'t'10运行结果： 4 6 83 #include<iostream.h> int fun(int n) if(n=0) return 1; return 2*fun(n-1);void main() int a=5; cout<<fun(a)<<endl; 运行结果： 324. #i nclude<iostream.h>void mai n() char *cp="word&q

3、uot;for (int i=0 ; i<4; i+ ) cout<<cp+i << 't'运行结果：word ordrd二、.根据程序功能填空。1. 程序把10个数存储到一维数组a中，并求该数组中最大值。#in clude<iostream.h>void mai n() int ma x;int a10=76,55,95,87,85,83,65,90,77,85;int *p=;max=*p;for( ; p< ; p+)if( *p>max ) max= ;cout<<"max= "<

4、;<max<<e ndl;45两位2 .下面程序的功能是输出1至100之间每位数字的乘积大于每位数的和的数。例如,数字的乘积为 4X 5=20,和为4+5=9。#in clude<iostream.h>void mai n() int n, k=1, s=0, m;for(n=1; n<=100; n+) k=1; s=0;while( ) k*=m%10;s+=m%10; if( ( z=3*x+2*y-5 ) % 7 )3333 程序对输入的 n求s = 1 + 1/2 + 1/3 + 1/n 。#in clude<iostream.h>vo

5、id mai n()double s; int i, n;cout<<" n= " cin»n;s = 0;for (i=1; i+)s= _ ;cout<<"s="<<s<<e ndl;4 函数create从键盘输入整数序列，以输入0为结束。按输入顺序建立单向链表并返回表头。struct no dei nt data; node * n ext;node * create()node *head, *p, *q;p = new no de;head = p;cin> >p_>d

6、ata;while( )q = p;q_>next = p; cin> >p->data;if (head = p) head = NULL;else ;retur n head;5以下程序求方程的全部整数解：3x + 2y - 7z = 5( 0< x, y, z < 100 )#in clude<iostream.h>void mai n() int x, y, z ;for( x=0; x<=100; x+ )for( y=0; y<=100; y+ ); /求出z的值if() /检查z的范围cout << &quo

7、t;x=" << x << " y=" << y << " z=" << z << endl ;三、程序设计1.编写函数输出以下形状的图形，其中构成图形的数字和输出的行数通过参数传送。12 2 23 3 3 3 34 4 4 4 4 4 4代码:1 package work;3 public cla.ss Homework 1 4public static void printsharp (int J list, int row) 6 if(list *length <

8、; row)7 System. out. print In (导数数爼卜标异常哦 1；return;LOLI1213M15161716L9>0int column = row * 2 - 1;for(int i = 0;i<row;i+)for(int j=0；j<column;j+)if(2 * i + 1 > j)System.out * print(listi + n n); else System,out,print (" ");System, out.printin ( "*');23public static void

9、main(String args) 25xnt( a = (1,2,3,4;HomeWorkl,printSharp(a,4);27曲>9运行结果:9 9?55555555567S91.011121,3141516171.819202on232425262728293031 2. 请编程序，输入两个正整数啊a和b ( a<b)，输出a、b之间所有整数的因数(除 1和本身)。每行输出数据不超过the factors of 6 :10个。例如，若输入 a为6，b为8,则输出格式要求如下:2 3the factors of 7 : no factorthe factors of 8 :2

10、 4代码:1 package work;2 public class HcmeWor)c2 public static void printFactorjin七 a, int b)( if(a >= b) System.outTprintln(,1 个参数貝能小于第二个参粧哦1 *); return;boolean flag = true;for(int i=a;i<=b;i+)flag = tru建；System, out. print In (*' thE factors of n + i 十" if (K=3)(System, out .printin (r

11、io factor" acntinue;for(int j=2;j<i;j+)= 0) flag = false;System.out.print(j + ” ” 'if(flag)(System* out print_("rLO factorT,);System. out： - print丄n ("");public static void main(String args) HomeWork2.printFactor2f 13);)运行结果the factors of 2:nofactorthefactorsof3:nofactorth

12、efactorsof4 :2thefactorsof5:nofactorthefactorsof6:2 3thefactorsof7:nofactorthefactorsof8:2 4thefactorsof9:3thefactorsof10:2 5thefactorsof11:nofactorthefactorsof12:2 34 6thefactorsof13:nofactor3请编程序，找出1至99之间的全部同构数。 同构数是这样一组数： 它出现在平方数的右边。 例如：5是25右边的数，25是625右边的数，所以5和25都是同构数。代码:-! public class HomeWork3

13、 4白1 package work;5if (start >= end6System . out .print In ("开始位置不龍大姑束伍置i 7return;89intwq口 areNuTD = 0;10boolean flag = false；11for(int i = start + 1; i < end;i+)12squareNEini =:;13flag = false；14while(squareNum < end15if(getSquaresquareNum < end)IEflag = true;17IQif(flagI19System.o

14、ut.print(squar&Num +2021squa reNum = getSquare(squarNum)I23ISystem, out * printin (ht '*;24252®public static int getSquare(int a)27return a * a;2B2 9ptiblic static void main (String args)30Homework3 TprintQmorph2sm(l, 99;31TO Ipublic static void printOmorphism(int start,int end)运行结果:2 4

15、 163 9 814 165 256 367 498 649 814. 编写一个程序，实现如下功能:(1) 从键盘输入a op b。其中a, b为数值；op为字符，限制为+、-、*、/。(2) 调用函数count(op,a,b)，计算表达式a op b的值。由主函数输出结果。代码：JLWpm cLVdX , iUL Apt. ,± L LL JL Oil i i10I public class Homework4 12*public static int count(String opr int a, int b1314151617ScriptEngineManager manage

16、r = new ScriptEngineManager(); Script Engine engine = manager .getEngineByName (,p js11); int result = 0;try result = (int) ftgin.eval(a + op + b);-1 Q,19| catoh ScriptException e / TODO Atito-generated catch block 巴 pi11 njt-StackTrace ();IT <Ti一4 -J return result;public staticisoperator(string

17、op)(2527String Operator! 1 = "十“, 7" ;Listotri ng> 1= Arrays. a$Li st (Op«: ator);if(list.contains(op)(2829303132return tru«1retuxn. false;33*35pub 1 i-c 审七attza void main (Stiring ( higs)bod«&n flag = trua;Scanner in nw Scanner(System 止门)r36373839404142S ys tern. cut

18、:, print In (H 晴输入弟 叶悭器！ T ： int a = in.nextint()；乞¥吕七em- out .pnintln (11 请笫入拖口 W ! n); string op = ,s p, ?while(flag!op = in next(;if(1 Homeworks-isCera tor(op)43 酬System.out.printin < "«A的半足操作寄* ifl戟If输入t;else45t la.q = f alse ji47j1484 950System, C5«t. print In请辂入 JfL：亍聲数!

19、Ti; int b = in.nextInt();System, out pi:LrrtJLii (fioni亡口工k4 匸oujit fopjib) J ;运行结果：谴输入第-个整数!请输入撩件符！请输入第二个赠龜!20|405. 编写一个程序，实现如下功能：(1) 输入k ( <100)个整数到数组 x100中；(2) 计算k个数的平均值及大于平均值的元素个数。代码：packaLga work;3 public class HomeWorkS (4spublic sta 七 iu irit； genera to rRadoifraflurri (in 七 num)5int 1 J =

20、 new intnum;6Int rNurn = 0 ；7for (int i =0; i <num;.: +) 8rNum = (int > Ma th.匸吕门口皿弭* 100)91i = rNum；10)11rsturn 1;1314*public statics double avgNum (int 1 15double sum 0;16double avgNum = 0*0;17for(in七 1:1)(10sum += i;19)20avgNum = sum / 1.length;21乍rireturn avgNum;23i2public static int getTh

21、anAvgCountdouble avg,int ()25±nt count = 0;26for(int i=0;i<l>length;i+)27if(1il > avg)28r qcount+;T30i?丄jreturn Count；3234public static void main(StringJ args)(int j = HomeWork5 - genera(100);double avgNuni = HomeWorkS . avgATLwn(j );System, out print In 卩均数是：初 + avgNum); int count = g

22、etThnAvgCaunt (avgNumi- j );System, out. print In: + count);40 41 )运行结果：leirnprikaieu rpsjinfvwui BiJ j«v 严购数是：45 ” 大于T均魏的个故址：4E6. 定义函数void revers ion (i nt ary,i nt size);，调用函数逆置数组ary的元素。例如实参数组原来为1, 5, 3, 2, 6, 8, 9, 4 reversion 后变成为4 , 9, 8, 6, 2, 3, 5, 1 。代码:1 package work;3publicalss Homewo

23、rk6public static voidreversion(Int arr吕it(start end| | start > end)6return;7)eif(start < end)9swap(a匚工'导匕art* end);10)11reversion “+startf '-tnd;12ft哉)L,public static voidswap(int( A, int a1 airrt temp;_htemp = AaJ ;Aa = Ab);18Ab = temp;19)20*public static voidprintArr(intarr)21for(int

24、 i:arr)(System.out.print (i +;r inh startfint end)public static void main(String args int arrtreversion (ri,Ofarr. ength - 1)；printArr (arr);运行结果：P " r” L IV v vs n V4,9,0,6,2,3,5,1,b (即7. 数组a包含50个整数，把 a中所有的后项除以前项之商取整后存入数组bi=ai/ai-1,并且b0=a0/a49），最后按每行5个元素的格式输出数组 b代码：1 package work;3 public clas

25、s HomeHorlc? ptiblic static int: generat口:rRmciomNuTn f iirvt :num 6int 1 J = mw int num /int rNum = 0;f or (int i = 0; i<nijjn; i+4) |匸N壮m = <int)Mathceil (Math. randoinO *100) j10111213141 i 1 - rlMwmj retiirli 1 ；publicint getAddiitionNum (int 1 a) int b J = naw int a. length) /17int dertom

26、lnator = 0;for (int i-0; Ka* length; i+!i 19try denomirtator = a i-lj ； ) catch (Except!口n ej sien 口 rninmt.cir = ata - length. =1 ;23Ib± 三且丄 / denominator;25 26 retum h;2039 1巧00 e5o111006300n0101412public static? vfiidl main (String (int. a = greneia tojrJ?a:af0nLMLUir (SO); int b - gretiddi

27、tlonWujn(a); for (int i=0; Kb-length; i+)if(i % 5 = 0> System out .print In (H p;)1System.out.prin+;运行结果:8. 编程输出所有不超过 100的其平方具有对称性质的正整数（也称回文数）。输出格式如下:nu mber square1 12 43 911 12122 48426 676代码：package work;J public class HomeWorkB |4“5£9101112131415161718*1920 fl* f|34p6public static boolea

28、u isPalindrome(int a) boolean flag = true;if(a < 10)return f l.agrString I = new Integer(d).toString();forint i=0;i <b*length();1+)if(b«charAt(i)!= b,charAt(b»Length()- i - 1)tflag = false;break; return flag;public static int getSquare(int a) return a * a;public static void main(String I arg?)int square = 0;±orint i=l;丄<=丄00;二+)sq