C语言程序的设计教程 课后复习题参考答案.docx

1、C语言程序的设计教程 课后复习题参考答案C语言程序设计教程课后习题参考答案习题11. （1）编译、 .exe（2）函数 主函数（或main函数）（3）编辑 编译 2. （1）-（5）：DDBBC （6）-（10）：ABBBC3. （1）答：C语言简洁、紧凑，使用方便、灵活；C语言是高级语言，同时具备了低级语言的特征；C语言是结构化程序设计语言，具有结构化的程序控制语句；C语言有各种各样的数据类型；C语言可移植性好；生成目标代码质量高，程序执行效率高。（2）编辑、编译、执行（3）一个C程序由一或多个函数组成，一函数若干条语句构成，每条语句的末尾必须以分号结束。（4）标识符，关键字，运算符，分隔符

2、，常量，注释符等4. 从键盘输入一个双精度小数，打印出它的余弦值。#include #include main( ) double x; scanf(“%lf”, &x); printf(“%lfn”, cos(x) );第2章1.（1）BDE、ACFG（2）D (3) C (4) C 2(1)错（2）错（3）错（4）对（5）错3. （1）a=3,b=-27（2）a=11,b=6,c=6（3）3（4）1 0 1 0 1 1 0（5）-9 9 8（6）1）20 2）8 3）70 4）0 5）0 6）04.（1）#include main( ) double r, h ,v; r = 2.5; h

3、 = 3.5; v = 3.14*r*r*h; printf(“v=%lfn”, v);(2)#include main( ) char ch; ch = getchar( ); printf(“%cn”, ch + 32);(3)#include main( ) printf(“ *n”); printf(“ *n”); printf(“ *n”); printf(“*n”);(4)#include main( ) double x; scanf(“%lf”, &x); printf(“%d , %lfn”, (int)x, x (int)x );(5)#include main( ) do

4、uble a=3, b=5; double result = (-2 * a + ( 4*a b )/( 2*a + b ) )/( (a - 4*b)/(a + b) ); printf(“%lfn”, result);习题31.（1）D(2)AD(3)C(4)B(5)A(6)-(10):BDACB2.(1)3.141593,3.1416,3.142(2)c=K(3)| 123.46|,|123 |(4)x= 1.23,y= 50.00(5)03.(1)scanf(%f”, c); 改为：scanf(“%f”, &c); f = (9/5)*c+32; 改为：f = (9.0/5)*c +

5、32; printf(“摄氏温度%f度相当于华氏温度%f度”, &c, &f); 改为： printf(“摄氏温度%f度相当于华氏温度%f度”, c, f);(2) 补充定义:int h; h = 500/60 改为: h = 500 / 60; m = 500% 60 改为： m = 500%60; printf(“500分钟是%d小时%d分钟,”&h, &m); 改为： printf(“500分钟是%d小时%d分钟” , h, m);4.（1）#includemain( ) char x,y; scanf(“%c%c”, &x, &y); printf(“%dn”, (x-0) + (y-

6、0) );（2）#include main( )char x, y; char tmp; printf(“Input two characters:”); scanf(“%c%c”, &x, &y); printf(“Before swap: x=%c, y=%cn”, x, y); tmp = x; x = y; y = tmp; printf(“After swap: x=%c, y=%cn”, x, y);(3)#include main( ) char ch; ch = getchar( ); printf(“%cn”, ch - 32);第4章1.（1）-（5）：CAACA2.(1)

7、BBB(2)AAABBBCCC(3)end(4)d=20(5)s=2,t=3(6)first third(7)y=0 y=5 y=10 y=53.(1)yz xz x=A & ch=a&ch2&x-1&x=2 (4)t=x; x=y; y=t;4.(1)#include main( ) int x, y , z, t; scanf(“%d%d%d”, &x, &y, &z); if ( xy ) t=x; x=y; y=t; if( x z ) t = x; x = z; z= t; if( y z ) t = y; y= z; z = t; printf(“%d %d %dn”, x, y

8、,z);(2)#include main( ) int score; scanf(“%d”, &score); if ( score 100 ) printf(“成绩不合理n”); else if( score=90 ) printf(“优秀n”); else if( score=80 ) printf(“良好n”); else if( score = 70 ) printf(“中等n”); else if( score = 60 ) printf(“及格n”); else printf(“不及格n”);(3)#include main( ) int n; int g,s,b,q;/各位上的数

9、值 scanf(“%d”, &n); g = n%10; /个位 s = n/10%10; /十位 b = n/100%10; /百位 q = n/1000%10; /千位 if( n 10 ) /一位数 printf(“%dn”, 1);/位数 printf(“%dn”, g); /各位上的数值 else if ( n 100 ) /两位数 printf(“%dn”, 2);/位数 printf(“%d %dn”, g,s); else if ( n 1000 ) /三位数 printf(“%dn”, 3);/位数 printf(“%d %d %dn”, g, s, b); else if

10、( n 10000 ) /四位数 printf(“%dn”, 4);/位数 printf(“%d %d %d %dn”, g, s, b, q); (4)#include main( ) int n; scanf(“%d”, &n); if( n % 3=0 & n%5=0 & n%7=0 ) printf(“能同时被3、5、7整除n”); else if( n%3=0 & n%5=0) printf(“能被3和5整除n”); else if( n%3=0 & n%7=0 ) printf(“能被3和7整除n”); else if( n%5=0 & n%7=0 ) printf(“能被5和7整

11、除n”); else if( n%3=0 | n%5=0 | n%7=0 ) if( n%3=0 ) printf(“能被3整除n”); else if( n%5=0 ) printf(“能被5整除n”); else printf(“能被7整除n”); else printf(“不能被3、5、7中任一个数整除n”);(5)#include main( ) int carType;/车型。1代表夏利；2代表富康；3代表桑塔纳 double xiali = 2.1; /每公里价格 double fukang = 2.4; double sangtana = 2.7; double distance

12、; /距离 double totalMoney;/总的收费 printf(请输入您乘坐的车型：1代表夏利；2代表富康；3代表桑塔纳：); scanf(%d, &carType); printf(请输入您乘车的总路程：); scanf(%lf, &distance); if( carType = 1)/夏利 if( distance 3 ) totalMoney = 7.0; else totalMoney = 7 + xiali * (distance 3); else if( carType = 2 ) /富康 if( distance 3 ) totalMoney = 8.0; else

13、totalMoney = 8 + fukang * (distance 3); else if( carType = 3 ) /富康 if( distance 3 ) totalMoney = 9.0; else totalMoney = 9 + sangtana * (distance 3); printf(（四舍五入）您的车费为：%.0lfn, totalMoney );(6)#include main( ) double a, b, c; scanf(“%lf%lf%lf”, &a, &b, &c); if( a+bc & b+ca & c+ab ) if( a=b & b=c ) pr

14、intf(“等边三角形n”); else if( a=b | b= c | c=a ) printf(“等腰三角形n”); else printf(“一般三角形n”); else printf(“不能构成三角形n”);第5章1. (1)C(2)C(3)K=36(4)C(5)B2.(1) 3次(2) x=1 & x=200&x210(3) e = 0(4) 6次(5) 103.(1) 20,10(2) 16,0(3) 7BAB4BAB1BC(4) ABABABC(5) * * *4.(1) a != b (2) n / 10(3) scanf(“%d”, &a);5. (3) 行 int fac

15、 = 1, sum = 0;6.(1)#include main( ) char ch; int alpha=0, space=0, digit=0, other=0; while( (ch=getchar( ) ) != n ) if( ch=A&ch=a&ch=0 & ch=9) digit+; else if( = ch ) space+; else other+; printf(“%d %d %d %dn”, alpha, digit, space, other );(2)#include main( ) int m20, m10; for(m20=1; m205; m20+) for

16、(m10 = 1; m1010; m10+) if( 20*m20+10*m10 = 100 ) printf(“%d, %dn”, m20, m10 ); (3)#include main( ) int x, y, z; for(x=0; x10; x+) for(y=0; y10; y+) for(z=0; z10; z+) if( x*100+y*10+z + y*100+z*10+z = 532 ) printf(“%d %d %dn”, x, y, z);(4)#include main( ) int row, spaceCount,starCount; int n; scanf(%

17、d, &n); for( row = 1; row = n; +row) for( spaceCount = 1; spaceCount = n - row ; + spaceCount) printf( ); /打印出某行上星号前的空格 for( starCount = 1; starCount = 2* row - 1; +starCount ) printf(*); /打印出某行上的所有星号 printf(n); /换行 /打印下半部分 for(row=1; rown; row+) for( spaceCount = 1; spaceCount = row ; + spaceCount)

18、 printf( ); /打印出某行上星号前的空格 for( starCount = 1; starCount = 2*(n- row) - 1; +starCount ) printf(*); /打印出某行上的所有星号 printf(n); /换行 (5)#include main( ) int n; int g,s,b,q; int t; scanf(%d, &n); g = n % 10; s = n / 10 % 10; b = n /100 % 10; q = n/ 1000%10; g = (g+5)%10; s = (s+5)%10; b = (b+5)%10; q = (q+5

19、)%10; /第1位和第4位交换 t = g; g = q; q = t; /第2位和第3位交换 t = s; s = b; b = t; printf(%d%d%d%dn, q,b,s,g);第6章1. （1）无返回值（2）double 2（3）无限循环（死循环）（4）result = 7202.（1）#include void print1( int n );main( ) int n; scanf(%d, &n); print1( n );void print1( int n ) int row, col; for(row = 1; row=n; row+) for(col=1; col

20、=row; col+) printf(#); printf(n); （2）#include void print2( int n );main( ) int n; scanf(%d, &n); print2( n );void print2( int n ) int row, col; for(row = 1; row=n; row+) for(col=1; col=2*row-1; col+) printf(); printf(n); （3）#include int yearOld(int byear, int bmonth, int bday, int nyear, int nmonth,

21、 int nday);main( ) int nowy,nowm,nowd; int by, bm, bd; int age; printf(请输入生日:); scanf(%d%d%d, &by, &bm, &bd); printf(请输入现在日期:); scanf(%d%d%d, &nowy, &nowm, &nowd); age= yearOld(by,bm,bd,nowy,nowm,nowd); printf(age=%dn, age);int yearOld(int byear, int bmonth, int bday, int nyear, int nmonth, int nday

22、) int age; age = nyear - byear; if( nmonthbday) age-; return age;（4）#include int sum( int n );main( ) int n,s; scanf(%d, &n); s = sum( n ); printf(s=%dn, s);int sum( int n ) int s=0; while ( n ) s += n % 10; n /= 10; return s;（5）#include double sumfac( int n );main( ) int n; scanf(%d, &n); printf(%.

23、0lfn, sumfac( n ) );double sumfac( int n ) double f=1.0, s = 0.0; int i; for(i=1;i=n; i+) f *= i; s += f; return f;（6）#include int gcd(int m , int n);main( ) int m, n; scanf(%d%d, &m, &n); printf(%dn, gcd(m ,n) );int gcd(int m, int n) int t,r; if( m n ) t = m; m= n; n = t; r = m % n; while( r ) m =

24、n; n = r; r = m % n; return n;（7）#include int gcd(int m , int n);int lcm(int m, int n);main( ) int m, n; scanf(%d%d, &m, &n); printf(%dn, lcm(m ,n) );int gcd(int m, int n) int t,r; if( m n ) t = m; m= n; n = t; r = m % n; while( r ) m = n; n = r; r = m % n; return n;int lcm(int m, int n) return m*n/

25、gcd(m,n);（8）#include double mypower(double x, int y);main( ) double x; int y; scanf(%lf%d, &x, &y); printf(%lfn, mypower(x,y) );double mypower(double x, int y) int i; double f=1.0; for(i=1; i=y; i+) f *= x; return f;第7章1.（1）6 （2）5 （3）不能 （4）int a32=1,2, 3,4, 5,6 ;（5）6 9（6）abc G2.（1）#include void reverse( int a , int n ); int main( ) int array10=0; int i; printf(“请输入10个整数：”); for( i=0; i10; i+) scanf(“%d”, &arrayi);