2023年高級(jí)語言程序設(shè)計(jì)C++平時(shí)作業(yè)答案

一、分析程序，寫輸出結(jié)果#include<iostream.h>#include<math.h>voidmainO{intm,k,i；for(m=l；m<=10;m+=2){k=m/3;for(i=2；i<=k;i++)if(m%i)cout?m?))運(yùn)營結(jié)果：79#include<iostream.h>voidfun()：voidmain(){inti;for(i=1;i<5;i++)fun();cout<<endl;)voidfun(){staticinta;intb=2;a+=2;cout?a+b?>\t*；packagework;publicclassHomeWork3{publicstaticvoidprintOmorphism(intstart,intend){if(start>=end){System,o口七.printin(“開始位置不能大「結(jié)束位置！*,);return;}intsquareNum=0;booleanflag=false;11for(inti=start+1;i<end;i++){squareNum=i;flag=false;while(squareNum<end){if(getSquare(squareNum)<end){flag=true;TOC\o"1-5"\h\z}if(flag){System.out.print(squareNum+"n);)squareNum=getSguare(squareNum);)System.out.printin("“);})26+publicstaticintgetSquare(inta){returna*a;)、publicstaticvoidmain(String[]args){Home^ork3.printOmorphism(1,99);31}QOI運(yùn)營結(jié)果：241639814165256367498649814.編寫一個(gè)程序，實(shí)現(xiàn)如下功能:（1）從鍵盤輸入aopbo其中a,b為數(shù)值;op為字符，限制為+、-、*、/。（2）調(diào)用函數(shù)811標(biāo)（?！?6）,計(jì)算表達(dá)式@opb的值。由主函數(shù)輸出結(jié)果。代碼:XAUpM^UJOVC1A?ACLXpu?OULXpLXVli/101ipublicolassHomeWork4{publiostaticintcount(Stringopfinta,intb){ScriptEngineManagermanager=newScriptEngineManager();14ScriptEngineengine=manager.getEngineByNamejsM);TOC\o"1-5"\h\zlbintresult=0;16try(result=(int)engine.eval(a+op+b);}catch(ScriptExceptione)(19//TODOAuto-generatedcatchblocke.prin|tStackTrace();21)returnresult;23}FpublicstaticbooleanisOperator(Stringop)(StringOperator[]■26List<String>list=Arrays.asList(Operator);if(list.contains(op)){returntrue;29)returnfalse;31)32publicstaticvoidmain(String[]args){34booleanflag=true;Scannerin=newScanner(System.in);36System.ou亡.printin(“沛輸入第一個(gè)夠敢！”)；inta?in.nextlnt();System.out.printin!");Stringop=while(flag){op=in.next();if(!HomeWork4.Isoperator(op)){System.out.printin("輸入的不足掾作符，詁旗新粕入！)else(flag=false;)}System.out.println(”請輸入第：個(gè)能敢！");49intb=in.nextlnt();System.out.printin(HomeWork4.count(op,a,b));運(yùn)營結(jié)果：請輸入第二不整數(shù)！請輸入操作符!★請輸入第二個(gè)整數(shù)！20I402.編寫一個(gè)程序，實(shí)現(xiàn)如下功能:(1)輸入k(<100)個(gè)整數(shù)到數(shù)組x[100]中;(2)計(jì)算k個(gè)數(shù)的平均值及大于平均值的元素個(gè)數(shù)。代碼：packagework;publicclassHomeWork5{publicstaticint[]generatorRadomNum(intnum){int1[]=newint[num];intrNum=0;for(inti=0;i<num;i++)(rNum=(int)Math.floor(Math.random()*100);1[i]=rNum;TOC\o"1-5"\h\z}return1;}publicstaticdoubleavgNum(int[]1){doublesum=0;doubleavgNum=0.0;for(inti:l){sum+=i;)avgNum=sum/1.length;returnavgNum;22}23publicstaticintgetThanAvgCount(doubleavg,int[]1){intcount=0;26for(inti=0;i<l.length;i++){if(l[i)>avg){count++;TOC\o"1-5"\h\z29)30}returncount;32)|3334publicstaticvoidmain(String()args){intj[]=HomeWork5.generatorRadomNum(100);36doubleavgNum=HomeWork5.avgWu/n(j);System.out.printIn(""+avgNum);intcount=getThanAvgCount(avgNum,j);System.out.printin("大于平均數(shù)的個(gè)數(shù)足：“+count);}}40運(yùn)營結(jié)果：vri5”D“5RjpavaMppnvaiiviijv.\rivyid平均數(shù)是：45?97大廣平均數(shù)的個(gè)數(shù)是：466.定義函數(shù)。voidreversion(intary[],intsize)；逆置數(shù)組ary的元素。例如實(shí)參數(shù)組本來為[1,5,3,2,6,8,9,4],調(diào)用函數(shù)rcversion后變成為[4,9,8,6,2,3,5,1]0代碼:packagework;publicclassHomeWork6{publicstaticvoidreversion(intarr[],intstart,intend){if(start==end||start>end){return;TOC\o"1-5"\h\z}if(start<end){swap(arr,start,end);}reversion(arr,++start#-end);}4publicstaticvoidswap(int[]A,inta,intb)(inttemp;temp=A[a];A[a]=A[b];A[b]=temp;TOC\o"1-5"\h\z}publicstaticvoidprintArr(int[]arr){for(inti:arr){System,out.print(i+**,*');}}publicstaticvoidmain(String[]args){26intarr[]=11,5,3,2,6,8,9,4);reversion(arr,0,arr.length-1);printArr(arr);}}運(yùn)營結(jié)果：■、、，???????QJUJi??、z???、，??n、zI”，，'4,9,8,6,2,3,5,1,47.數(shù)組a包含50個(gè)整數(shù)，把a(bǔ)中所有的后項(xiàng)除以前項(xiàng)之商取整后存入數(shù)組b（即b[i]=a[i]/a[i-1],并且b[0]=a[0]/a[49]），最后按每行5個(gè)元素的格式輸出數(shù)組b。代碼:1packagework;3publicclassHomework7(4publicstaticint[]generatorRadomNum(intnum){6int1(]=newint[num];intrNum=0;for(inti=0;i<num;i++){rNum=(int)Math.ceil(Math.random()*100);1[i]=rNum;}retuxm1;13)14publicstaticint(]getAdditionNum(int[Ja)(16intb[]=newint[a.length];intdenominator=0;for(inti=0;i<a.length;i++){try{denominator?a[i-1];)catch(Exceptione){denominator=a[a.length-1];TOC\o"1-5"\h\z23}?24b[i]=a[i]/denominator;}returnb;}publicstaticvoidmain(String[]args)(inta[]=generatorRadomNum(50);31intb(]=getAdditionNum(a);for(int1=0;i<b.length;if(i%5==0){System.out.printin(HW);)|System.out.print(b[i]+"\t");})|運(yùn)營結(jié)果:1011020020061100310080110503000200310021110120101401制.編程輸出所有不超過100的其平方具有對(duì)稱性質(zhì)的正整數(shù)（也稱回文數(shù)）。輸出格式如下:numbersquarc149111212248426676代碼：packagework;publicclassHomeWork8{-publicstaticbooleanisPalindrome(inta){booleanflag=true;6if(a<10)(returnflag;8}Stringb=newInteger(a).toString();10for(inti=0;i<b.length();i++){if(b.charAt(i)!=b.charAt(b.length()-i-1)){flag=false;break;TOC\o"1-5"\h\z))returnflag;)-publicstaticintgetSquare(inta){returna*a;),publicstaticvoidmain(String(]args)(intsquare=0;for(inti=l;i<=100;i++)(24square=getSquare(i);if(isPalindrome(square)){System.out.printin(i+"\t"+square);TOC\o"1-5"\h\z}})}運(yùn)營結(jié)果:

〈terminated〉HomeWork8[JavaApplicatk11243911121|22484266769.編寫程序,打印如下楊輝三角。11121133114641代碼：packagework;publicclassHomework9{publicstaticvoidmain(String[]args){finalintNMAX=10;int[][]odds=newint[NM/kX+1][];for(intn=0;n<=NMAX;n++)odds[n]=nowint[n+1];for(intn=0;n<odds.length;n++)for(intk=0;k<odds[n].length;k++)(14intlotteryOdds=1;for(inti=1;i<=k;i++)lotteryOdds=lotteryOdds*(n-i+1)/i;odds|[n](k]=lotteryOdds;TOC\o"1-5"\h\zIfor(int[]row:odds)22(for(intodd:row)24System,out.printfodd);System.out.printIn();))}運(yùn)營結(jié)果:

111iiiT"\r，qii1121i331i464115101051i615201561i72135352171i8285670562881i93684126126843691110451202102522101204510110.定義一個(gè)函數(shù)，計(jì)算長度為k的整型數(shù)組元素的平均值及大于平均值的元素個(gè)數(shù)。調(diào)用函數(shù)的語句為：count(a,k,ave,num)；其中a是數(shù)組名，k是數(shù)組元素個(gè)數(shù),av。返回的平均值，num返回大于平均值的元素個(gè)數(shù)。運(yùn)營結(jié)果：468103.Sinclude<iostream.h>intfun(intn){if(n==0)return1；return2*fun(n-1)；)voidmain(){inta=5;cout<<fun(a)<<end1;}運(yùn)營結(jié)果：324.Sine1ude<iostream.h>voidmain(){char*cp=z，word”;for(inti=0;i<4;i++)cout?cp+i<<'\t'；)運(yùn)營結(jié)果：wordcordd二、.根據(jù)程序功能填空。.程序把10個(gè)數(shù)存儲(chǔ)到一維數(shù)組a中，并求該數(shù)組中最大值。#include<iostream.h>voidmainO{intmax;inta[10]={76,55,95,87,85,83,65,90,77,85);int*p=;°max=*p;for(;p<；p++)0if(*p>max)max=；acout<<，，max="?max?end1；).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，45兩位數(shù)字的乘積為4X5=20,和為4+5=9。#include<iostrearn.h>voidmain(){intn,k=1,s=0,m;for(n=1;n<=100;n++){k=l;s=0：?whi1e()s+=m%10；)if(k>s)cout<<n?>\t*;))3.程序?qū)斎氲膎求s=1+1/23+1/33+…+1/n3。#include<iostream.h>voidmain(){doubies;inti,n；cout<V〃n=cin?n；s=0;for(i=1;;i++)s=；cout?"s="?s<<endl；).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I可表頭。structnode{intdata;node*next;}；node*createO(。node*head,*p,*q；。p=newnode;head=p;?cin?p->data；while()(°q=p;時(shí)；q->next=p;cin>>p->delta;)。if(head==p)head=NULL；ooelse;,returnhead;).以下程序求方程的所有整數(shù)解：3x+2y—7z=5。(0Wx,y,zW100)#inc1ude<iostream.h>voidmainO{intx,y,z;for(x=0;x<=100；x++)for(y=0;y<=100;y++){if((z=3*x+2*y-5)%7)；。//求出z的值if（）//檢查z的范圍<<z<<endl;<<z<<endl;<<z<<endl;cout?"x="?x?"y="<<y<<"z=

<<z<<endl;三、程序設(shè)計(jì)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。代碼:1packagework;23publicclassHomeWorkl{5a5a6789LOLIL2L3L4L5L6L7L8L92021222324925262728}5a6789LOLIL2L3L4L5L6L7L8L920212223249255a6789LOLIL2L3L4L5L6L7L8L92021222324925262728})intcolumn=row*2-1;for(inti=0;i<row;i++){for(intj=0;j<column;j++){if(2*i+1>j){System.out.print(list[i]+"");)else{System.out.print(nn);))System.out.printin(nn);)publicstaticvoidmain(String[]args){int[]a={1,2,3,4);HomeWorkl.printsharp(a,4);29運(yùn)營結(jié)果:2233334444445555555552.請編程序，輸入兩個(gè)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和自身)。每行輸出數(shù)據(jù)不超過10個(gè)。例如，若輸入a為6,b為8,則輸出格式規(guī)定如下:thefactorsof6:23thefactorsof7:nofactorthefactorsof8:24代碼:1packag