C primer plus(第五版)编程练习第七章

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第一题：编写一个程序，该程序读取输入直到遇到#字符，然后报告读取的空格数据、读取的换行符数目以及读取的所有其他字符数目。

解：

代码如下：

#include <stdio.h>

int main(void)

{

int sp = 0;

int en = 0;

int ot = 0;

char ch;

printf("Input a text:\n");

while((ch = getchar()) != '#')

{

if(ch == '\n')

en++;

else if(isblank(ch))

sp++;

else

ot++;

}

printf("there are %d spaces,%d enter and %d other characters\n",sp,en,ot);

return 0;

}

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第二题：编写一个程序，该程序读取输入直到遇到#字符。使程序打印每个输入的字符以及它的十进制ASCII码，每行打印8个字符/编码对。建议：利用字符技术和模运算符（%）在每8个循环周期时打印一个换行符。

解：本题直接获取字符立刻打印的话，会在回车时就输出结果，并可以继续输入，这里采用了先存储，后处理的方法避免这种情况。

代码如下：

#include <stdio.h>

#include <string.h>

int main(void)

{

int i;

int n =0;

char text[1000];

printf("Input a text:\n");

while((text[n] = getchar()) != '#')

n++;

while(getchar() != '\n')

continue;

n =0;

i = strlen(text);

while(n < i)

{

if(text[n] != '#')

{

putchar(text[n]);

printf("/%d ",text[n]);

n++;

if((n % 8) == 0)

printf("\n");

}

else

break;

}

printf("\n");

return 0;

}

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第三题：编写一个程序，该程序读取整数，直到输入0.输入终止后，程序应该报告输入的偶数（不含0）总个数，偶数的平均值，输入的奇数综个数以及奇数的平均值。

解：

代码如下：

#include <stdio.h>

int main(void)

{

int n,i,j,s1,s2,a1,a2;

i = 0;

j = 0;

s1 = 0;

s2 = 0;

printf("Input int numbers:\n");

while(scanf("%d",&n) && n != 0)

{

if(n % 2 != 0)

{

i++;

s1 += n;

}

else

{

j++;

s2 += n;

}

printf("there are %d odd number,average is %d\n",i,s1 / i);

printf("there are %d even number,average is %d\n",j,s2 / j);

return 0;

}

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第四题：利用if else语句编写程序读取输入，直到#。用一个感叹号代替每个句号，将原有的每个感叹号用两个感叹号代替，最后报告进行了多少次替代。

解：

代码如下：

#include <stdio.h>

#define P '.'

#define E '!'

#define DE "!!"

int main(void)

{

char ch;

int n = 0;

printf("Input a text,end with #:\n");

while((ch = getchar()) != '#')

{

if(ch == P)

{

putchar(E);

n++;

}

else if(ch == E)

{

printf(DE);

n++;

}

else

putchar(ch);

}

printf("replace %d times\n",n);

return 0;

}

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第五题：用switch重做第三题

解：

代码如下：

#include <stdio.h>

int main(void)

{

int n,i,j,s1,s2,a1,a2;

i = 0;

j = 0;

s1 = 0;

s2 = 0;

printf("Input int numbers:\n");

while(scanf("%d",&n) && n != 0)

{

switch(n % 2)

{

case 1:

i++;

s1 += n;

break;

case 0:

j++;

s2 += n;

}

printf("there are %d odd number,average is %d\n",i,s1 / i);

printf("there are %d even number,average is %d\n",j,s2 / j);

return 0;

}

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第六题：编写一个程序读取输入，直到#，并报告序列ei出现的次数。提示：记住前一个字符和当前字符，并用类似"Receive your eieio award"的输入来测试。

解：

代码如下：

#include <stdio.h>

int main(void)

{

char ch_pre,ch_cur;

ch_pre = ' ';

int n = 0;

printf("Input a text:\n");

while((ch_cur = getchar()) != '#')

{

if(ch_pre == 'e' && ch_cur == 'i')

n++;

ch_pre = ch_cur;

}

printf(" %d times ei exist \n",n);

return 0;

}

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第七题：编写程序，要求输入一周中的工作小时数，然后打印工资总额、税金以及净工资。作如下假设：

a，基本工资等级=10.00美元/小时

b，加班（超过40小时）=1.5倍的时间

c，税率前300美元为15%，下一个150美元为20%，余下的位25%

用#define定义常量，不必关心本例是否符合当前的税法。

解：

代码如下：

#include <stdio.h>

#define H_WAGE 10.00

#define H_OVER 40.0

#define H_RATE 1.5

#define D_1 300.0

#define D_2 450.0

#define T_1 0.15

#define T_2 0.2

#define T_3 0.25

int main(void)

{

float total,tax,salary,hour;

printf("Input how many hours you worked last week:");

scanf("%f",&hour);

if(hour <= H_OVER)

total = hour * H_WAGE;

else

total = (H_OVER + (hour - H_OVER) * H_RATE) * H_WAGE;

if(total <= D_1)

tax = total * T_1;

else if(total > D_1 && total <= D_2)

tax = D_1 * T_1 + (D_2 - total) * T_2;

else

tax = D_1 * T_1 + (D_2 - D_1) * T_2 + (total - D_2) * T_3;

salary = total - tax;

printf("last week you worked %.2f hours,your total salary is %.2f,reduce tax %.2f,you own %.2f\n",hour,total,tax,salary);

return 0;

}

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第八题：修改第七题的假设a，使程序提供一个可选择工资等级的菜单。用switch选择工资等级。程序运行的开头应该像这样：

**************************************************************************

Enter the number corresponding to the desired pay rate or action:

1) $8.75/hr 2) $9.33/hr

3) $10.00/hr 3) $11.20/hr

5) quit

**************************************************************************

如果选择1到4，那么程序应该请求输入工作小时数。程序应该一直循环运行，直到输入5.如果输入1到5之外的选项，那么程序应该提醒用户核实的选项是哪些，然后再循环。用#define为各种工资等级和税率定义常量。

解：使用两个自定义函数，处理对工资等级的选择问题和计算问题，主函数循环执行这两个自定义函数。

代码如下：

#include <stdio.h>

#define H_WAGE1 8.75

#define H_WAGE2 9.33

#define H_WAGE3 10.00

#define H_WAGE4 11.20

#define QUIT 5.0

#define H_OVER 40.0

#define H_RATE 1.5

#define D_1 300.0

#define D_2 450.0

#define T_1 0.15

#define T_2 0.2

#define T_3 0.25

float chose(void);

void com(float h_wage);

int main(void)

{

float h_wage;

while((h_wage = chose()) != QUIT)

com(h_wage);

printf("Good Bye!\n");

return 0;

}

float chose(void)

{

float h_wage;

int choice;

{

printf("**************************************************************************\n");

printf("Enter the number corresponding to the desired pay rate or action:\n");

printf("1) $8.75/hr 2) $9.33/hr\n");

printf("3) $10.00/hr 4) $11.20/hr\n");

printf("5) quit\n");

printf("**************************************************************************\n");

scanf("%d",&choice);

while(getchar() != '\n'); //scanf()函数在读取数据时，如果不符合格式则会读取失败，该输入仍然在缓冲区中，如果是循环读取，那么就会造成无限循环。这时就需要清空缓存。

switch(choice)

{

case 1:h_wage = H_WAGE1;

break;

case 2:h_wage = H_WAGE2;

break;

case 3:h_wage = H_WAGE3;

break;

case 4:h_wage = H_WAGE4;

break;

case 5:h_wage = QUIT;

break;

default:printf("bad choice,try again\n");

}

}while(choice < 1 || choice > 5);

return h_wage;

}

void com(float h_wage)

{

float hour,total,tax,salary;

printf("Input how many hours you worked last week:");

scanf("%f",&hour);

while(getchar() != '\n');

if(hour <= H_OVER)

total = hour * h_wage;

else

total = (H_OVER + (hour - H_OVER) * H_RATE) * h_wage;

if(total <= D_1)

tax = total * T_1;

else if(total > D_1 && total <= D_2)

tax = D_1 * T_1 + (D_2 - total) * T_2;

else

tax = D_1 * T_1 + (D_2 - D_1) * T_2 + (total - D_2) * T_3;

salary = total - tax;

printf("last week you worked %.2f hours,your total salary is %.2f,reduce tax %.2f,you own %.2f\n",hour,total,tax,salary);

}

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第九题：编写一个程序，接受一个整数输入，然后显示所有小于或等于该函数的素数。

解：

代码如下：

#include <stdio.h>

#include <stdbool.h>

int main(void)

{

int n,i,j;

bool isprime;

printf("Input a int number.\n");

scanf("%d",&n);

printf("prime numbers:\n");

for(i = 2;i <= n;i++)

{

for(j = 2,isprime = true;(j * j) <= i;j++)

{

if(i % j == 0)

isprime = false;

}

if(isprime)

printf(" %d",i);

}

printf("\n");

return 0;

}

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第十题：1988年United States Federal Tax Schedule是近期最基本的。它分为4类，每类有两个等级。下面是其摘要；美元数为应征税的收入。

种类

税金

单身

前17,850美元按15%，超出部分按28%

户主

前23,900美元按15%，超出部分按28%

已婚，共有

前29,750美元按15%，超出部分按28%

已婚，离异

前14,875美元按15%，超出部分按28%

例如，有20000美元应征税收入的单身雇佣劳动者应缴税金0.15X17850美元+0.28+（20000美元-17850美元）。编写一个程序，让用户指定税金种类和应征税收入，然后计算税金。使用循环以便用户可以多次输入。

解：

代码如下：

#include <stdio.h>

#define LEVEL1 17850.0

#define LEVEL2 23900.0

#define LEVEL3 29750.0

#define LEVEL4 14875.0

#define QUIT 5.0

#define RATE1 0.15

#define RATE2 0.28

float chose(void);

void com(float income);

int main(void)

{

float level;

while((level = chose()) != QUIT)

com(level);

printf("Good Bye!\n");

return 0;

}

float chose(void)

{

float level;

int choice;

{

printf("**************************************************************************\n");

printf("Enter the number chose your status:\n");

printf("1) single 2) houseman\n");

printf("3) married,common 4) married,divorced\n");

printf("5) quit\n");

printf("**************************************************************************\n");

scanf("%d",&choice);

switch(choice)

{

case 1:level = LEVEL1;

break;

case 2:level = LEVEL2;

break;

case 3:level = LEVEL3;

break;

case 4:level = LEVEL4;

break;

case 5:level = QUIT;

break;

default:printf("bad choice,try again\n");

}

}while(choice < 1 || choice > 5);

return level;

}

void com(float level)

{

float salary,tax;

printf("Input how many dollers is your salary:");

scanf("%f",&salary);

if(salary <= level)

tax = salary * RATE1;

else

tax = level * RATE1 + (salary - level) * RATE2;

printf("tax is %.2f\n",tax);

}

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第十一题：ABC Mail Order Grocery 朝鲜蓟的售价是1.25美元/磅，甜菜的售价是0.65美元/磅，胡萝卜的售价是0.89美元/磅。在添加运输费用之前，他们为100美元的订单提供5%的打折优惠。对5磅或一下的订单收取3.50美元的运输费和装卸费用；超过5磅而不足20磅的订单收取10.00美元的运输和装卸费用；20磅或以上的运输，在8美元的基础上，每磅加收0.1美元。编写程序，在循环中使用switch语句，以便对输入a的响应是让用户所需的朝鲜蓟磅数，b为甜菜的磅数，c为胡萝卜的磅数，而q允许用户退出订购过程。然后程序计算总费用、折扣和运输费用（如果有运输费的话），以及总磅数。随后程序应该显示所有的购买信息：每磅的费用、订购的磅数、该订单每种蔬菜的费用、订单的总费用、折扣，如果有的话加上运输费用，以及所有费用的总数。

解：因为数据过多而导致程序庞大，其实核心不难，关键点在于处理读取和过滤用户输入。

代码如下：

#include <stdio.h>

#define A 1.25

#define B 0.65

#define C 0.89

#define QUIT 5.0

#define CR_D 100.0

#define CR_D_RATE 0.05

#define CR_P1 5.0

#define CR_P2 20.0

#define CR_P_D1 3.5

#define CR_P_D2 10.0

#define CR_P_D3 8.0

#define CR_P_D_R 0.1

float chose(void);

int main(void)

{

float type,pan,pan1,pan2,pan3,pans,sum1,sum2,sum3,sum,pay,discount,transport;

pan1 = 0.0;

pan2 = 0.0;

pan3 = 0.0;

pans = 0.0;

sum1 = 0.0;

sum2 = 0.0;

sum3 = 0.0;

discount = 0.0;

while((type = chose()) != QUIT)

{

printf("How many pands do you want:\n");

if(scanf("%f",&pan))

{

while(getchar() != '\n'); //过滤非首个数字和回车的垃圾输入缓存。

if(type == A)

{

sum1 += type * pan;

pan1 += pan;

}

else if(type == B)

{

sum2 += type * pan;

pan2 += pan;

}

else

{

sum3 += type * pan;

pan3 += pan;

}

pans = pan1 + pan2 +pan3;

}

pay = sum = sum1 + sum2 + sum3;

if(sum >= CR_D)

discount = sum * CR_D_RATE;

pay -= discount;

if(pans > 0.0)

{

if(pans <= CR_P1)

transport = CR_P_D1;

else if(pans < CR_P2)

transport = CR_P_D2;

else

transport = CR_P_D3 + pans * CR_P_D_R;

pay += transport;

}

if(pan1 > 0.0)

printf("globe artichoke $%.2f/pound,%.2f pounds is ordered,there $%.2f will be added for it\n",A,pan1,sum1);

if(pan2 > 0.0)

printf("beet $%.2f/pound,%.2f pounds is ordered,there $%.2f will be added for it\n",B,pan2,sum2);

if(pan3 > 0.0)

printf("carrot $%.2f/pound,%.2f pounds is ordered,there $%.2f will be added for it\n",C,pan3,sum3);

printf("the summary is $%.2f\n",sum);

if(discount > 0.0)

printf("the discount is $%.2f\n",discount);

printf("%.2f pounds goods you will pay $%.2f for it\n",pans,transport);

printf("so ,finally,you will pay $%.2f for them.\n",pay);

printf("Good Bye!\n");

return 0;

}

float chose(void)

{

float type;

char choice;

{

printf("**************************************************************************\n");

printf("Enter the letter chose your wants:\n");

printf("a) globe artichoke $%.2f/pound b) beet $%.2f/pound\n",A,B);

printf("c) carrot $%.2f/pound \n",C);

printf("q) quit\n");

printf("**************************************************************************\n");

choice = getchar();

while(getchar() != '\n'); //过滤非首个字母和回车的垃圾输入缓存。

switch(choice)

{

case 'a':type = A;

break;

case 'b':type = B;

break;

case 'c':type = C;

break;

case 'q':type = QUIT;

break;

default:printf("bad choice,try again\n");

}

}while(choice != 'a' && choice != 'b' && choice != 'c' && choice != 'q');

return type;

}

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附：用python来重做了第十一题，只简化了10余行代码：

#!/usr/bin/env python

A =1.25

B =0.65

C =0.89

QUIT =5.0

CR_D =100.0

CR_D_RATE =0.05

CR_P1 =5.0

CR_P2 =20.0

CR_P_D1 =3.5

CR_P_D2 =10.0

CR_P_D3 =8.0

CR_P_D_R =0.1

def chose():

while True:

print "**************************************************************************"

print "Enter the letter chose your wants:"

print "a) globe artichoke $%.2f/pound b) beet $%.2f/pound" % (A,B)

print "c) carrot $%.2f/pound" % B

print "q) quit"

print "**************************************************************************"

choice = raw_input()[0]

if choice == 'a':

type = A

pou = input("How many pands do you want:")

break

elif choice == 'b':

type = B

pou = input("How many pands do you want:")

break

elif choice == 'c':

type = C

pou = input("How many pands do you want:")

break

elif choice == 'q':

type = QUIT

pou = 0

break

else:

print "bad choice,try again"

return (type,pou)

def com():

pou1 = 0

pou2 = 0

pou3 = 0

sum1 = 0

sum2 = 0

sum3 = 0

while True:

type,pou = chose()

if type == QUIT:

break

else:

if type == A:

sum1 += type * pou

pou1 += pou

elif type == B:

sum2 += type * pou

pou2 += pou

else:

sum3 += type * pou

pou3 += pou

return (pou1,pou2,pou3,sum1,sum2,sum3)

def main():

pou1,pou2,pou3,sum1,sum2,sum3 = com()

pous = pou1 + pou2 +pou3

pay = sum = sum1 + sum2 +sum3

discount = 0

if sum >= CR_D:

discount = sum * CR_D_RATE

pay -= discount

if pous > 0:

if pous <= CR_P1:

transport = CR_P_D1

elif pous <= CR_P2:

transport = CR_P_D2

else:

transport = CR_P_D3 + pous * CR_P_D_R

pay += transport

if pou1 > 0:

print "globe artichoke $%.2f/pound,%.2f pounds is ordered,there $%.2f will be added for it" % (A,pou1,sum1)

if pou2 > 0:

print "beet $%.2f/pound,%.2f pounds is ordered,there $%.2f will be added for it" % (B,pou2,sum2)

if pou3 > 0:

print "carrot $%.2f/pound,%.2f pounds is ordered,there $%.2f will be added for it" % (C,pou3,sum3)

print "the summary is %.2f" % sum

if discount > 0:

print "the discount is $%.2f" % discount

print "%.2f pounds goods you will pay $%.2f for it" % (pous,transport)

print "so ,finally,you will pay $%.2f for them." % pay

print "Good Bye!"

if __name__ == "__main__":

main()

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