c语言-课后题答案（Clanguage-after-schoolquestionsanswer）

c语言ー课后题答案(Clanguage-after-schoolquestionsanswer)Clanguageafter-schoolexerciseanswer:ThefirstchapterPleaserefertotheexampleofchapter1.,writeaCprogramtooutputthefollowinginformation:Verygood!Solution:#include<stdio.h>Void(main)[Printf(〃**************************")；Printf("\n");Printf("Verygood\n!");Printf("\n");Printf("**************************")；writeaCprogram,a,B,Cinputthreeoutputvalue,themaximumvalue.Solution:#include<stdio.h>Void(main)!Inta,B,C,max;Printf("pleaseinputthreenumbersa,B,c:\n);Scanf(%d,%d,%d,&a,&b,&c);Max=a;If(max<b)max=b;If(max<c)max=c;Printf(maximumnumber:%d,Max);ThesecondchapterIf1.ofourGDPannualgrowthratewas10%,10yearsafterthecalculationofourGDPgrowthpercentagecomparedtonow.ThecalculationformulaforP=(1+r)n,Rannualgrowthrate;nyears;Paspercentageandcomparedtonow.Solution:#include<stdio.h>#include<math.h>Void(main)!DoubleP,r=0.1,n=10;(P=pow(1+r),n);Printf(%lf\n,P);)pleaseprogramwillbetranslatedinto"China"passworddecodingruleisusedintheoriginallettersbehindthefourthlettersinsteadoftheoriginalletter.Forexample,theletter"A"behindthefourthletterisE",E"insteadof"A".Therefore,"China"shouldbetranslatedas"Glmre".Pleasewriteaprogram,withthemethodoftheinitialvalueofCl,C2,C3,C4,C5valuesoffivevariableswere'C','H','i','n','a',Cl,C2afteroperation,C3,C4,C5,'1'and，G,respectivelyinto'm','r','e',andoutput.Solution:ttinclude<stdio.h>Void(main){Charcl='C',c2='h',c3=，i',c4='n',c5='a';Cl+=4;C2+=4;C3+=4;C4+=4;C5+=4;Printf("passwordis%c%c%c%c%c\n",Cl,C2,C3,C4,C5);)Thethirdchapterusingthescanffunctionoftheinputdata,thefollowinga=3,b=7,x=8.5,y=71.82,cl='A',c2=，a'.Askhowtoenteronthekeyboard?Solution:#include<stdio.h>Void(main)(Inta,floatx,B;Y;charCl,c2;Scanf(a=%d,b=%d,&a,&b);Scanf(x=%f,y=%e,&x,&y);Scanf(cl=祝,c2=%c,&cl,&c2);)A=3,b=7X=8.5,y-71.82C1=A,c2=a5.circleradiusr=l.5,cylinderh=3,findtheperimeterandareaofacircle,spheresurfacearea,volume,volumeofcylindricalball.Theinputdatawithscanf,theoutputresults,theoutputrequirementsoftext,take2decimalplaces.Pleaseprogram.Solution:#include<stdio.h>Void(main)(浮R、H、Cl、沙县、Sb、Va、VB；scanf("%、%、&R,F和H)；Cl=2X3.14XR；SA=3.14XrXr；某人=4・隐形刺客;VA=4X3.14XrXrXr/3；VB=SAXh；printf(“Cl=%o2f\n"，Cl)；printf("SA=%〇2f\NSB=%〇2f\NVA=%〇2f\NVB=%。2f\n”,沙县,Sb,Va,VB)；6o输入一个华氏温度,要求输出摄氏温度。公式为C=5(f-32)/9,输出要求有文字说明,取位2小数。解:#包括くstdio.h>无效main()(floatf,c；scanf( ,和F)；C=5*(f-32)/9；printf(“C=%〇2FC)；)70编程序,用getchar函数读入两个字符给Cl、C2,然后分别用putchar函数和printf函数输出这两个字符。思考以下问题:(1)变量ClC2应定义为字符型或整形、?或二者皆可?(2)要求输出ClC2值的和ASCH码，应如何处理?用putchar函数还是printf函数?(3)整形变量与字符变量是否在任何情况下都可以互相代替?如:charCl>C2:与intCl、C2:是否无条件地等价?解:#包括〈stdio.h>无效main()(charCl>C2；Cl=getchar()；C2=getchar()；putchar(Cl)；putchar('\n)；putchar(C2)；putchar('\n)；)#包括くstdio.h>无效main()icharCl、C2；Cl=getchar()；C2=getchar()；printf(“%dCl=C2=%d\n”，Cl,C2)；printf("Cl=C2=%%CCN"、Cl、C2)；)第四章三.写出下面各逻辑表达式的值。设A=3,B=4,C=5〇A+B>C&B==C一个IIB+C和B-C(3)!(a>b)&!CI|1(4)!(x=a)&(y=b)和0(5)!(A+B)+C和B+C/2解:0110150有3个整数一、B、C,由键盘输入,输出其中最大的数,请编程序。解:#包括くstdio.h>无效main()a,b,c,温度,最大;printf(“请输入3个整数:")；scanf(u%d,%d,%dv,&,&B,C)；温度=(a>b)?答:B；最大值=(温度>C)?温度:C；printf("3个整数的最大数是%d\n",最大);)6〇给ー个百分制成绩,要求输出等级ABC、、、会、eo90分以上为'a',80~90分为B,70~79分为'C',60分以下为会。解:#包括〈stdio.h>无效main（）{浮动评分;字符级;printf（“请输入学生成绩:"）；scanf（"犷,得分）；（评分〉100时丨|评分V0）{printf（"\n输入有误,请重新输入:"）;scanf（ ,得分）；）开关（（int）（得分10））!案例10：案例9:等级=a;打破;案例8：等级=B；休息;案例7:等级为C;打破;案例6:等级=会”;打破;案例5：案例4：案例3：案例2：案例1：案例0：等级=e；打破;）printf（"成绩是％5.If»相应的等级是枇'n”,评分等级）;）7〇给定一个不多于5位的正整数,要求:（1）求出它是几位数;（2）分别输出每一位数字;（3）按逆序输出各位数字。例如原数为321,应输出123〇解:#包括くstdio.h>无效main（）（长整型数;在个人,十,百,千,ten_thousand»地方分别代表个位、十位、百位、千位、万位和位数:/**/printf（“请输入ー个整数（0〜99999）：v）；scanf（"%ld",与民）；如果（民数记》9999）的地方=5；如果（民数记》999）的地方=4；如果（民数记》99）的地方=3；如果（民数记》9）的地方=2；别的地方=1;printf（"=%d\n",地方）；ten_thousand=民/10000；千=民/1000%10；百=民/100%10；十二民％100/10；个人=民%10；开关（的地方）｛案例5；printf（"%d%d%d,%d,%d”,ten_thousand、千、百、十、个人）；printf（"\n反序数字为:"）printf（a%d%d%d%d%d\n,,,个人,十,百,千,ten_thousand）；打破;案例4：printf（"%d,%d,%d,%d”,干,百,十,个人）；printf（"\n反序数字为:"）；printf（（＜%d%d%d%d\n”,个人,十,百,千）;打破;案例3：printf（"％d,%d,强d”,百,十,个人）；printf（"\n反序数字为:"）；printf（“%d%d%d\n”,个人、十、100）；打破;案例2：printf（"％d,%d”,十个人）；printf（"\n反序数字为:"）；printf（"%d%d\n",个人,十）；打破；案例1：printf案’％d”,个人）；printf（"\n反序数字为:"）；printf（"%d",个人）；打破；））8〇企业发放的奖金根据企业的当年利润决定。当利润我低于或等于100000元时,奖金可提成10%；利润大于100000元,小于200000元（100000く我W20000〇）时,低于100000元的部分按10%提成,Morethan100000yuan,theCommission7.5%;profitmorethan200000yuan,lessthan400000yuan(200000<I=400000),lessthan200000yuanstillbythewayofCommission(below),morethan200000yuanpartofthe5%commission.Theprofitofmorethan400000yuan,lessthan600000yuan(400000<I=600000),higherthan400000yuanpartofthe3%commission.Theprofitofmorethan600000yuan,lessthan1000000yuan(600000<I=1000000),higherthanthe600000partofthe1.5%Commission;profitmorethan1000000yuan(I>1000000),partofmorethan1000000yuanby1%commission.InputfromthekeyboardfortheIyearprofit,totalprizemoneyshouldbepaid.Requirement：theprogramusingtheifstatement;(2)theprogramwiththeswitchstatement.Solution:programwiththeifstatement.ttinclude<stdio.h>Void(main)Longi;Floatbonus,bonl,bon2,bon4,bon6,bonlO;Bonl=l00000*0.1;/*profitat100thousandyuanbonus.Bon2=bonl+100000*0.075;/*profitat200thousandyuanbonus.Bon4=bon2+200000*0.05;/*profitat400thousandyuanbonus.Bon6=bon4+200000*0.03;/*profitat600thousandyuanbonus.Bonl0=bon6+400000*0.015;/*profitat1millionyuanbonus.Printf("pleaseentertheprofiti:;Scanf(%ld,&i);If(iく=100000)bonus=i*0.1;/**/profitby0.1royaltymoneyinlessthan100thousandyuanElseif(i<=200000)Bonus=bonl+(i-100000)*0.075;in100thousandto200thousandyuanbonus/*profitElseif(i<=400000)Bonus=bon2+(i-200000)*0.05;in200thousandto400thousandyuanbonus/*profitElseif(i<=600000)Bonus=bon4+(i-400000)*0.03;in400thousandto600thousandyuanbonus/*profitElseif(i<=1000000)Bonus=bon6+(i-600000)*0.015;in600thousandto1millionyuanbonus/*profitElseBonus=bonlO+(i-1000000)*0.01;whenmorethan1millionyuanbonus/*profitPrintf("bonusis%10.2f\n",bonus);)programwiththeswitchstatement.#include<stdio.h>Void(main)(Longi;Floatbonus,bonl,bon2,bon4,bon6,bonlO;Intc;Bonl=100000*0.1;Bon2=bonl+100000*0.075;Bon4=bon2+200000*0.05;Bon6=bon4+20000〇・〇.03;Bonl0=bon6+400000*0.015;Printf("pleaseentertheprofiti:;Scanf(%ld,&i);C=i/100000;If(c>10)c=10;Switch(c)(Case0:bonus=l*0.1;break;bonus=bonl+(i-100000)*0.075;break;bonus=bon2+(i-200000)*0.05;break;案例4：案例5:奖金=bon4+(i-400000)X0.03；休息;案例6：案例8：案例9:奖金=bon6+(i-600000)XO.015;休息;案例10:奖金=bonlO+(i-1000000)*0.01；)printf("奖金是%10.2f”,奖金)；)9〇输入4个整数,要求按由小到大的顺序输出。解:#包括くstdio.h>无效main()(intt,A,B,C,D；printf(“请输入4个整数:")；scanf("%d,%d,%d,刎”，&,&B,C,D)；printf("\n=%d,B=%d,C=%d,D=%d\n”，A,B,C,D)；如果(a>b){T=ー个;A=B；B=T；}如果(A>C){T=ー个;A=C；C=T；}TOC\o"1-5"\h\z如果(a>d) { T =ー个;a d； a T； }如果(B、C) { T = A； B = C； C = T； }如果(B、D) { T = b； b = D； D = T； }如果(C、D) { t = C； C = D； D = T； }printf("排序结果如下:\n")；printf("%d,%d,%d%d\nw,A,B,C,D)；)10o有4个圆塔,圆心分别为(2,2)、(-2,2).(-2,-2)、(2-2),圆半径为1。这4个塔的高度分别为10m。塔以外无建筑物今输入任一点的坐标,求该点的建筑高度(塔外的高度为零)。解:#包括〈stdio.h>无效main()浮XI=2,Y1=2,X2=2,Y2=2,X3=-2,Y3=-2,X4=-2,Y4=2,X,Y,DI,D2,D3,D4；printf(“请输入ー个点(x,y)：")；scanf(u%f,,ネロX,Y)；DI=(x-xl)*(x-xl)+(Y)*(Y)；/**/求该点到各中心点的距离D2=(x-x2)*(x-x2)+(Y+y2)*(Y+y2)；D3=(X+X3)*(x+X3)+(y-y3)*(y-y3)；D4=(X+X4)*(x-x4)+(Y+Y4)*(Y+Y4)；如果(DI和D2>1>1>1和D3和D4〉1)H=0；/・判断该点是否在塔外・/printf(“该点高度为%d\n",H)；)llo求方程的解。根据代数知识,应该有以下几种可能:=0,不是二次方程,而是一次方程。(2)、有两个相等的实根。（3）、有两个不等的实根。（4）、有两个共辄复根。编写程序,运行时分别给出不同的A,B,C,值,相应于上面4种情况,分析输出结果。解:包括くstdio.h>包括くく数学。”包括くく程序。”无效main（）{浮,C、盘、XI、X2、实部、imagpart；scanf（ %%,F,F,F,和A,和B,和C）；printf（“方程”）；如果（晶圆厂（一）<=le-6）printf（"不是ー个二次、n”）；盘=b***CB-4；如果(晶圆厂(盘)<=le-6)printf("有两行根:%8.4f\n"，B/(2*))；如果(晶圆厂(盘)>le-6){xl=(b+sqrt(disc))/(2*a);x2=(b-sqrt(disc))/(2*a);printf("hasdistinctrealroots:%8.4fand%8.4f\n",xl,x2);)else1realpart=b/(2*a);imagpart=sqrt(disc)/(2*a);printf("hascomplexroots:\n");printf("%8.4f+%8.4fi\n",realpart,imagpart);printf("%8.4f-%8.4fi\n",realpart,imagpart);第五章1.求100〜200间的全部素数.解:includestdio.h><includemath,h><voidmain()(intm,n=0;doublek;for(m=101;m<=200;m=m+2)[k=sqrt(m);for(i=2<=k;i++)if(m%i==0)break;if(i>=k+1){printfく‘%d",m);n=n+1;)if(n%10==0)printf("ヽn");)printf("ヽn");)2.输入一行字符,分别统计出其中英文字母、空格、数字和其他字符的个数.解:#includestdio.h>くvoidmain()charc;inti=0,y=0,k=0,1=0;while((c=getchar())!='\n'){if(c>='a'&&cく='z'IIc>='a'&&cく=z)++;elseif(c>=’〇'&&cく='9')j++;elseif(c=='')c++;else1++;)printf("二・d,j=%d,k=%d,1=%d\n",i,j,k,1);)3.输出所有〃水仙花数'所谓〃水仙花数〃是指一个三位数,其各位数字立方和等于该本身.例如:153是ー个水仙花数,因为153二1へ3+5へ3+3へ3.解:includestdio.h><includemath,h><voidmain()inti,j,k,n;printf(〃水仙花〃数是:〃)；for(n=100;n<1000;n++)i=n/100;j=n/10-*10;k=n%10;if(n==i*i*i+j*j*j+k*k*k)printf(〃%4d”,n);)printf(〃、n");)4.猴子吃桃问题.猴子第一天摘下若干个桃子,当即吃了一半,还不过瘾,又多吃了一个.第二天早上又将剩下的桃子吃掉一半，又多吃ー个.以后每天早上都吃了前一天剩下的一半零一个.到第10天早上想再吃时,见只剩下ー个桃子了.求第一天共摘多少桃子.解:#includestdio.h><voidmain()(intday,xl,x2;day=9;x2=1;while(i>0).{xl=(x2+1)*2;x2=xl;day;printf(total=%d\n”,xl);5.ー球从100米高度自由下落,每次落地后返回原高度的一半,再落下.求它在第10次落地时,Afteratotalofhowmanymeters?Thetenthrallyhigh?Solution:#include<stdio.h>Void(main)IInti,doubleh=100,s=100n;Scanf(%d,&n);For(i=l;i<=n;i++)IH*=0.5;If(i==l)continue;S=2*h+s;Printf(h=%f,s=%f\n,h,s);Thefollowing6,printingpatternsMissMissSolution:#include<stdio.h>#include<math.h>Void(main)Inti,J,k;For(i=0;i<=3;i++)For(j=0;jく=2-i;j++)("printf");For(k=0;k<=2*i;k++)Printf("*");Printf("\n");)For(i=0;iく=2;i++){For(j=0;j<=i;j++)("printf");For(k=0;kく=4-2*i;k++)Printf("*");Printf("\n");7.twotabletennisteamcompetition,eachperson3.AteamforA,B,C3,X,Y,teamBZ3.Thedrawhasdecidedtomatchthelist.Someonetoinquireaboutthelistofplayersthegame,AsaidhedidnotX,CsaidhedidnotX,Yratio,programmedtoidentify3teamthreeplayersagainstthelist.Solution:#include<stdio.h>Void(main)!CharI,J,K;/*iAJBistheopponent,opponents,KrivaltoC*For(i=>X,;iく='Z';i++)For(j=X，;j<=，T;j++)iIf(I=j!)For(k='X’;k仁'Z';k++){if(I!=k&&j!=k)If(I='X'&&k='X'&&k='Z'!!!)printf("orderisA—%c\tB—%c\tC—%c\n",I,J,K);Thesixthchaptergivenaclassof10students,askedtoenterthese10gradesofstudents,andthencalculatedtheaveragescoresofthem.Solution:ttinclude<stido.h>(main)IFloata[10],sum,avg;Inti;Sum=0.0;Printf("Pleaseinputthestudentsscore:");For(i=0;i<10;i++){scanf(%f,&a[i]);Sum=sum+a[i];)Avg=sum/10;Printf("Theaverageis:%f\n",AVG);)givenaclassof10students,storedinaone-dimensionalarray,whichrequirestofindthehighestscoreofthestudentachievementandthenumberofstudents.Solution:#include<stdio.h>Void(main)Intch[10];Inti,max=O,xh;Printf〈pleaseenterthe10studentachievement:\n);For(i=0;i<10;i++)(Scanf(%d,&ch[i]);If(ch[i]>max)!Max=ch[i];Xh=i;))Printf("thehighestscore:%d\n,Max);Printf("thestudentnumber:%d\n,XH);)thereare3students,4classes,allstudentsaskedtoentertheresultsofthecourse,andcalculatedtheaveragescoresforeachcourse.Solution:includesstdio.h><includesstdlib.h><definerenshu3definekecheng4voidmain()Ifloatchengji[4][4];inti,j;printfC\n请按人输入(一次输入一个人所有课程的成绩)成绩(%d人,％d门课程)：“renshu,kecheng);for(i=0;i<renshu;i++)Iprintf("\n第％d人"，i+1).for(j-0,j<kecheng;j++)scanf("%f",&chengji[the][j]).for(j=0,j<kecheng;j++)chengji[renshu][j]=0.for(i=0;i<renshu;i++)chengji[renshu][j]+=chengji[the][j].chengji[renshu][j]=chengji][j][renshu/renshu;)for(i-0;iく=renshu;i++)(for(j=0,j<kecheng;j++)printf("%8.2f”,chengji[the][j]).printf("ヽn");printf("ヽn");4.已知5个学生的4门课的成绩,要求求出每个学生的平均成绩,然后对平均成绩从高到低将各学生的成绩记录排序.解:includesstdio.h>くincludesstdlib.h><voidmain()Iinta[5][4]={{94,78,87,76},{66,87,75,69},{100,98,89,77},{82,58,72,84},{82,73,67,54}};inti,j,sum;floataverage(b[5],t;for(i=0;i<5;i++)Isum=0;for(j=0,j<4;j++)sum=sum+to[the][j].b[i]=sum/4.0.)printf(averaged=%4.2f\n”,i+1,b[i]).)for(j=0,j<4;j++)for(i=j+1,<5;i++)if(b[i]>b[j]){t=b[i].b[i]=b[j].b[j]=t;)for(i=0;i<5;i++)printf("%.2f\n",b[i]).5.将一个数组的值按逆序重新存放,例如,原来顺序为:8,6,5,4,1.要求改为:1,4,5,6,8.解:#includesstdio.h>くvoidmain()!inti,[10],temp.for(i=0;i<10;i++)scanf("%d”,&[i]).for(i=0;i<10;i++)printf("$4d",[]);for(i=0;i<5;i++)!temp=a[i].[i]=a[9-);[9-]=temp;for(i=0;i<10;i++)printf("$4d",[]);printf("ヽn");)6.有个15数按由小到大顺序存放在ー个数组中,输入一个数,要求用折半查找法找出该数组中第几个元素的值.如果该数不在数组中,则输出〃无此数〃.解:#includesstdio.h>くvoidmain()!int[15]14,13,12,11,10,9,8,7,6,5,4,3,2,1,0={};intstart,end,mid,thefind_flag;intx;printf<inputx").scanfC%d”,&x）;start=0;end=14;find_flag=0;do{mid=（start+end）/2;如果（X==一个［中］）{printf（"找数,这是ー个［%］”中）;打破;）如果（x〉ー个［中］）=中叶1；其他的开始=中+1;}而（启动くく结束）;?〇输出以下图案:****************・火・・文**解:#包括くstdio.h>无效main（）（inti,j,k；chara[5][5]；为（i=0；