新概念C语言教程答案参考(自做)中国电力.docx

第一篇算法与C程序结构第1单元c语言程序设计初步1 .选择题:(1) B； (2) D2 .找错;(l)%d->%f；(2)%d->%f； float->int；(3);->,； %d->%f；(4)float->int；(5) float->int；3 .探索验证:4 .开发练习:1.程序编程:#include<stdio.h>int main(void)float r;printf("请输入圆的半径:")；scanf("%f',&r);printf("该圆的面积:%.3f',3.14*r*r);return 0;)结果捉图：“E:电脑作业、编程储存、DebuCppll. exe，2.1该便13.847Press any key to continue程序编程:#include<stdio.h>int main(void)int a,b,c,d;printf("请输入第一个人的成绩:");scanf("%d",&a);printf("请输入第二个人的成绩:")；scanf("%d",&b);printf("请输入第三个人的成绩:")；scanf("%d",&c);printf("请输入第四个人的成绩:");scanf("%d",&d);printff该小组的平均成绩:%.2f',(float)(a+b+c+d)/4.0);return 0;)结果捉图:二三四平 入入入入组士贝W贝主贝责狛 成成成成. 拘拘旳旳靑2 6 169 5 8 7to continue第2单元有选择功能的C语言程序1 .用c语言描述下列命题(1) (a<b)|(a<c)(2) a%b != 0(3) (a>c)&&(b>c)(4) a%2= 1(5) (a<c)|(b<c)(6) float a,b; a>0,b<0a <=0(7) inta;代码分析1 .写出下列表达式的值 1 1 0 0 02 .A3 .无4 .判断输入的字符时什么?5 .自己可以编程验证6 .自己可以编程验证探索验证1 .通过判断语句运行2 .什么都不存储,可自行编程验证:如：int a;char b;float c,d;scanfft%d%c%f%f,&a,&b,&c,&d);printf("d%c%f%fn”,a,b,c,d);开发练习:2.11.程序编程:#include<stdio.h>int main(void)double a,b,c;printf("请分别输入三角形的三边:")；scanf("%li%lf%lf',&a,&b,&c);if(a+b)>c)&&(a+c)>b)&&(b+c)>a)(if(a=b)&&(a=c)printf("该三角形是等边三角形、n");elseif(a=b)&&(a*a+b*b!=c*c)|(a=c)&&(a*a+c*c!=b*b)|(b=c)&&(b*b+c*c!=a *a)printf("该三角形是等腰三角形'n");elseif(a!=b)&&(a*a+b*b=c*c)|(a!=c)&&(a*a-i-c*c=b*b)|(b!=c)&&(b*b+c*c=a*a)printf("该三角形是直角三角形'n");elseif(a=b)&&(a*a+b*b=c*c)|(a=c)&&(a*a+c*c=b*b)|(b=c)&&(b*b+c*c= a*a)printf("该三角形是等腰直角三角形'n");elseprintf("该三角形是一般的三角形'n");elseprintf("该三边不能组成三角形")；return 0;)结果捉图:C' "E电脑作业、编程储存、DebugCppl2. exe* 请分何!入=!疆竜:45 6.2 3.4 该三角形是一般的三角形Press any key to continue程序编程:#include<stdio.h>int main(void)int thisStudent;fbr(thisStudent=l;thisStudent<5;thisStudent+)ifif(thisStudent!=l)+(thisStudent=3)+(thisStudent=4)+(thisStudent!=4)=3) (if(thisStudent= 1)printf("做好事的甲同学n");else ifithisStudent=2)printf("做好事的乙同学n");else if(thisStudent=3)printf("做好事的丙同学n");else ifithisStudent=4)printf("做好事的丁同学n");return 0;结果捉图:I c; "E电脑作业、编程储存、DebugCppl3. exe* 做好事的丙同軍IPress any key to continue 2.21.程序编程:#include<stdio.h>int main(void)int yew;printf(”请输入年份:");scanf(”d”,&yew);if(yew%100=0)if(yew%400=0)printff%d 是闰年n",yew);else if(yew%4=0)printf(n%d 是闰年'n”,yew);elseprintd 不是闰年n”,yew);return 0;)结果捉图:c( "E: '电脑作业、编程储存、DebuCppl3. exe*属款星惠,20052005不是闰年Press any key to continue程序编程:#include<stdio.h>int main()int mouth,ps;printff请依次输入月份、票数:")；scan d%d”,&mouth,&ps);if(7<=mouth)& & (9>=mouth)if(ps>=20)print优惠率为:百分之十”);elseprint优惠率为:百分之五”);elseif（ps>=20）printf（"优惠率为:百分之二十"）;elseprintf（"优惠率为:百分之十"）；return 0;结果捉图:c电脑作业、编程储存、DebugCppl6. exe*请依次甄人月份、室数:頭2优惠享J: 上）;"之一十Press any key to continue第3单元重复结构代码分析1-5EACAD 6-8 无答案 BB开发练习3 .! 代码 3.1：#include<stdio.h>int main()int a,b,c;printff请输入两个整数:n);scanf(*'%d%d,&a,&b);while(c!=0)c=a%b;a=b;b=c;printfi”最大公因子为:%dn",a);推导:#include<stdio.h>int main()int u,v,r;printf(nn请输入两个正整数:");scanf(H%d,%dn,&u,&v);printf(nutvtrtn");printf(n%dt%dtH,u,v);while(v!=O)r=u%v;printf(,'%dn",r);u=v;v=r;printf("dt%d ,u,v);printf(nn%dnH,u);return 0;)3.1代码3.2#include<stdio.h>int main()int a,b,c;printf("请输入两个整数:")；scanf(n%d%dM,&a,&b);if(b>0)doc=a%b;a=b;b=c;while(c!=O);printf("最大公因子为:%dnw,a);return 0;)推论:#include<stdio.h>int main()int a,b,c;printf("请输入两个正整数:");scanfC%d%d",&a,&b);printf(natbtctn*');printf(H%dt%dtH,a,b);doc=a%b;prin 氓"dn”,c);a=b;b=c;printf(”dt%dt”,a,b);while(c!=0);printf(nn 最大公因子:dn”,a);return 0;3.1代码3.3#include<stdio.h>int main(void)int fitlast= 1 ,fitthis= 1 ,fitnext=O;int m=2;while(m<12)fitnext=fitlast+fitthis;fitlast=fitthis;fitthis=fitnext;m+;printf,nfib(%d)=%d',m,fitnext);return 0;)推论:#include<stdio.h>int main(void)int fiblast= 1 ,fibthis= 1 ,fibnext= 1;int m=2;printf(ufibnexttfiblasttfibthistmtn,');printfl(M%dt%dt%dt%dtnH,fibnext,fiblast,fibthis,m);while(m<12)fibnext=fiblast+fibthis;printf("%dtn,fibnext);fiblast=fibthis;fibthis=fibnext;printf(H%dt%dtH,fiblast,fibthis);m-H-；printff%dn”,m);)printf(Hnfib(%d)=%dn",m,fibnext);)3.1代码3.4#include<stdio.h>int main(void)int fitlast=l ,fitthis=l ,fitnext=O;int m=2;dofitnext=fitlast+fitthis;fitlast=fitthis;fitthis=fitnext;m-H-; while(m<12);printfi,nfib(%d)=%dH,m,fitnext);return 0;)推论:#include<stdio.h>int main(void)int fiblast= 1 ,fibthis= 1 ,fibnext= 1;int m=2;printf(nfibnexttfiblasttfibthistmtnn);printf(,%dt%dt%dt%dtnM,fibnext,fiblast,fibthis,m);dofibnext=fiblast+fibthis;printf(n%dtM,fibnext);fiblast=fibthis;fibthis=fibnext;m+;printfC%dn”,m);while(m<12);printfK，,nfib(%d)=%dnH,m,fibnext);return 0;|3.1代码3.5#include<stdio.h>int main(void)int fitlast= 1 ,fitthis= 1 ,fitnext= 1;int m;fbr(m=2;m<12;m+)fitnext=fitlast+fitthis;fitlast=fitthis;fitthis=fitnext;)printffnfib(%d尸d”,m,fitnext);return 0;推论:#include<stdio.h>int main(void)int fiblast= 1 ,fibthis= 1 ,fibnext= 1;int m=2;printf(ufibnexttfiblasttfibthistmtnn);printfu%dt%dt%dt%dtnH,fibnext,fiblast,fibthis,m);fbr(m=3;m<=12;m+)fibnext=fiblast+fibthis;fiblast=fibthis;fibthis=fibnext;printf(H%dt%dtH,fiblast,fibthis);printfC%dnM,m);printfi(nnfib(%d)=%dn",m,fibnext);return 0;)3.1代码3.6#include<stdio.h>int main()int peachNumber= 1 ,days= 10;while(days>l)peachNumber=(peachNumber+1)*2;days;)printffn 地d 天的桃子数为:d 个。”,days,peachNumber);return 0;)推论:#include<stdio.h>int main()int peachNumber=l ,days=10;printff'peachNumbertdaystn");printfi(n%dt%dtn",peachNumber,days);while(days>l)peachNumber=(peachNumber4-l )*2;printff%dt”,peachNumber);-days;printfC'%dnn,days);)printfif'Vn 地d 天的桃子数为:%d 个。”,days,peachNumber);return (0);)3.1代码3.9#include<stdio.h>#include<stdlib.h>#include<math.h>double func(double x)double y;y=x*x-x-l;return y;int main(void)double x 1 ,x2,error,root;printff'input error:n);scanf(M%ir,&error);printff'input xlandx2:");scanf(H%lf%ir,&xl,&x2);if(func(x l)*func(x2)>0)printf(”此方程在lf%lf 无解”,x l,x2);exit(0);else(while(fabs(x 1 -x2)>error) root=(x I+x2)/2;ifi(func(root)*func(x l)<0)x2=root;elsex 1 =root;print方程在的If上的根为:%lfnM,xl,x2,(xl+x2)/2);return 0;)推论:#include<stdio.h>#include<stdlib.h>#include<math.h>double func(double x)double y;y=x*x-x-l;return y;)int main(void)double xl,x2,error,root;printff'input error:*');scanf(M%lf',&error);printfif'input xlandx2:");scanf(M%lf%lf,&xl,&x2);printf("xlttx2tn");printf("%lft%lftn",x l,x2);if(func(x l)*fiinc(x2)>0)printf(”此方程在lf%lf 无解”,x l,x2);exit(0);elsewhile(fabs(x 1 -x2)>error) root=(xl+x2)/2;ifi(func(root)*func(x2)<0)x 1 =root;printf(H%lft%lftnH,xl,x2);else(x2=root;printf(H%lft%lftn",x l,x2);printf("n 方程在上的根为:%lfnM,xl,x2,(xl+x2)/2);return 0;)探索验证:3.1讨论:#include<stdio.h>int main()int i=5,s=0;s=+i + +i；printf(,(3)%dt%dn,i,s);i=5,s=0;s=i+ + +i;printfi("(4)%dt%dn",i,s);i=5,s=0;s=(i+)+(i+);printf(,(5)%dt%dn",i,s);i=5,s=0;s=i+i;printfC'(6)%dt%dn",i,s);i=5,s=0;s=i+ + i+;printf(H(7)%dt%dn,i,s);i=5,s=0;s=i-H- +i；printf(,(8)%dt%dn,i,s);i=5,s=0;s=(+i)+(iH);printf(n(9)%dt%dn,i,s);i=5,s=0;s=i+(-H-i);printf(H(10)%dt%dn",i,s);i=5,s=0;s=+i + i+;printf(n(ll)%dt%dnH,i,s);i=5,s=0;s=(i-H-)+i;printf(H( 12)%dt%dnn,i,s);return 0;3.1练习 !. (4)编程验证:#include<stdio.h>int main(void)int a=0,i；for(i=l;i<5;i-H-)switch(i)case O:case 3:a+=2;printf(n%dn'a);case 1 :case 2:a+=3;printf(M%dnM,a);default:a+=5;printf(M%dn",a);printf(n%dnn,a);return 0;)推论:#include<stdio.h>int main( void) int a=0,i;fbr(i=l;i<5;i-H-)switch(i)case 0:case 3:a+=2;printfC%dn'a);case 1 xase 2:a+=3;printf(n%dn",a);default:a+=5;printf(,%dn,a);printf(n%dnT,a);return 0;3.1 练习1的(5)编程验证:#include<stdio.h>int main()int k=5,n=0;while(k>0)switch(k)default:break;case 1 :n+=k;case 2case 3 n+=k;printf(n%dnM,n);return 0;)3.1开发练习:i (i)程序编程:#include<stdio.h>int main() int nlast=0,nthis= 1 ,nnext= 1;int a4,a3,a2,al;int n=0;printfT'ntnlasttnthistnextn”)；doprintf(H%dt%dt%dt%dn,',n,nlast,nthis,nnext); nlast=nthis;)while(n<3);a4=nthis;a3=0;a2=0,a l=a4;while(n<20) n=n+l; nlast=nthis;nthis=a 1 +a2+a3+a4;a4=(a4+a3); a3=a2;a2=al;al=a4;nnext=a 1 +a2+a3+a4;printf(n%dt%dt%dt%dnn,n,nlast,nthis,nnext);printff 牛牛的数量为:%d",nthis);return 0;)或#include <stdio.h>int main(void)|int a=l,b=l,c=l,d=O,i； /*d 为总牛数/printf(,atbtctdn,');printf("dt%dt%dt%dn”,a,b,c,d);fbr(i=4;i<=20;i+)d=a+c;a=b;b=c;c=d;printf("%dt%dt%dt%dnM,a,b,c,d)；printff第20年母牛的总数为dn”,d);return 0;或#include<stdio.h>int main()int a=l,b=l,c=l,d,i;fbr(i=4;i<=20;i+)d=a+c;a=b;b=c;c=d;)printf(n%dn,d);return 0;结果捉图:cC *C: Docusent s and Sett ings'Adsinist rator桌面C语言Debugf gdsf. exe”Xnnlast nthis next1011211131114 1235 2346 3467 4698691399131910131928111928411228416013416088146088129158812918916129189277171892774061827740659519406595872205958721278牛牛的数量为:872Press any key to continue.1 (2)程序编程:#include<stdio.h> int main()double sum=1.0;int t,i=l;printfl(nttsumttttinM);while(i<50)t=(i+l)%2;sum=sum+sum+t;i+；printf(H%dt%lfttt%dnH,t,sum,i);)printf("lf',sum);return 0;结果捉图:1 (3)程序编程:#include<stdio.h>int main()double sum=2.0;int i=l;int n;printff请输入切的刀数:")；scanf(H%d",&n);printf(wsumttinH);printtft%dn”,sum,i);while(i<n)sum=sum+i4-l;printff%的”,sum);卄printff%dn”,i);printffn最多可以分为块'n”,sum);return 0;结果捉图:2 . (1)程序编程:#include<stdio.h>int main()double sum=1.5;int i;(printff%出,sum);sum=sum*10/9;printff%dn”,i);)printff他第一次托起的髙度:%lfmnM,sum);return 0;结果捉图:*C:Docu>ents and SettingsAdsinistrator桌面、C语言、Debugghfg. exe”二 9kl1.50000011.66666721.85185232.05761342.28623752.54026362.8225157他第一次托起的高度：3.136127mPress any key to continue.-d2. (2)程序编程:#include<stdio.h> int main()(double sum=200.0;int ij;fbr(i=l;i<=5;i+)if(i=l)fbr(j=2ij<=12;j+-»-)sum=sum/l .0032;printf(,'i=%dtj=%dt,',ij);printf(Msum=%lfnM,sum);)else(sum=sum+200;for(j=l;j<=12y+)sum=sum/1.0032;printf(Hi=%dtj=%dt",ij);printf(nsum=%lfn",sum);printf("当初他共存了If钱。、n",sum);return 0;或:#include<stdio.h>int main()int i;double sum=200;fbr(i=l;i<60;i-H-)(if(i%12=0)sum=(sum4-200)/l .0032;elsesum=sum/1.0032;printf("当初他存的钱为:%lfnM,sum);return 0;结果捉图:*C:Docu>ents and Setti.-口|xi=lj=12sum=193.093314i=2j=lsum=391.839428i=2j=2sum=390.589542i=2j=3sim=389.343642i=2j=4sur=388.101717i=2j=5sum=386.863753i=2j=6sum=385.629737i=2j=?sum=384.399659i=2j=8sum=383.173503i=2j=9sum=381.951259i=2j=10sum=380.732914i=2j=llsum=379.518455i=2j=12sun=378.307870i=3j=lsum=576.463188i=3j=2sum=574.624389i=3j=3sum=572.791457i=3j=4sum=570.964371i=3j=5sun=569.143113i=3j=6sum=567.327664i=3j=7sum=565.518007i=3j=8sum=563.714122i=3j=9sum=561.915990i=3j=10sum=560.123595i=3j=llsum=558.336917i=3j=12sum=556.555938i=4j=lsum=754.142681i=4j=2sum=751.737122i=4j=3sum=749.339237i=4j=4sun=746.949000i=4j=5sum=744.566388i=4j=6sum=742.191375i=4j=?sum=739.823939i=4j=8sum=737.464054i=4j=9sum=735.111696i=4j=10sum=732.766842i=4j=llsum=730.429468i=4j=12sum=728.099549i=5j=lsun=925.139104i=5j=2sum=922.188102i=5j=3sum=919.246514i=5j=4sum=916.314308i=5j=5sun=913.391455i=5j=6sun=910.477926i=5j=7sun=907.573690i=5j=8sum=904.678718i=5j=9sun=901.792980i=5j=10sum=898.916448i=5j=llsum=896.049091i=5j=12sum=893.190880隹初他共存了893.190880钱。 Press any key to continue_3EI老师答案:728编程:#include<stdio.h>#include<math.h>int main()int i;double x=0,sum=0;fbr(i=5；i>l；i-)|x=(x+200)/pow( 1.0032,12);printff%.2f ,x);return 0;)结果:aJ2. (3)程序编程:#include<stdio.h>int main()int i,sum=l;fbr(i=l;i128;i*=2)sum=sum*2;printf("八戒一路偷吃了d个仙桃n”,sum/);return 0;结果捉图:2. (4)程序编程:#include<stdio.h>int main()int i=l,sum=500;double s=500.0,k;doprintff在倒数的d个站应建在离沙漠边沿If处,储油量为d的加油点nM,i, 1000-s,sum);k=2巧+1;s=s+500.0/k;i+；sum=500*i;while(s<=1000);return 0;)或:#include<stdio.h>int main()int i=l,k=l,m=500;double s=500,sum=500;while( sum<= 1000)printfC倒数第%d站距上一站的距离为比储存油量为dn”,i凡m);k=k+2;s=500/k;sum=sum+s;i+;m=m+500;return 0;)结果捉图:点点点点点点 L-t-t- rr -L. rr rk.»M4n1 p HH 勺勺勺勺9 9 CD 9-0 0 0 ¢2 0 0 0 0 5 0 5 0 0 5 112 2 0 5 句勺勺灯灯3 3 b車量量量壽記 由由由由H9J 鳶储储储储飜 歹攵攵攵上上 0 3 3 2 6攵攵 0 3 3 6 0 12 0 3 3 7 2 6 2 0 3 3 4 9 6 1 0 3 3 0 4 4 3 0 3 3 9 3 9 3 > - 8 4 0 3 3 16 - 0 3 3 6 0 0 2 5 3 2 116 2 运边边边边边边. 离离离离离离离nt 在在在在在在在co J娃筵)slvrgrar熒 应应应应应应应t 百占占占占占占y 文立文立立 e 个个个个个个k 1234567 y 囈数数数数数数s 可隹隹隹隹隹隹eS 在在在在在在在pr法二(老师编程答案)#include <stdio.h >void main()(intk=l;/*倒数第k站/double d=500; /加油站距B点的距离,倒数第1站为500km */double x=500;/*每单边上消耗油量(也等于每趟的距离),倒数第1站为500L*/double v=500; /:每个加油站的贮油量，倒数第1站为500L*/while(d<=1000)printff倒数第d站距下站的距离为:%lft贮油量为:%lfn",k,x,v);k+;v=500.k;x=500.0/(2*k-l);d+=x;结果捉图:c< *C: Docu*ents and Sett ings'Adsinist rator桌面C语言Debuggdf gh. exe数数数数数数数esi站距下站的距离却 2站距下站的距离为: 3站距下站的距离为: 4站距下站的距离为: 5站距下站的距离为: 6站距下站的距离为: 7站距下站的距离为: any key to continue500.000000166.666667100.00000071.42857155.55555645.45454538.461538为为为为为为为 量量量量量量量 由由由由由由由 ，”r < M.- . « n- - n; Mrl 贮贮贮贮贮贮贮500.0000001000.0000001500.0000002000.0000002500.0000003000.0000003500.0000003. (1)(答案是老师编写的)程序编程:#include <math.h>#include<stdio.h>void main()float xl,xO,f,fl;xl=1.5;doxO=xl;f=(2*x0-4)*x0+3)*x0-6;fl=(6*xO-8)*xO+3;xl=x0-f7fl; while(fabs(x l-xO)>= le-5);printf(MTHe root of equation is %5.2fnM,xl);结果捉图