C Primer Plus 第7章 C控制语句：分支和跳转 7.12 编程练习及答案-优快云博客

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1、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目、读取的换行符数目以及读取的所有其他字符数目。

#include <stdio.h>
#define STOP '#'
int main(void)
{
    int space=0,line=0,other=0;
    char a;
    printf("Please input a string end by #:");
    while ((a=getchar())!= STOP)
     {
         if (a==' ')
             space++;
             else if (a=='\n')
             line++;
             else
                other++;
     }
     printf("space: %d,line: %d,other: %d",space,line,other);
     return 0;
}

2、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：利用字符计数和模运算符（%）在每8个循环周期时打印一个换行符。

#include <stdio.h>
#define STOP '#'
int main(void)
{
    char ch ;
    int i,count;

    while ((ch=getchar())!=STOP)
    {
      count++;
      printf("%c--%d ",ch,ch);
      if(count%8==0)
      printf("\n");
    }
    printf("\n");
    return 0;
}

3、编写一个程序，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（不包括0）总个数、偶数平均值，输入的奇数总个数以及奇数平均值。

#include<stdio.h>

int main(void)
{
 int i_even = 0, sum_even = 0, i_odd = 0, sum_odd = 0, num;
 printf("Please input numbers (0 to quit):");
 while(scanf("%d",&num)==1 && num!=0)
 {
  if (num % 2 == 0) {i_even++; sum_even += num;}
  else {i_odd++; sum_odd += num;}
 }
 printf("even number's count: %d\n",i_even);
 printf("even number's sum  : %d\n",sum_even);
 printf("odd  number's count: %d\n",i_odd);
 printf("odd  number's sum  : %d\n",sum_odd);
 return(0);
}

4、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，将原有的感叹号用两个感叹号代替，最后报告进行了多少次替代。

#include<stdio.h>
int main(void)
{
    char ch ;
    int a=0,b=0;
    printf("Please input a string end by #:");
    while ((ch=getchar())!='#')
    {
        if(ch=='.')
        {
            putchar('!');
            a++;
        }
        else if(ch=='!')
        {
            putchar('!');
            putchar('!');
            b++;
        }
        else putchar(ch);
    }
    printf("\nthe times of '.' replaced by '!':  %d\n",a);
    printf("the times of '!' replaced by '!!': %d\n",b);
    return(0);
}

5、利用switch重写题3。

#include<stdio.h>

int main(void)
{
 int i_even = 0, sum_even = 0, i_odd = 0, sum_odd = 0, num;
 printf("Please input numbers (0 to quit):");
 while(scanf("%d",&num)==1 && num!=0)
 {
 switch (num % 2)
 {    --switch下面一定要加大括号！！！
   case 0:
    i_even++; sum_even += num;
    break;
   default:
    i_odd++;
    sum_odd += num;
 }
 }
 printf("even number's count: %d\n",i_even);
 printf("even number's sum  : %d\n",sum_even);
 printf("odd  number's count: %d\n",i_odd);
 printf("odd  number's sum  : %d\n",sum_odd);
 return(0);
}

6、编写一个程序读取输入，直到#，并报告序列ei出现的次数。


#include<stdio.h>
int main(void)
{
    int count=0;
    char ch,prev=0;

    printf("Please input a string end by #:");
    while ((ch=getchar())!='#')
    {
        if(prev=='e'&&ch=='i')  /*前一个值和当前值同时满足*/
            count++;
        prev=ch;  /*记住前一个的值*/
    }
    printf("ei has appeared %d times\n",count);
    return 0;
}

7、编写程序要求输入一周中的工作小时数，然后打印工资、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%

余下的为25%

用#define定义常量，不必关心本例是否符合当前的税法。

#include<stdio.h>
//a.基本工资
#define BASIC 10.00 //基本工资等级=BASIC美元/小时
//b.加班
#define TIME 40  //加班（超过TIME小时） =
#define ADD  1.5  //ADD倍的时间
//c.税率
#define LIMIT1 300  //前LIMIT1美元为RATE1
#define RATE1 0.15
#define LIMIT2 150  //下一个LIMIT2美元为RATE2
#define RATE2 0.20
#define RATE3 0.25 //余下的位RATE3

int main(void)
{
 double hours,gross,tax;
 printf("input the work hours of a week:");
 scanf("%lf",&hours);
 if (hours > 40) hours = 40 + (hours - 40) * 1.5;
 gross = hours * BASIC;
 printf("gross income:\t\t%lf\n",gross);
 if (gross <= LIMIT1) tax = gross * RATE1;
 else if (gross <= LIMIT2) tax = LIMIT1 * RATE1 + (gross - LIMIT1) * RATE2;
 else tax = LIMIT1 * RATE1 + LIMIT2 * RATE2 + (gross - LIMIT1 - LIMIT2) * RATE3;
 printf("tax:\t\t\t%lf\n",tax);
 printf("net income:\t\t%lf\n",gross - tax);
 return(0);
}

8、修改练习7中的假设a，使程序提供一个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) $lO.OO/hr 4) $11.20/hr

5) quit

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

如果选择1-4，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1-5以外的选项，那么程序应该提醒用户合适的选项是哪些，然后再循环。用#define定义工资等级和税率常量。

#include<stdio.h>
int get_int(void);
//b.加班
#define TIME 40  //加班（超过TIME小时） =
#define ADD  1.5  //ADD倍的时间
//c.税率
#define LIMIT1 300  //前LIMIT1美元为RATE1
#define RATE1 0.15
#define LIMIT2 150  //下一个LIMIT2美元为RATE2
#define RATE2 0.20
#define RATE3 0.25 //余下的位RATE3
int main(void)
{
    double basic,hours,gross,tax;
    printf("Enter the number corresponding to the desired pay rate or action:\n");
    printf("1) $8.75/hr\t\t\t2) $9.33/hr\n");
    printf("3) $10.00/hr\t\t\t4) $11.20/hr\n");
    printf("5) quit\n");
    switch(get_int())
    {
        case 1: basic=8.75; break;
        case 2: basic=9.33; break;
        case 3: basic=10.00; break;
        case 4: basic=11.20; break;
        default: printf("quit.\n"); return 0; //退出程序
    }
    printf("You select $%.2lf.\n",basic);
    printf("Input the work hours of a week:");
    scanf("%lf",&hours);
    if (hours>40) hours=TIME+(hours-TIME)*1.5;
    gross=hours*basic;
    printf("Gross income:\t\t%.lf\n",gross);
    if(gross<LIMIT1) tax=gross*RATE1;
    else if(gross<LIMIT2) tax=LIMIT1*RATE1+(gross-LIMIT1)*RATE2;
    else tax=LIMIT1*RATE1+LIMIT2*RATE2+(gross-LIMIT1-LIMIT2)*RATE3;
    printf("tax:\t\t\t%lf\n",tax);
    printf("net:\t\t\t%lf\n",gross-tax);
    return 0;
}

int get_int(void)  //得到一个合法的整数，滤除非法数
{
    int num;
    char str[40];
    while(scanf("%d",&num)!=1)
    {
        gets(str);
        printf("error! %s is not a number,please input a number.",str);
    }
    while(getchar()!='\n');
    return num;

}

9、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这个数的素数。

#include <stdio.h>
int isprime(int);
int main(void)
{
    int num,i;
    printf("input a positive number:");
    scanf("%d",&num);
    printf("all the prime <= %d:\n",num);
    for (i=2;i<=num;i++)
        if(isprime(i))
        printf("%d",i);
    printf("\n");
    return 0;

}

int isprime(int n)  //如果n是素数，返回1，否则返回0
{
    int div;
    for (div=2;div*div<=n;div++)
        if(n % div==0)
        return 0;
    return 1;
}

10、

10. 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l5%，超出部分按28% ┃

┗━━━━━━┻━━━━━━━━━━━━━━━━━━┛

例如，有20 000美元应征税收入的单身雇佣劳动者应缴税金0.15×17 850美元+0.28×（20 000美元-17 850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。

#include<stdio.h>
#define SINGLE 17850
#define HOST 23900
#define MARRIED_SHARE 29750
#define MARRIED_DIVORCE 14875
#define RATE1 0.15
#define RATE2 0.28


int main(void)
{
 double type,pay,tax;
 char ch;
 while(1)
 {
  printf("Select the type of marriage:\n");
  printf("1)SINGLE\n2)HOST\n3)MARRIED_SHARE\n4)MARRIED_DIVORCE\n5)quit\n");
  while((ch = getchar()) == '\n') continue; //滤掉回车
  switch(ch)
  {
  case '1': type = SINGLE; break;
  case '2': type = HOST; break;
  case '3': type = MARRIED_SHARE; break;
  case '4': type = MARRIED_DIVORCE; break;
  case '5': printf("quit\n"); return(0); //退出程序
  default : printf("input error\n"); continue;
  }
  printf("you have select %c\n",ch);
  printf("input the pay:");
  scanf("%lf",&pay);
  if (pay <= type) tax = pay * RATE1;
  else tax = type * RATE1 + (pay - type) * RATE2;
  printf("wax is %.2lf\n",tax);
 }
}

11. 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磅或以下的定单收取3.50美元的运输和装卸费用；超过5磅而不足20磅的定单收取1O.OO美元的运输和装卸费用：加磅或以上的运输，在8美元基础上每磅加收0.1美元。编写程序，在循环中使用switch语句，以便对输入a的响应是让用户输入所需的朝鲜蓟磅数，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。

#include <stdio.h>
#include <ctype.h>  //需要调用tolower()函数
#define ARTICHOKE 1.25
#define BEET 0.65
#define CARROT 0.89
#define DISCOUNT_LIMIT 100
#define DISCOUNT_RATE 0.05
#define FREIGHT_FEE1 3.50
#define FREIGHT_LIMIT1 5
#define FREIGHT_FEE2 10.00
#define FREIGHT_LIMIT2 20
#define FREIGHT_FEE3 8
#define FREIGHT_RATE 0.1
int main(void)
{
    char ch ;
    double artichoke=0,beet=0,carrot=0;  //磅数
    double sum,discount,freight;
    //列示单价
    printf("Please select your vegetable :a,b,c,q\n");
    printf("a.artichoke price:$%.2f\n",ARTICHOKE);
    printf("b.beet price:$%.2f\n",BEET);
    printf("c.carrot price:$%.2f\n",CARROT);
    printf("q.end\n");
    printf("(price as dollars per pound)\n");
    //选购菜品
    while(tolower(ch=getchar())!='q')
    {
        switch(ch)
        {
            case 'a':
                 printf("How many pounds of artichokes do you want?");
                 scanf("%lf",&artichoke);
                 printf("Please select your vegetable :a,b,c,q:");
                 continue;    //continue使得while循环得以继续
            case 'b':
                printf("How many pounds of beets do you want?");
                scanf("%lf",&beet);
                printf("Please select your vegetable: a,b,c,q:");
                continue;
            case 'c':
                printf("How many pounds of carrots do you want?");
                scanf("%lf",&carrot);
                printf("Please select your vegetable: a,b,c,q:");
                continue;
            default:break;
        }
    }
    //列示清单
    printf("%10s%10s%10s%10s\n"," ","artichoke","beet","carrot");
    printf("%10s%10.2lf%10.2lf%10.2lf\n","price",ARTICHOKE,BEET,CARROT);
    printf("%10s%10.2lf%10.2lf%10.2lf\n","pound",artichoke,beet,carrot);
    printf("%10s%10.2lf%10.2lf%10.2lf\n","charge",ARTICHOKE * artichoke,
           BEET * beet,CARROT * carrot);
    //合计金额       
    sum=ARTICHOKE * artichoke+BEET * beet+CARROT * carrot;
    //计算折扣
    if(sum>=DISCOUNT_LIMIT) discount=sum*DISCOUNT_RATE;
    else discount=0;
    printf("discount : %.2f\n",discount);
    //计算运费
    if(artichoke+beet+carrot<=5) freight=FREIGHT_FEE1;
    else if (artichoke+beet+carrot<20) freight=FREIGHT_FEE2;
    else freight = 8 + (artichoke + beet + carrot) * 0.1;
    printf("freight : %.2f\n",freight);
    //合计部费用
    sum=sum-discount+freight;
    printf("sum : %2f\n",sum);
    return 0;
}

转载于:https://my.oschina.net/idreamo/blog/700681