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"精编C语言上机练习参考答案：编程乐趣无限"

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The C programming language is a powerful and versatile tool for software development. With its ability to directly access and manipulate memory, C allows for efficient and precise coding that can tackle complex tasks with ease. This makes it a popular choice for system programming, embedded systems, and high-performance applications. In the provided exercise, the task is to write a simple program that displays a message on the screen. The output should read: "Programming in C is fun! I love C language." This exercise serves as a basic introduction to C programming and demonstrates the use of the printf() function to output text to the screen. To achieve this output, we need to use the printf() function in C, which is a standard library function used for formatted output. The function takes a format string as its argument, which specifies the text to be displayed, and any optional arguments that are formatted according to the placeholders in the format string. In this case, the format string is "---------------------------------\nProgramming in C is fun! I love C language.\n---------------------------------", where "\n" represents a newline character that moves the cursor to the next line. This format string is passed as an argument to the printf() function, which then displays the specified text on the screen. Overall, this exercise serves as a simple yet essential practice in using the printf() function in C to output text to the screen. It highlights the fundamental principles of C programming and sets the stage for more complex and challenging tasks in the future. With a solid understanding of basic concepts and functions like printf(), aspiring programmers can start their journey towards mastering the C programming language and unlocking its full potential.

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第 16 页

（1）如果 a 和 b 的值为零，则没有解；

（2）如果 a 的值为零，则只有一个解（x=-c/b）;

（3）如果 b

-4ac 为负数，则没有实数根；

（4）否则，有两个实数根：

acbb

��

�

acbb

��

�

输入 a, b 和 c 的值，输出求根的情况，如有实数根，则输出实数根的值。

Program

#include <stdio.h>

#include <math.h>

main ( )

{ float a, b, c, f;

printf ( "Please input the value of a, b and c in the quadratic equation:\nax*x + bx

+ c = 0\n" );

printf ( "a = " ); scanf ( "%f", &a );

printf ( "b = " ); scanf ( "%f", &b );

printf ( "c = " ); scanf ( "%f", &c );

printf ( "About %.2fx*x + %.2fx + %.2f = 0: ", a, b, c );

if ( a == 0 && b == 0 )

printf("There is no solution!\n");

else if ( a == 0 )

printf("There is only one root: %.2f.\n",- c/b);

else

{ f = b * b - 4 * a * c;

if ( f <0 )

printf("There are no real root!\n");

else

printf("There are 2 real roots: %.2f, %.2f.\n",

(-b + sqrt(f))/2/a, (-b - sqrt(f))/2/a );

Output

Please input the value of a, b and c in the quadratic equation:

ax*x + bx + c = 0

a = -3¿ /* Blue is input */

b = 4¿ /* Blue is input */

c = 5¿ /* Blue is input */

About -3.00x*x + 4.00x+5.00 = 0: There are 2 real roots: -0.79, 2.12.

5-9 编写程序，读取 x 的值，求下面函数的值

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��

�

xfor

第 17 页

（1）用 else if 语句实现

（2）用嵌套 if 语句实现

（3）用条件运算符 ? : 实现

Program (1)

#include <stdio.h>

main ( )

{ float x;

printf ( "Please input the value of x: " );

scanf ( "%f", &x );

if ( x > 0 )

printf ( "y = 1\n" );

else if ( x == 0 )

printf ( "y = 0\n" );

else printf ( "y = -1\n" );

Program (2)

#include <stdio.h>

main ( )

{ float x;

printf ( "Please input the value of x: " );

scanf ( "%f", &x );

if ( x != 0 ) /*写成 if(x)是否正确？*/

{ if ( x > 0 ) /*写成 if(x)是否正确？*/

printf ( "y = 1\n" );

else

printf ( "y = -1\n" );

else printf ( "y = 0\n" );

Program (3)

#include <stdio.h>

main ( )

{ float x;

printf ( "Please input the value of x: " );

scanf ( "%f", &x );

printf ( "y = %d\n", x > 0 ? 1 : x < 0 ? -1 : 0 );

Output

Please input the value of x: 9¿ /* Blue is input */

y = 1

5-10 某个服装展示厅宣布以下所卖物品季节性打折：

第 18 页

折扣

购买总额

机加工制品

手工制品

0~100

无折扣

101~200

7.5%

201~300

7.5%

10.0%

大于 300

10.0%

15.0%

编写程序，计算某顾客应付的款项。

要求：同时用 switch 和 if 语句实现。

Program

#include <stdio.h>

main ( )

{ float price, discount;

char category;

printf ( "Category (m/h): " ); scanf ( "%c", &category );

printf ( "Price: " ); scanf ( "%f", &price );

if ( category == 'm' )

{ switch ( (int) (price-1)/100 )

{ case 0: discount = 0; break;

case 1: discount = 0.05; break;

case 2: discount = 0.075; break;

default: discount = 0.1;

else

{ switch ( (int) (price-1)/100 )

{ case 0: discount = 0.05; break;

case 1: discount = 0.075; break;

case 2: discount = 0.1; break;

default: discount = 0.15;

printf ("The net amount to be paid is %.2f\n", price*(1-discount));

Output

Category (m/h): h¿ /* Blue is input */

Price: 250.60¿ /* Blue is input */

The net amount to be paid is 225.54

5-11 编写程序，输入 x，计算并输出下列分段函数 f(x)的值（保留 2 位小数）。

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��

)(

xfy

提示：请调用 sqrt( )函数求平方根，调用 pow( )函数求幂。

第 19 页

Program

#include <stdio.h>

#include <math.h>

main()

{ float x, y;

printf("Please input the value of x: ");

scanf("%f", &x);

if(x<0)

y=pow(x,5)+2*x+1/x;

else

y=sqrt(x);

printf("The value of y is: %.2f. \n", y);

Output

Please input the value of x: -2¿ /* Blue is input */

The value of y is: -36.50

5-12 编写程序，输入一个实数（例如 456.78），求不小于该数的最小整数和不大于该数的最大整数。（提

示：注意负数）。

Program

#include <stdio.h>

main()

{ float f;

int n;

printf("Please input a real number: "); scanf("%f", &f);

n=f;

if(f>=0)

{ printf("The largest integer (not larger than %.2f) is: %d\n", f, n);

if (f-n>0)

printf("The smallest integer (not less than %.2f) is: %d\n", f, n+1);

else

printf("The smallest integer (not less than %.2f) is: %d\n", f, n);

else

{ if (f-n<0)

printf("The largest integer (not larger than %.2f) is: %d\n", f, n-1);

else

printf("The largest integer (not larger than %.2f) is: %d\n", f, n);

printf("The smallest integer (not less than %.2f) is: %d\n", f, n);

Output (1)

Please input a real number: 6.78¿ /* Blue is input */

The largest integer (not larger than 6.78) is 6

The smallest integer (not less than 6.78) is 7

Output (2)

第 20 页

Please input a real number: 6¿ /* Blue is input */

The largest integer (not larger than 6.00) is 6

The smallest integer (not less than 6.00) is 6

Output (3)

Please input a real number: -6.78¿ /* Blue is input */

The largest integer (not larger than -6.78) is -7

The smallest integer (not less than -6.78) is -6

5-13 输入一个整数（考虑负数），判断它是奇数还是偶数（认为 0 是偶数），并输出它的绝对值（不要使

用 labs( )函数）。

（1）使用 if-else 语句；

（2）使用 if 语句，但不使用 else 语句；

（3）使用 switch 语句。

Program (1)

#include <stdio.h>

main()

{ int number, m;

printf("Please input a number: ");

scanf("%d", &number);

m= number<0?-1*number:number;

if (m%2==0) /*如果条件是 if(number%2==0)，是否正确？*/

{ printf("%d is an even number, and the absolute number is %d.\n", number, m);

else

{ printf("%d is an odd number, and the absolute number is %d.\n", number, m); /*如果

条件改为 if(m%2)，if 语句是否需要调整？*/

Program (2)

#include <stdio.h>

main()

{ int number, m;

printf("Please input a number:");

scanf("%d", &number);

m=number;

if (number<0)

m*=-1;

if (number%2==0)

printf("%d is an even number, and the absolute number is %d.\n", number, m);

if (number%2==1||number%2==-1)

printf("%d is an odd number, and the absolute number is %d.\n", number, m);

Program (3)

#include <stdio.h>

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