c語言-課后題答案（Clanguage-after-schoolquestionsanswer）

c語言ー課后題答案(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、沙縣、Sb、Va、VB；scanf("%、%、&R,F和H)；Cl=2X3.14XR；SA=3.14XrXr；某人=4?隱形刺客;VA=4X3.14XrXrXr/3；VB=SAXh；printf(“Cl=%o2f\n"，Cl)；printf("SA=%〇2f\NSB=%〇2f\NVA=%〇2f\NVB=%。2f\n”,沙縣,Sb,Va,VB)；6o輸入一個(gè)華氏溫度,要求輸出攝氏溫度。公式為C=5(f-32)/9,輸出要求有文字說明,取位2小數(shù)。解:#包括くstdio.h>無效main()(floatf,c；scanf( ,和F)；C=5*(f-32)/9；printf(“C=%〇2FC)；)70編程序,用getchar函數(shù)讀入兩個(gè)字符給Cl、C2,然后分別用putchar函數(shù)和printf函數(shù)輸出這兩個(gè)字符。思考以下問題:(1)變量ClC2應(yīng)定義為字符型或整形、?或二者皆可?(2)要求輸出ClC2值的和ASCH碼，應(yīng)如何處理?用putchar函數(shù)還是printf函數(shù)?(3)整形變量與字符變量是否在任何情況下都可以互相代替?如:charCl>C2:與intCl、C2:是否無條件地等價(jià)?解:#包括〈stdio.h>無效main()(charCl>C2；Cl=getchar()；C2=getchar()；putchar(Cl)；putchar('\n)；putchar(C2)；putchar('\n)；)#包括くstdio.h>無效main()icharCl、C2；Cl=getchar()；C2=getchar()；printf(“%dCl=C2=%d\n”，Cl,C2)；printf("Cl=C2=%%CCN"、Cl、C2)；)第四章三.寫出下面各邏輯表達(dá)式的值。設(shè)A=3,B=4,C=5〇A+B>C&B==C一個(gè)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個(gè)整數(shù)一、B、C,由鍵盤輸入,輸出其中最大的數(shù),請編程序。解:#包括くstdio.h>無效main()a,b,c,溫度,最大;printf(“請輸入3個(gè)整數(shù):")；scanf(u%d,%d,%dv,&,&B,C)；溫度=(a>b)?答:B；最大值=(溫度>C)?溫度:C；printf("3個(gè)整數(shù)的最大數(shù)是%d\n",最大);)6〇給ー個(gè)百分制成績,要求輸出等級ABC、、、會(huì)、eo90分以上為'a',80~90分為B,70~79分為'C',60分以下為會(huì)。解:#包括〈stdio.h>無效main（）{浮動(dòng)評分;字符級;printf（“請輸入學(xué)生成績:"）；scanf（"獷,得分）；（評分〉100時(shí)丨|評分V0）{printf（"\n輸入有誤,請重新輸入:"）;scanf（ ,得分）；）開關(guān)（（int）（得分10））!案例10：案例9:等級=a;打破;案例8：等級=B；休息;案例7:等級為C;打破;案例6:等級=會(huì)”;打破;案例5：案例4：案例3：案例2：案例1：案例0：等級=e；打破;）printf（"成績是％5.If?相應(yīng)的等級是枇'n”,評分等級）;）7〇給定一個(gè)不多于5位的正整數(shù),要求:（1）求出它是幾位數(shù);（2）分別輸出每一位數(shù)字;（3）按逆序輸出各位數(shù)字。例如原數(shù)為321,應(yīng)輸出123〇解:#包括くstdio.h>無效main（）（長整型數(shù);在個(gè)人,十,百,千,ten_thousand?地方分別代表個(gè)位、十位、百位、千位、萬位和位數(shù):/**/printf（“請輸入ー個(gè)整數(shù)（0?99999）：v）；scanf（"%ld",與民）；如果（民數(shù)記》9999）的地方=5；如果（民數(shù)記》999）的地方=4；如果（民數(shù)記》99）的地方=3；如果（民數(shù)記》9）的地方=2；別的地方=1;printf（"=%d\n",地方）；ten_thousand=民/10000；千=民/1000%10；百=民/100%10；十二民％100/10；個(gè)人=民%10；開關(guān)（的地方）｛案例5；printf（"%d%d%d,%d,%d”,ten_thousand、千、百、十、個(gè)人）；printf（"\n反序數(shù)字為:"）printf（a%d%d%d%d%d\n,,,個(gè)人,十,百,千,ten_thousand）；打破;案例4：printf（"%d,%d,%d,%d”,干,百,十,個(gè)人）；printf（"\n反序數(shù)字為:"）；printf（（＜%d%d%d%d\n”,個(gè)人,十,百,千）;打破;案例3：printf（"％d,%d,強(qiáng)d”,百,十,個(gè)人）；printf（"\n反序數(shù)字為:"）；printf（“%d%d%d\n”,個(gè)人、十、100）；打破;案例2：printf（"％d,%d”,十個(gè)人）；printf（"\n反序數(shù)字為:"）；printf（"%d%d\n",個(gè)人,十）；打破；案例1：printf案’％d”,個(gè)人）；printf（"\n反序數(shù)字為:"）；printf（"%d",個(gè)人）；打破；））8〇企業(yè)發(fā)放的獎(jiǎng)金根據(jù)企業(yè)的當(dāng)年利潤決定。當(dāng)利潤我低于或等于100000元時(shí),獎(jiǎng)金可提成10%；利潤大于100000元,小于200000元（100000く我W20000〇）時(shí),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:獎(jiǎng)金=bon4+(i-400000)X0.03；休息;案例6：案例8：案例9:獎(jiǎng)金=bon6+(i-600000)XO.015;休息;案例10:獎(jiǎng)金=bonlO+(i-1000000)*0.01；)printf("獎(jiǎng)金是%10.2f”,獎(jiǎng)金)；)9〇輸入4個(gè)整數(shù),要求按由小到大的順序輸出。解:#包括くstdio.h>無效main()(intt,A,B,C,D；printf(“請輸入4個(gè)整數(shù):")；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=ー個(gè);A=B；B=T；}如果(A>C){T=ー個(gè);A=C；C=T；}TOC\o"1-5"\h\z如果(a>d) { T =ー個(gè);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("排序結(jié)果如下:\n")；printf("%d,%d,%d%d\nw,A,B,C,D)；)10o有4個(gè)圓塔,圓心分別為(2,2)、(-2,2).(-2,-2)、(2-2),圓半徑為1。這4個(gè)塔的高度分別為10m。塔以外無建筑物今輸入任一點(diǎn)的坐標(biāo),求該點(diǎn)的建筑高度(塔外的高度為零)。解:#包括〈stdio.h>無效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(“請輸入ー個(gè)點(diǎn)(x,y)：")；scanf(u%f,,ネロX,Y)；DI=(x-xl)*(x-xl)+(Y)*(Y)；/**/求該點(diǎn)到各中心點(diǎn)的距離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；/?判斷該點(diǎn)是否在塔外?/printf(“該點(diǎn)高度為%d\n",H)；)llo求方程的解。根據(jù)代數(shù)知識,應(yīng)該有以下幾種可能:=0,不是二次方程,而是一次方程。(2)、有兩個(gè)相等的實(shí)根。（3）、有兩個(gè)不等的實(shí)根。（4）、有兩個(gè)共輒復(fù)根。編寫程序,運(yùn)行時(shí)分別給出不同的A,B,C,值,相應(yīng)于上面4種情況,分析輸出結(jié)果。解:包括くstdio.h>包括くく數(shù)學(xué)?！卑à绦??！睙o效main（）{浮,C、盤、XI、X2、實(shí)部、imagpart；scanf（ %%,F,F,F,和A,和B,和C）；printf（“方程”）；如果（晶圓廠（一）<=le-6）printf（"不是ー個(gè)二次、n”）；盤=b***CB-4；如果(晶圓廠(盤)<=le-6)printf("有兩行根:%8.4f\n"，B/(2*))；如果(晶圓廠(盤)>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間的全部素?cái)?shù).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.輸入一行字符,分別統(tǒng)計(jì)出其中英文字母、空格、數(shù)字和其他字符的個(gè)數(shù).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.輸出所有〃水仙花數(shù)'所謂〃水仙花數(shù)〃是指一個(gè)三位數(shù),其各位數(shù)字立方和等于該本身.例如:153是ー個(gè)水仙花數(shù),因?yàn)?53二1へ3+5へ3+3へ3.解:includestdio.h><includemath,h><voidmain()inti,j,k,n;printf(〃水仙花〃數(shù)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.猴子吃桃問題.猴子第一天摘下若干個(gè)桃子,當(dāng)即吃了一半,還不過癮,又多吃了一個(gè).第二天早上又將剩下的桃子吃掉一半，又多吃ー個(gè).以后每天早上都吃了前一天剩下的一半零一個(gè).到第10天早上想再吃時(shí),見只剩下ー個(gè)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.ー球從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次落地時(shí),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請按人輸入(一次輸入一個(gè)人所有課程的成績)成績(%d人,％d門課程)：“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個(gè)學(xué)生的4門課的成績,要求求出每個(gè)學(xué)生的平均成績,然后對平均成績從高到低將各學(xué)生的成績記錄排序.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.將一個(gè)數(shù)組的值按逆序重新存放,例如,原來順序?yàn)?8,6,5,4,1.要求改為: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.有個(gè)15數(shù)按由小到大順序存放在ー個(gè)數(shù)組中,輸入一個(gè)數(shù),要求用折半查找法找出該數(shù)組中第幾個(gè)元素的值.如果該數(shù)不在數(shù)組中,則輸出〃無此數(shù)〃.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==一個(gè)［中］）{printf（"找數(shù),這是ー個(gè)［%］”中）;打破;）如果（x〉ー個(gè)［中］）=中葉1；其他的開始=中+1;}而（啟動(dòng)くく結(jié)束）;?〇輸出以下圖案:****************?火??文**解:#包括くstdio.h>無效main（）（inti,j,k；chara[5][5]；為（i=0；