新概念C语言教程答案参考(自做)中国电力.pdf

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：#includeint 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：C；电脑作业编程储存DebugCppl 1.exe*该阊的面枳：13.847Press any key to continueME程序编程:#includeint 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：c 电脑作业 编程储存DebugCppl 1.exe一二三四平入入入入汩责责责责成成成成-m-m-m-mI26169587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)(ab)|(ac)&(bc)(4)a%2=1(5)(ac)|(b0,b0(7)int a;a=0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;scanfC%d%c%f%f,&a,&b,&c,&d);printf(u%d%c%f%fn,a,b,c,d);开发练习：2.11.程序编程：#includeint main(void)double a,b,c;printf(请分别输入三角形的三边：);scanf(%lf%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+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也)elseprintf(该三边不能组成三角形)return 0;2.程序编程：#includeint main(void)int thisStudent;fbr(thisStudent=l;thisStudent5;thisStudent+)if(thisStudent!=l)+(thisStudent=3)+(thisStudent=4)+(thisStudent!=4)=3)if(thisStudent=l)printf(做好事的甲同学n)；else if(thisStudent=2)printf(做好事的乙同学n)；else if(thisStudent=3)printf(做好事的丙同学n)；else if(thisStudent=4)printf(做好事的丁同学3”)；)return 0;)结果捉图：cr*E:电脑作业 编程储存DebugCpp 13,exe，做好事的丙同学Press any key to continue2.21.程序编程：#includeint main(void)int yew;printf(请输入年份：);scanf(n%du,&yew);if(yew%100=0)c E:电脑作业编程储存DebugCppl3.e*e 日日请输入年 槛1 20052005不是闰年Press any key to continueif(yew%400=0)printf(%d 是闰年n,yew);else if(yew%4=0)printf(%d 是闰年n,yew);elseprintf(%d 不是闰年n,yew);return 0;)结果捉图：x2.程序编程：#includeint main()int mouth,ps;printf(请依次输入月份、票数：)scanf(n%d%dn,&mouth,&ps);if(7=mouth)if(ps=20)printf(优惠率为：百分之十)elseprintff优惠率为：百分之五）elseif(ps=20)printff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D ebugC ppl6.exe*优惠率为：百分之一十Press any key to con tin ueH B g第3单元重复结构代码分析1-5EACAD 6-8 无答案 BB开发练习3.1 代码 3.1：#includeint main()int a,b,c;printf(请输入两个整数：)；scanf(%d%d,&a,&b);while(c!=0)c=a%b;a=b;b=c;)printf(最大公因子为：%dn,a);return 0;推导:#includeint main()int u,v,r;printf(”n 请输入两个正整数：”);scanf(n%d,%dH,&u,&v);printf(nutvtrtnM);printfC%dt%dt”,u,v);while(v!=O)r=u%v;printf(%dn,Jr);u=v;v=r;printf(u%dt%dtn,u,v);printf(un%dnH,u);return 0;3.1代码3.2#includeint main()int a,b,c;printf(“请输入两个整数：”);scanf(%d%dn,&a,&b);if(b0)doc=a%b;a=b;b=c;while(c!=O);print,最 大 公 因 子 为:%dnn,a);return 0;推论：#includeint main()int a,b,c;printf(”请输入两个正整数：);scanf(%d%dH,&a,&b);printf(uatbtctn);printf(n%dt%dta,b);doc=a%b;printf(%dnu,c);a=b;b=c;printf(u%dt%dtn,a,b);while(c!=0);printf(un 最大公因子:%dn”,a);return 0;3.1代码3.3#includeint main(void)int fitlast=l,fitthis=l,fitnext=O;int m=2;while(m12)fitnext=fitlast+fitthis;fitlast=fitthis;fitthis=fitnext;m+;)printf(nnfib(%d)=%dM,m,fitnext);return 0;)推论：#includeint main(void)int fiblast=1 ,fibthis=1 ,fibnext=1;int m=2;printf(nfibnexttfiblasttfibthistmtnu);printf(n%dt%dt%dt%dtnn,fibnext,fiblast,fibthis,m);while(m12)fibnext=fiblast+fibthis;printf(%dtu,fibnext);fiblast=fibthis;fibthis=fibnext;printf(n%dt%dtn,fiblast,fibthis);m;printf(%dnu,m);)printf(unfib(%d)=%dn,m,fibnext);return 0;3.1代码3.4#includeint main(void)int fitlast=1 ,fitthis=1 ,fitnext=O;int m=2;dofitnext=fitlast+fitthis;fitlast=fitthis;fitthis=fitnext;m-H-；while(m12);printf(nnfib(%d)=%d,m,fitnext);return 0;推论：#includeint main(void)int fiblast=1 ,fibthis=1 ,fibnext=1;int m=2;printf(nfibnexttfiblasttfibthistmtnn);printf(,%dt%dt%dt%dtn,fibnext,fiblast,fibthis,m);dofibnext=fiblast+fibthis;printf(%dtn,fibnext);fiblast=fibthis;fibthis=fibnext;printf(n%dt%dtH,fiblast,fibthis);m+;printf(dn”,m);while(m12);printf(nnfib(%d)=%dn,m,fibnext);return 0;)3.1代码3.5#includeint main(void)int fitlast=l,fitthis=l,fitnext=l;int m;fbr(m=2;ml 2;m+)fitnext=fitlast+fitthis;fitlast=fitthis;fitthis=fitnext;printf(nnfib(%d)=%dM,m,fitnext);return 0;推论：#includeint main(void)int fiblast=1 ,fibthis=1 ,fibnext=1;int m=2;printf(nfibnexttfiblasttfibthistmtnu);printf(n%dt%dt%dt%dtnH,fibnext,fiblast,fibthis,m);for(m=3;m=12;m+)fibnext=fiblast+fibthis;printf(%dtn,fibnext);fiblast=fibthis;fibthis=fibnext;printf(n%dt%dtn,fiblast,fibthis);printf(,%dn,m);printf(unfib(%d)=%dn,m,fibnext);return 0;)3.1代码3.6#includeint main()int peachNumber=1 ,days=10;while(daysl)peachNumber=(peachNumber+1)*2;-days;)printffn 地1天的桃子数为：d 个。”,days,peachNumber);return 0;)推论：#includeint main()int peachNuinber=1 ,days=10;printf(npeachNumbertdaystnu);printf(,%dt%dtn,peachNumber,days);while(daysl)peachNumber=(peachNumber+l)*2;printf(,%dtn,peachNumber);days;printf(%dnu,days);printffn 地(1天的桃子数为:%(1个。”,days,peachNumber);return(0);)3.1代码3.9#include#include#includedouble func(double x)double y;y=x*x-x-l;return y;int main(void)double x 1 ,x2,error,root;printf(ninput error:”)；scanf(n%lf&error);printf(ninput xlandx2:u);scanf(n%lf%lf;&xl,&x2);if(func(x l)*func(x2)0)(printf(“此方程在 无解”,x l,x2);exit(0);elsewhile(fabs(x 1 -x2)error)root=(x I+x2)/2;if(func(root)*func(x l)0)x2=root;elsexl=root;printff,方 程 在 上 的 根 为:%lfn*,xl,x2,(xl+x2)/2);return 0;推论：#include#include#includedouble func(double x)double y;y=x*x-x-l;return y;int main(void)double xl,x2,error,root;printfVinput error:11);scanf(n%lf,&error);printf(n input xlandx2:u);scanf(,%lf%lf,&xl,&x2);printf(nxlttx2tnu);printf(n%mt%lftnH,x 1 ,x2);if(func(x l)*fiinc(x2)0)printf(此方程在 无解”,x l,x2);exit(0);elsewhile(fabs(x 1 -x2)error)root=(xl+x2)/2;if(func(root)*func(x2)0)x 1 =root;printf(H%lft%lfttn,x 1 ,x2);else(x2=root;printf(n%lft%lftn,xl,x2);)printffn 方 程 在 上 的 根 为：%lfn,xl,x2,(xl+x2)/2);return 0;)探索验证：3.1讨论：#includeint main()int i=5,s=0;s=+i+i;printf(3)%dt%dn,i,s);i=5,s=0;s=i-H-+i;printfC(4)%dt%dnM,i,s);i=5,s=0;s=(i+)+(i+);printf(5)%dt%dn,i,s);i=5,s=0;s=i+i;printf(6)%dt%dn,i,s);i=5,s=0;s=i+i+;printf(7)%dt%dn,i,s);i=5,s=0;s=i+i;printf(M(8)%dt%dni,s);i=5,s=0;s=(+i)+(i+);printf(9)%dt%dn,i,s);i=5,s=0;s=i+(+i);printf(n(10)%dt%dn,i,s);i=5,s=0;s=+i+i+;printf(u(11)%dt%dn,i,s);i=5,s=0;s=(i+)+i;printf(u(12)%dt%dn,i,s);return 0;3.1练习 1.(4)编程验证:#includeint main(void)int a=0,i;for(i=l;i5;i+)switch(i)case O:case 3:a+=2;printf(n%dn,a);case 1 :case 2:a+=3;printf(H%dn,a);default:a+=5;printf(M%dn,a);)printf(dn”,a);return 0;)推论：#includeint main(void)int a=0,i;fdr(i=l;i5;i-H-)switch(i)case 0:case 3:a+=2;printf(H%dn,a);case 1 :case 2:a+=3;printf(M%dn,a);default:a+=5;printf(dn”,a);printf(H%dnn,a);return 0;3.1练 习 1 的(5)编程验证：#includeint main()int k=5,n=0;while(k0)switch(k)default:break;case 1 :n+=k;case 2:case 3:n+=k;k-；printf(n%dnn,n);return 0;3.1开发练习：i(i)程序编程：#includeint main()int nlast=0,nthis=1 ,nnext=1;int a4,a3,a2,al;int n=0;printfTntnlasttnthistnextn”)；don=n+1;printf(n%dt%dt%dt%dnu,n,nlast,nthis,nnext);nlast=nthis;while(n3);a4=nthis;a3=0;a2=0,a l=a4;while(n20)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(u%dt%dt%dt%dnn,n,nlast,nthis,nnext);printfC牛牛的数量为:%dn,nthis);return 0;或#include int main(void)int a=l,b=l,c=l,d=O,i;/*d 为总牛数*/printf(,atbtctdnn);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%dnu,a,b,c,d);printf(“第 20年母牛的总数为dn”,d);return 0;或：#includeint main()int a=l,b=l,c=l,d,i;for(i=4;ient s and$6七 七111854(1:11115七 七0 桌面(语言口61)118845.6X广-llXn1n la st n th is next0 112 1113 1114 12 35 2 3 46 3 4 67 4 6 98 6 9 139 9 13 1910 13 19 2811 19 28 4112 28 41 6013 41 60 8814 60 88 12915 88 129 18916 129 189 27717 189 277 40618 277 406 59519 406 595 87220 595 872 1278牛牛的数量为：872Press any key to c o n tin u e.1 (2)程序编程:#includeint main()double sum=1.0;int t,i=l;printf(nttsumttttinn);while(i50)t=(i+D%2；sum=sum+sum+t;i+;printf(,%dt%lfttt%dnM,t,sum,i)；printf(n%lf,sum);return 0;结果捉图:1 (3)程序编程：#includeint main()double sum=2.0;int i=l;int n;printfC请输入切的刀数:*);scanf(%d,&n);printf(nsumttinn);printf(n%lft%dnu,sum,i);while(iinistraator桌面C语言D ebugfgdsf.exe情输入切的刀数：6-|n|xsum2.0000004.0000007.00000011.00000016.00000022.00000023456最多可以分为2 2.000000块Press any key to continue，:2.(1)程序编程:#includeint main()double sum=1.5;int i;for(i=l;ient s and S ett in g sA d*in istr a to r 桌面C语言D ebugghf g.exeBBS1.500000 11.666667 21.851852 32.057613 42.286237 52.540263 62.822515 7他第一次托起的高度：3.136i27mPress any key to co n tin u e.2.(2)程序编程：#includeint main()(double sum=200.0;int i,j;fbr(i=l;i=5;i+)(if(i=l)fbr(j=2;j=12;j+)sum=sum/1.0032;printf(,i=%dtj=%dt,ij);printf(,sum=%lfn,sum);)elsesum=sum+200;for(j=l；j=12;j+)sum=sum/1.0032;printf(Mi=%dtj=%dtn,i,j);printf(,sum=%lfn,sum);)printf(当初他共存了If钱。n,sum);return 0;或：#includeint main()int i;double sum=200;fdr(i=l;ient；s and Sett i.-|E!i=l 3=12 sum=193.093314i=2 j=l sur=391.839428i=2 j=2 sun=390.589542i=2 j=3 sun=389.343642i=2 j=4 sur=388.101717i=2 j=5 sum=386.863753i=2 j=6 sum=385.629737i=2 j=7 sum=384.399659i=2 j=8 sum=383.173503i=2 j=9 sum=381.951259i=2 j=10 S U R=380.732914i=2 j=ll sur=379.518455i=2 j=12 sun=378.307870i=3 j=l sun=576.463188i=3 j=2 sum=574.624389i=3 j=3 sun=572.791457i=3 j=4 sum=570.964371i=3 j=5 sun=569.143113i=3 j=6 sun=567.327664i=3 j=7 sur=565.518007i=3 j=8 sum=563.714122i=3 j=9 sun=561.915990i=3 j=10 sun=560.123595i=3 j=ll sun=558.336917i=3 j=12 sum=556.555938i=4 j=l sur=754.142681i=4 j=2 S U R=751.737122i=4 j=3 sun=749.339237i=4 j=4 sun=746.949000i=4 j=5 sum=?44.566388i=4 j=6 sum=742.191375i=4 j=7 sum=739.823939i=4 j=8 sum=737.464054i=4 j=9 sun=735.111696i=4 j=10 S U R=732.766842i=4 j=ll sur=730.429468i=4 j=12 sun=728,099549i=5 j=l sum=925.139104i=5 j=2 sun=922.188102i=5 j=3 sum=919.246514i=5 j=4 sum=916.314308i=5 j=5 sun=913.391455i=5 j=6 sun=910.477926i=5 j=7 sun=907.573690i=5 j=8 sum=904.678718i=5 j=9 sum=901.792980i=5 j=10 sum=898.916448i=5 j=ll sum=896.049091i=5 j=12 sum=893.190880苜初他共存了 893.19088。钱。Press any key to continueFl老师答案：728编程:#include#includeint main()int i;double x=0,sum=0;for(i=5;il;i-)x=(x+200)/pow(1.0032,12);)printf(n%.21f,x);return 0;结果：Cr*E:电脑作业 编程储存DebugCpp 16.exe*727 55Press any key to co n tin u e.2.(3)程序编程：#includeint main()int i,sum=l;fbr(i=l;ients and$611111&$4(11111$门10 桌面(语言口61)1188118.6*62 24 48 81 6 1 632 326 4 6 41 2 8 1.2 8八戒一路偷吃了1 2 7个仙桃Press any key to continue2.(4)程序编程：#in c l u d e in t m a in()in t i=l,s u m=5 0 0;d o u b l e s=5 0 0.0,k;d o p r in t f(在倒数的 d个站应建在离沙漠边沿肝处，储油量为(1的加油点 nn,i,1 0 0 0-s,s u m);k=2*i+l;s=s+5 0 0.0/k;i+；s u m=5 0 0*i;w h il e(s =1 0 0 0);r e t u r n 0;成.#in c l u d e in t m a in()in t i=l,k=l,m=5 0 0;d o u b l e s=5 0 0,s u m=5 0 0;w h il e(s u m HRAHHDM-.的的CDC9CS3CSSD000505005112205为为为为为枳出B3B-3316-3360025321162运边边边边边边离离离离离离离nt仕在在在在在g建建建建建建。应应应应应应tkn占占占占占占V-立立立立交交交eTTr个个个个k123456?y-Pn/Yrn,/tT-iVra田gJ-gggggP，法二(老师编程答案)#include void main()(int k=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尸每一单边上消耗油量(也等于每一趟的距离)，倒数第1站为500L*/double v=500;/*；每一个加油站的贮油量，倒数第1 站为500L*/while(dent s and$611:11185耻1*1鼠51门10：桌面(语言口61)1188(18116/6-|n|x58S莪数数数数数数i站距下一站的距离为:2站距下一站的距离为:3站距下一站的距离为:4站距下一站的距离为:5站距下一站的距离为:6站距下一站的距离为:7站距下一站的距离为: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#includevoid main()float xl,xO,f,fl;xl=1.5;do(xO=xl;f=(2*x0-4)*x0+3)*x0-6;fl=(6*xO-8)*xO+3;xl=x0-f7fl;while(fabs(x 1 -x0)=le-5);printf(nTHe root of equation is%5.2fnn,xl);结果捉图:3.(2)程序代码:#include#define M 10e-5int main(void)double f2=3.0,s=1.0;int i=l;while(l/f2)M)i=(-l)*i;s=s+i/f2;f2=f2+2;printf(圆周率的值为恰1 1”#*4);return 0;结果捉图：c:*C:D ocuaents and$611854(1:11151门1 0 1 桌面(语言口61)118(：11.”6Press any key to co n tin u e.日 回3.(3)(百度得到的答案，自己还不明白)问题分析与算法设计利用“正多边形逼近”的方法求出兀值在很早以前就存在，我们的先人祖冲之就是用这种方法在世界上第一个得到精确度达小数点后第6位的兀值的。利用圆内接正六边形边长等于半径的特点将边数翻番，作出正十二边形，求出边长，重复这一过程，就可获得所需精度的兀的近似值。假设单位圆内接多边形的边长为2 b,边数为i,则边数加倍后新的正多边形的边长为：x=Q-2-2*4-l-b*b2周长为：y=2*i*x i:为加倍前的正多边形的边数程序编程：#include#includeint main()double e=0.1,b=0.5,d;long int i;/*i:正多边形边数*/for(i=6;产=2)/*正多边形边数力0倍*/(d=l.O-sqrt(l.O-b*b);/*计算圆内接正多边形的边长*/b=O.5*sqrt(b*b+d*d);iR2*i*bi*e le15)break;/*精度达 le-15 则停止计算*/e=b;/*保存本次正多边形的边长作为下一次精度控制的依据*/printf(pai=%.151fn”,2*i*b);/*输出工值和正多边形的边数*/printffThe number of edges of required polygon:%ldn,i);return 0;)结果捉图：4.(1)4.(2)3.2 代码 3.11#include#define NUMBER 100int main()int m,n;printf(HnThe primers from 3 to%d is:nM,NUMBER);for(m=3;iTi=NUMBER;m-H-)int flag=1;for(n=2;nm/2;n+)if(m%n=0)flag=0;break;if(flag=0)continue;printff%dt”,m);return 0;)3.2 代码 3.12#includeint main()int men,women;double children;printf(nmentwomentchildrennn);men=0;while(men+8)women=0;while(women+11)children=36-men-wonien;if(4*men+3*women+children/2=36)printf(n%dtM,men);printf(n%dt,women);printf(n%dnn,int(children);return 0;)3.2代码313#includeint main()int a=O,b=O,c=O,d=O,e=O,i=1;fbr(a=1 ;a=5;a-H-)for(b=l;b=5;b+)if(b=a)continue;fbr(c=1 ;c=5;c+)if(c=a)|(c=b)continue;fbr(d=1 ;d=3)continue;/D*if(d=l)&(e!=2)continue;/E*if(a=5)&(c=3)continue;/c*if(e=1 )+(b=2)+(a=5)+(c!=l)+(d=1)=2)printffWW 第(1种可能：”,i+);printffA 排第d 名”,a);printffB 排第d 名”,b);printffC 排第1名卜)；printfCD 排第d 名”,d);printffE 排第d 名，e);return 0;探索验证：1 有判断语句和循环体结构决定2 可以在每个运算语句下加输出语句，从输出结果判断3.2开发练习：1.（1）程序编程：#includeint main()double m=30.0,n=6.0,t;t=m/n;printf(umttntttn);while(t2.0)m+=l;printf(%lftn,m);n-H-;printf(M%lftn,n);t=m/n;printf(,%lfn,t);prin 氓d”,int(m-30);return 0;或 4.#includeint main()int i;for(i=0;i=90;i+)(if(30+i 尸=(6+i)*2)printf(M%d,i);)return 0;结果捉图:c*C:Docu*ents and SettingsAdMin:strator桌面C语言Debugg(is.exe.nnt31.0000007.0000004.42857132.0000008.0000004.00000033.0000009.0000003.66666734.00000010.0000003.40000035.00000011.0000003.18181836.00000012.0000003.00000037.00000013.0000002.84615438.00000014.0000002.71428639.00000015.0000002.60000040.00000016.0000002.50000041.00000017.0000002.41176542.00000018.0000002.33333343.00000019.0000002.26315844.00000020.0000002.20000045.00000021.0000002.14285746.00000022.0000002.09090947.00000023.0000002.04347848.00000018Press any24.000000key to c o n tin u e.2.0000001.(2)程序编程:#includeint main()int a=O,b=O,sum 1 ,sum2,m,i,d,c;printf(”请输入的m值：)scanf(%d,&m);if(m%2=l)fbr(i=l;i=m/2;i-H-)(suml=m*m*m;c=m*m-2*i;a=a+m*m-2*i;d=m*m+2*i;b=b+m*m+2*i;sum2=m*m+a+b;printf(,%d+%d+,c,d);printf(u=%dnn,sum2);printtr，%dA3=%dn,m,sum 1);ifl(suml=sum2)printf(任何一个自然数m 的立方均可以写成m 个连续数的奇数之和n”);else if(m%2=0)(for(i=0;ient s and SettingsA dM iIlistrator桌面C语言DebugCppl(fsdas.exe79+83+77+85+75+87+73+89+=7299人3=729任何一个自然数n的立方均可以写成m个连续数的奇数之和Press any key to co n tin u e.更好的做法：程序编程：#includeint main()int sum 1 =0,sum2=0,m,c=0;printf(请输入的m值：)scanf(%d,&m);printf(u%dA3=M,m);/*if(m%2=l)*/vvhile(cm)(sum2=m*m+2*c-(m-1);sum 1 =sum 1 +sum2;printfC,%d+n,sum2);C+；printffW);printf(nnn);printf(n%dA3=%dnM,m,m*m*m);printR”个数之和为：%dnn,suml);if(suml=m*m*m)printff此自然数m 的立方可以写成m 个连续数的奇数之和n)else if(m%2=0)(while(cm)sum2=m*m+2*c-(m-1);sum 1 =sum 1 +sum2;printf(,%d+,sum2);C+；printfiW);printffW)；printfi(n%dA3=%dnn,m