《精通C程序设计教程》二~九章习题答案

1、精通C程序设计教程部分习题答案2005年4月说明：本答案没有仔细校对，如果发现错误，请自行更正; 蓝色的程序经过了上机验证.第二章 C源程序基本语法与顺序结构简单程序设计2.10 ABEFG2.11 ACFHJMNPQRT2.12 (1)3 (2)0 (3)0 (4)1 (5)1 (6)160 (7)4 (8)4 (9)8 (10)8(11)1 (12)2 (13)1 (14)0 (15)6 (16)0 (17)29(18)3 (19)1 (20)12.13 2.52.14 3.52.15 (1) 4 3 1 1 (2) 4 2 2 2 (3) 3 2 1 2 2.16 (1) 0 0 (2)

2、 100 100 (3) 2 2 (4) 1 1(5) 20 20 (6) 114 115 2.17 (1) y*y*y*y*(a*a-b*b) (2) sin(y/sqrt(x*x+y*y)(3) sqrt(1+3.14/2*tan(48.5/180*3.14)(4) a/x/(a/(a+y)+b/(a+b/(a+b/z)(5) p*q*(r+1)*(r+1)/(r+1)*(r+1)-1)(6) fabs(3-exp(-x)*log(fabs(1+x)2.20 #include “stdio.h”void main()long a;int b; double x;float y; scanf

3、(“%ld%d%lf%f”,&a,&b,&x,&y); printf(“a=%lo,b=%.0f,x=%0G,y=%Lgn”,a,(float)b,x,(long double)y);2.22 #include “stdio.h”#include “math.h”void main() double a,b,c,p,s; printf(“Input a b c:”);scanf(“%lf%lf%lf”,&a,&b,&c); p=0.5*(a+b+c); s=sqrt(p*(p-a)*(p-b)*(p-c); printf(“s=%.4fn”,s);2.23 解：第1月开始: a（尚未还清的借款

4、本金）第2月开始: a(1+p)-x第3月开始: (a(1+p)-x)(1+p)-x=a(1+p)2-x(1+p)-x第4月开始: (a(1+p)2-x(1+p)-x)(1+p)-x=a(1+p)3-x(1+p)2-x(1+p)-x第n+1月开始:a(1+p)n-x(1+p)n-1+(1+p)n-2+1)=0 于是#include stdio.h#include math.hvoid main() double a,p,x;int n; printf(Input a n p:);scanf(%lf%d%lf,&a,&n,&p); x=a*p*pow(1+p,n)/(pow(1+p,n)-1);

5、 printf(x=%.2fn,x);输入a n p为 120 0.004，输出为1050.91第三章 选择结构与循环结构3.3 a=2, b=1, c=23.4 x=10, y=63.5 BCD153.6 循环体共执行了3次，输出为： 12 93.7 abcd43.8 解：a的值就是循环体被执行的次数先分析内两层循环： 最内层循环次数j=1 1j=2 2 j=i i总计: 1+2+i=由于i从1循环到n-1，因此，总的循环次数是3.9 BF3.10#include main() char c,i , ;long s=0; for(i=1;i=10;i+) ; c=getchar();if(1

6、=c=1&c100) p+=(I-100)*0.01; I=100; if(I60) p+=(I-60)*0.015;I=60; if(I40) p+=(I-40)*0.03;I=40; if(I20) p+=(I-20)*0.05;I=20; if(I10) p+=(I-10)*0.075;I=10; p+=I*0.1; printf(“p=%fn”,p);/* switch语句求解 */#include “stdio.h”void main() double I, p=0; printf(“Input I=”);scanf(“%lf”,&I); if(I=100) p+=(I-100)*0

7、.01;I=100; swith(int)(I/10) case 10: case 9:case 8:case 7:case 6: p+=(I-60)*0.015;I=60;case 5:case 4: p+=(I-40)*0.03;I=40;case 3:case 2: p+=(I-20)*0.05; I=20;case 1: p+=(I-10)*0.075;I=10;default: p+=I*0.1; printf(“p=%fn”,p);3.19 解：#include “stdio.h”void main() int s1,s2,s3; float av; char level; whi

8、le(1) printf(“Input three scores of one student:n”); scanf(“%d%d%d”,&s1,&s2,&s3); if(s10|s20|s30) break; av=(s1+s2+s3)/3.0F; switch(int)(av/10) case 9:case 10: level=A;break; case 8:level=B;break; case 7:level=C;break; case 6:level=D;break; default:level=F; printf(“level=%cn”,level); 第四章 基本程序设计技术4.3

9、 解：#include “stdio.h”void main() double s,a;int k,n; printf(“Input n=”);scanf(“%d”,&n); s=0;a=1; for(k=0;k0#include stdio.hvoid main() int n,m,k,c; printf(Input n m=);scanf(%d%d,&n,&m); if(n-mm) m=n-m; c=1; for(k=1;k1e-6) s+=a;k+=2;a*=-x*x/k/(k-1); printf(sin(%.4f)=%.4fn,x,s);4.6 用公式计算近似值，计算时取1000个因

10、子解：#include stdio.hvoid main() int n;double pi=1; for(n=2;n0#include stdio.hvoid main() int n,k;double x,a,b,s=0; printf(Input x n=);scanf(%lf%d,&x,&n); a=x*x/(1+x*x);b=1; for(k=1;k=n+1;k+) s+=b;b*=a*(2*k)/(2*k+1); s*=x/(1+x*x);printf(s=%fn,s);程序输入x,n 为0.5 100，输出为0.4.8 输入x值（|x|1） 。要求计算到最后一项绝对值小于10-6

11、时停止计算解：#include stdio.h#include math.hvoid main() float x,s,a;int k; printf(Input x=);scanf(%f,&x); if(x=1) printf(Input is not correct.n);return; a=x;s=0;k=1; while(fabs(a)1e-6) s+=a/k;a*=x;k+; printf(s=%.4fn,s);4.10计算Fibonaui分数序列前n项之和#include stdio.hmain() int i,n,t;float a=2,b=1,s=0;scanf(%d,&n);

12、for(i=1;i=n;i+)s=s+a/b; t=a; a=a+b;b=t printf(s=%10.2n,s);4.13 韩信点兵：有兵一队，五伍剩一，六伍剩五，七伍剩四，十一伍剩十，问兵几何？解：#include stdio.hvoid main() int m; for(m=1;m=10000;m+) if(m%5=1&m%6=5&m%7=4&m%11=10) printf(m=%dn,m);1万以内的解有4个：即2111, 4421, 6721, 90414.14 解：#include stdio.hvoid main() int man,child,women; printf(%8

13、s%8s%8sn,man,women,child); for(man=1;man17;man+)for(women=1;women=1&man+women+child=30) printf(%8d%8d%8dn,man,women,child);4. 15 将一张面值100元的人民币兑换为5元，2元，1元和5角的零钞，要求零钞总数不超过40张且每种至少1张，求所有兑换方案。（要求每输出20个解之后暂停输出，按任意键之后，继续输出）解：#include stdio.h#include conio.hvoid main() int i5,i2,i1,i0_5,count=0; for(i5=1;i

14、5=20;i5+) for(i2=1;i237;i2+) for(i1=1;i1=1&i5+i2+i1+i0_5=40) printf(%4d%4d%4d%4dn,i5,i2,i1,i0_5); count+; if(count%20=0) printf(press any key to continue.n); getch(); 4. 16 解：#include stdio.hvoid main() long s;int a,b,c,d; for(a=1;a=9;a+)for(b=1;b=9;b+) for(c=1;c=9;c+)for(d=1;d=9;d+) s=+a*10000+(475

15、0+b-(3049+c*100)*(106+d*10); if(s=) printf(%4d%4d%4d%4dn,a,b,c,d);4. 17 解：#include stdio.hvoid main() int i,j,n; printf(Input n=);scanf(%d,&n); for(i=1;i=n;i+) for(j=1;j=40-i;j+) putchar(32); for(j=1;j=1;j-) printf(%d,j);putchar(n); 4.18 打印如下图形，打印行数n由用户输入。要求把图形放在屏幕中央。 a a b a a b c b a a b c d c b a

16、a b c d e d c b a解：#include stdio.hvoid main() int i,j,n;int w=3; printf(Input n=);scanf(%d,&n); for(i=1;i=n;i+) printf(%*s,40-i*w,); for(j=1;j=1;j-) printf(%*c,w,a+j-1);printf(n); 4.19 解：#include stdio.hvoid main() int n,i,j; printf(Input n=);scanf(%d,&n); if(n%2=0) printf(Input is not correct.n);r

17、eturn; n=n/2+1; for(i=1;i=n;i+) printf(%*s,n-i,); for(j=1;j=1;i-) printf(%*s,n-i+1,); for(j=1;j=2*i-1;j+) putchar(*);printf(n); 4. 20 解：#include stdio.hvoid main() int n,i,j; printf(Input n=);scanf(%d,&n); if(n%2=0) printf(Input is not correct.n);return; n=n/2+1; for(i=1;i=n;i+) printf(%*s,n-i,); fo

18、r(j=1;j=1;i-) printf(%*s,n-i+1,); for(j=1;j=2*i-1;j+) putchar(j=1|j=2*i-1?A+i-1:-);printf(n); 4.21 解：#include stdio.hvoid main() int i,j,n,a=1,b; printf(Input n=);scanf(%d,&n); for(i=1;i=n;i+) b=a; for(j=1;j=n;j+) printf(%4d,b);b+=j; printf(n); a+=i; 4.22 解：#include stdio.hvoid main() int i,j; for(i

19、=1;i=9;i+) printf(%*s,6*(i-1),); for(j=i;j=9;j+) printf(%d*%d=%-2d ,i,j,i*j);printf(n); 4. 23 解：#include stdio.h#include math.h#include conio.hvoid main() int m,k,i,count=0; for(m=3;m=1000;m+=2) k=(int)sqrt(m); for(i=3;ik) k=(int)sqrt(m+2); for(i=3;ik) printf(%4d%4dn,m,m+2);count+; if(count%20=0) pr

20、intf(Press any key to continue.n);getch(); 4. 24 解：#include stdio.hvoid main() int m,i,s; for(m=1;m=1000;m+) s=1; for(i=2;i=m/2;i+) if(m%i=0) s+=i;if(s=m) printf(%d=1,m); for(i=2;i=m/2;i+) if(m%i=0) printf(+%d,i); printf(n); 4. 25 解：gcd(a1,a2,an-1,an)=gcd(gcd(a1,a2,an-1),an) lcm(a1,a2,an-1,an)=lcm(l

21、cm(a1,a2,an-1),an)#include stdio.hvoid main() int a,lcm=0,gcd=0,t,b;long m; printf(Input natural numbers until input 0:n); while(1) scanf(%d,&a);if(a=0) break; b=a; while(gcd) t=a%gcd;a=gcd;gcd=t; gcd=a;a=b;m=(long)a*lcm;while(lcm) t=a%lcm;a=lcm;lcm=t; lcm=m=0?a:(int)(m/a); printf(gcd=%d,lcm=%dn,gcd

22、,lcm);4.26#include stdio.h#include math.hvoid main() long m,p; for(m=2;m=10000;m+) p=(long)(log(m)/log(10)+1);p=(long)pow(10,p); if(m*m%p=m) printf(%ld:%ld*%ld=%ldn,m,m,m,m*m); 4.27 解：#include stdio.hvoid main() int i,j,s1,s2,k,c=0; for(i=1;i=1000;i+) for(j=i+1;j=1000;j+) s1=0; for(k=1;k=i/2;k+) if(

23、i%k=0) s1+=k; s2=0; for(k=1;k=j/2;k+) if(j%k=0) s2+=k; if(s1=j&s2=i) printf(%4d %4dn,i,j); 第五章 数组5.4 解：/* 方法1：另辟数组空间进制复制*/#include stdio.h#define N 8void main() int aN,bN,i,j; printf(Input %d integer numbers:n,N); for(i=0;iN;i+) scanf(%d,&ai); j=0; for(i=0;iN;i+) if(ai%2=1) bj+=ai; /* 先复制奇数 */ for(i

24、=0;iN;i+) if(ai%2=0) bj+=ai; /* 再复制偶数 */ for(i=0;iN;i+) printf(%4d,bi); printf(n);/* 方法2：不另辟数组空间进制复制*/#include stdio.h#define N 8void main() int aN,i,j,t; printf(Input %d integer numbers:n,N); for(i=0;iN;i+) scanf(%d,&ai); j=0; for(i=0;iN;i+) if(ai%2=1) t=aj;aj+=ai;ai=t; for(i=0;iN) printf(Input is

25、too big.n);return; for(i=0;in;i+) ai0=aii=1; for(i=2;in;i+) for(j=1;ji;j+) aij=ai-1j+ai-1j-1; for(i=0;in;i+) for(j=0;j=i;j+) printf(%4d,aij); printf(n); 5.9 解：#include stdio.h#define N 10void main() int aN,i,j; printf(Input %d integer numbers:n,N); for(i=0;iN;i+) scanf(%d,&ai); for(i=j=0;i=0) aj+=ai

26、; for(i=0;ij;i+) printf(%6d,ai); printf(n);5.12 解：k=-1 bi=1 ai ck+ =a&si=A&si=0;i-) putchar(si); printf(n);/*指针法*/#include stdio.hvoid main() char s81,*p; printf(Input a string:);gets(s); for(p=s;*p;p+) if(*p=a&*p=A&*p=s;p-) putchar(*p); printf(n);6.10 解：#include stdio.hvoid main() char s81,*p;int c

27、ount=0; printf(Input a string:);gets(s); for(p=s;*p;p+) if(*p=0&*p=9) count+; printf(number of digits=%dn,count);6.11 解：/*下标法*/#include stdio.h#define N 8void main() int aN,i,j,x,m; printf(Input %d integer numbers:n,N); for(i=0;iN;i+) scanf(%d,a+i); printf(Input m=);scanf(%d,&m); for(i=1;i=m;i+) x=a

28、0;for(j=1;jN;j+) aj-1=aj; aN-1=x; for(i=0;iN;i+) printf(%4d,ai); printf(n);/*指针法*/#include stdio.h#define N 8void main() int aN,i,*p,x,m; printf(Input %d integer numbers:n,N); for(p=a;pa+N;p+) scanf(%d,p); printf(Input m=);scanf(%d,&m); for(i=1;i=m;i+) x=*a;for(p=a+1;pa+N;p+) p-1=*p; *(p-1)=x; for(p

29、=a;pa+N;p+) printf(%4d,*p); printf(n);6. 14 解：#include stdio.h#include stdlib.hvoid main() int *p,n,i,max,min;double av; printf(Input n=);scanf(%d,&n); if(n=0) printf(Input is not correct.n);return; p=(int *)malloc(n*sizeof(int); if(p=NULL) printf(Not enough memory.n);return; printf(Input %d scores:

30、n,n); for(i=0;in;i+) scanf(%d,p+i); av=max=min=p0; for(i=1;imax) max=pi; else if(pimin) min=pi; av+=pi; av/=n;printf(max=%d,min=%d,av=%.2fn,max,min,av); free(void *)p);6.16 解：#include stdio.h#include string.h#define MaxLen 80void main() char s1MaxLen+1,s2MaxLen+1;int n,i,len1,len2; printf(Input stri

31、ng 1:n);gets(s1); printf(Input string 2:n);gets(s2); printf(Input n=);scanf(%d,&n); len1=strlen(s1);len2=strlen(s2); if(nlen1) printf(Insert position is not correct.n);return; if(len1+len2MaxLen) printf(Not enough memory to contain the combined strings.n); return; s1len1+len2=0; for(i=len1-1;i=n;i-)

32、 s1i+len2=s1i; for(i=0;ilen1) if(len+len2-len1MaxLen) printf(Not enough space.n);return; else for(q=s+len;q=p+len1;q-) *(q+len2-len1)=*q; if(len2len1) for(q=p+len1;q=s+len;q+) *(q+len2-len1)=*q;for(i=0;i=p+maxlen;q-) q1=p;q2=q;len=q2-q1+1; while(q1=q2) if(lenmaxlen) maxlen=len;pmax=p;break; p+;if(s+

33、strlen(s)-p1) for(i=0;i=2;k-) n/=10; return (int)(n%10);void main() long n;int k; printf(Input n k:);scanf(%ld%d,&n,&k); printf(%dn,digit(n,k);7.10 解：#include stdio.h#include math.h#include conio.hisprime(int n) int k,m; m=(int)sqrt(n+1); for(k=2;km;void main() int n,p,c=0; for(n=4;n=1000;n+=2) for(

34、p=2;p=10) c+=55;else c+=48;bufi+=c; for(-i;i=0;i-) printf(%2c,bufi);void main() long x; printf(Input a long integer number:);scanf(%ld,&x); tohex(x);printf(n);/*递归算法*/#include stdio.hvoid tohex(unsigned long x) if(x=0) return; tohex(x/16); x%=16; if(x=10) x+=55;else x+=48; printf(%2c,(char)x);void m

35、ain() long x; printf(Input a long integer number:);scanf(%ld,&x); tohex(x);printf(n);7.12 解：/*非递归算法*/#include stdio.h#include string.hvoid reverse(char *s) char *p,*q,t; p=s;q=s+strlen(s)-1; while(p=q) return; t=*p;*p=*q;*q=t; reverse(p+1,q-1);void main() char s81; printf(Input a string:n);gets(s);

36、reverse(s,s+strlen(s)-1); printf(Reversed string is n);puts(s);7.13 解:#include stdio.hvoid gw(char *s,char *word) int maxlen,len;char *pmax; maxlen=0;len=0; while(1) if(*s=a&*s=A&*smaxlen) maxlen=len;pmax=s-len; len=0; if(*s=0) break; s+; for(;maxlen=1;maxlen-) *word+=*pmax+; *word=0;void main() char s256,word32; printf(Input a English sentence:n);gets(s); gw(s,word); printf(The longest word isn);puts(word);7.14 解： 1 m=0 = m0#include stdio.hlong comb(long n,long m) if(m=0) return 1L; return (n-m+1)*comb(n,m-1)/m;void main() long n,m; printf(Input n m:);s