2020年整合高級語言程序設(shè)計C++平時作業(yè)答案要點名師精品資料

1、一、分析程序，寫輸出結(jié)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 << " 運行結(jié)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運行結(jié)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; 運行結(jié)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'運行結(jié)果：word ordrd二、.根據(jù)程序功能填空。1. 程序把10個數(shù)存儲到一維數(shù)組a中，并求該數(shù)組中最大值。#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;45兩位2 .下面程序的功能是輸出1至100之間每位數(shù)字的乘積大于每位數(shù)的和的數(shù)。例如,數(shù)字的乘積為 4X 5=20,和為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?qū)斎氲?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 函數(shù)create從鍵盤輸入整數(shù)序列，以輸入0為結(jié)束。按輸入順序建立單向鏈表并返回表頭。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數(shù)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() /檢查z的范圍cout << &quo

7、t;x=" << x << " y=" << y << " z=" << z << endl ;三、程序設(shè)計1.編寫函數(shù)輸出以下形狀的圖形，其中構(gòu)成圖形的數(shù)字和輸出的行數(shù)通過參數(shù)傳送。12 2 23 3 3 3 34 4 4 4 4 4 4代碼: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 (導(dǎo)數(shù)數(shù)爼卜標(biāo)異常哦 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運行結(jié)果:9 9?55555555567S91.011121,3141516171.819202on232425262728293031 2. 請編程序，輸入兩個正整數(shù)啊a和b ( a<b)，輸出a、b之間所有整數(shù)的因數(shù)(除 1和本身)。每行輸出數(shù)據(jù)不超過the factors of 6 :10個。例如，若輸入 a為6，b為8,則輸出格式要求如下:2 3the factors of 7 : no factorthe factors of 8 :2

10、 4代碼:1 package work;2 public class HcmeWor)c2 public static void printFactorjin七 a, int b)( if(a >= b) System.outTprintln(,1 個參數(shù)貝能小于第二個參粧哦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);)運行結(jié)果the factors of 2:nofactorthefactorsof3:nofactorth

12、efactorsof4 :2thefactorsof5:nofactorthefactorsof6:2 3thefactorsof7:nofactorthefactorsof8:2 4thefactorsof9:3thefactorsof10:2 5thefactorsof11:nofactorthefactorsof12:2 34 6thefactorsof13:nofactor3請編程序，找出1至99之間的全部同構(gòu)數(shù)。 同構(gòu)數(shù)是這樣一組數(shù)： 它出現(xiàn)在平方數(shù)的右邊。 例如：5是25右邊的數(shù)，25是625右邊的數(shù)，所以5和25都是同構(gòu)數(shù)。代碼:-! public class HomeWork3

13、 4白1 package work;5if (start >= end6System . out .print In ("開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)運行結(jié)果:2 4

15、 163 9 814 165 256 367 498 649 814. 編寫一個程序，實現(xiàn)如下功能:(1) 從鍵盤輸入a op b。其中a, b為數(shù)值；op為字符，限制為+、-、*、/。(2) 調(diào)用函數(shù)count(op,a,b)，計算表達式a op b的值。由主函數(shù)輸出結(jié)果。代碼：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 審七attza void main (Stiring ( higs)bod«&n flag = trua;Scanner in nw Scanner(System 止門)r36373839404142S ys tern. cut

18、:, print In (H 晴輸入弟 葉慳器！ T ： int a = in.nextint()；乞¥呂七em- out .pnintln (11 請笫入拖口 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輸入t;else45t la.q = f alse ji47j1484 950System, C5«t. print In請輅入 JfL：亍聲數(shù)!

19、Ti; int b = in.nextInt();System, out pi:LrrtJLii (fioni亡口工k4 匸oujit fopjib) J ;運行結(jié)果：譴輸入第-個整數(shù)!請輸入撩件符！請輸入第二個贈龜!20|405. 編寫一個程序，實現(xiàn)如下功能：(1) 輸入k ( <100)個整數(shù)到數(shù)組 x100中；(2) 計算k個數(shù)的平均值及大于平均值的元素個數(shù)。代碼：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.匸呂門口皿弭* 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 卩均數(shù)是：初 + avgNum); int count = g

22、etThnAvgCaunt (avgNumi- j );System, out. print In: + count);40 41 )運行結(jié)果：leirnprikaieu rpsjinfvwui BiJ j«v 嚴(yán)購數(shù)是：45 ” 大于T均魏的個故址：4E6. 定義函數(shù)void revers ion (i nt ary,i nt size);，調(diào)用函數(shù)逆置數(shù)組ary的元素。例如實參數(shù)組原來為1, 5, 3, 2, 6, 8, 9, 4 reversion 后變成為4 , 9, 8, 6, 2, 3, 5, 1 。代碼:1 package work;3publicalss Homewo

23、rk6public static voidreversion(Int arr呂it(start end| | start > end)6return;7)eif(start < end)9swap(a匚工'導(dǎo)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);運行結(jié)果：P " r” L IV v vs n V4,9,0,6,2,3,5,1,b (即7. 數(shù)組a包含50個整數(shù)，把 a中所有的后項除以前項之商取整后存入數(shù)組bi=ai/ai-1,并且b0=a0/a49），最后按每行5個元素的格式輸出數(shù)組 b代碼：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壯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+;運行結(jié)果:8. 編程輸出所有不超過 100的其平方具有對稱性質(zhì)的正整數(shù)（也稱回文數(shù)）。輸出格式如下:nu mber square1 12 43 911 12122 48426 676代碼：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