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>
int main(void)
{
int space=0,newline=0,others=0;
char a;
printf("Please input a string end by #:");
while((a=getchar()) != '#')
if(a == ' ') space++;
else if (a == '\n') newline++;
else others++;
printf("space: %d,newline: %d,others: %d\n",space,newline,others);
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>
int main(void)
{
char a;
int i;
printf("Please input a string end by #:");
for(i = 1; (a = getchar()) != '#'; i++)
{
printf("%c--%d\t",a,a);
if(i%8 == 0) printf("\n");
}
printf("\n");
return(0);
}
3、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（不包括0）总个数、偶数的平均值，输入的奇数总个数以及奇数的平均值。
#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(1)
{
scanf("%d",&num);
if (num == 0) break;
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。
#include<stdio.h>
int main(void)
{
int a=0,b=0;
char ch;
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、用switch重做练习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(1)
{
scanf("%d",&num);
if (num == 0) break;
switch(num % 2)
{
case 0: i_even++; sum_even += num;break;
case 1: 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。此程序必须要记住前一个字符和当前的字符。用诸如“Receive your eieio award.”的输入测试它。
#include<stdio.h>
int main(void)
{
char former=0,present;
int count=0;
while((present=getchar()) != '#')
{
if((former == 'e') && (present == 'i')) count++;
former = present;
}
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//J,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>
#define BASIC 10.00
#define TIME 40
#define ADD 1.5
#define LIMIT1 300
#define RATE1 0.15
#define LIMIT2 150
#define RATE2 0.20
#define RATE3 0.25
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l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l到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为各种工资等级和税率定义常量。
#include<stdio.h>
int get_int(void);
#define TIME 40
#define ADD 1.5
#define LIMIT1 300
#define RATE1 0.15
#define LIMIT2 150
#define RATE2 0.20
#define RATE3 0.25
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 have select $%.2lf\n",basic);
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);
}
int get_int(void) //得到一个合适的整数，滤除非法数
{
int num;
char str[40];
while(scanf("%d",&num)!=1)
{
gets(str);
printf("error! %s is not a number. input again.\n",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该数的素数。
#include<stdio.h>
int isprime(int);
int main(void)
{
int num,i;
printf("input a positive number:");
scanf("%d",&num);
printf("all the primes <= %d:\n",num);
for(i=2;i<=num;i++)
if( isprime(i) )
printf("%d\t",i);
printf("\n");
return(0);
}
int isprime(int n)
{
int div;
for(div = 2; div * div <= n; div++)
if (n % div == 0)
return 0;
return 1;
}
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>
#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;
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");
While( ( ch = tolower( getchar() ) ) != 'q') //tolower：大写转小写
{
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;
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;
}
}
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);
sum = ARTICHOKE * artichoke + BEET * beet + CARROT * carrot;
if (sum > DISCOUNT_LIMIT) discount = sum*DISCOUNT_RATE;
else discount = 0;
printf("discount: %.2f\n",discount);
if (artichoke + beet + carrot <= 5) freight = 3.50;
else if (artichoke + beet + carrot <20) freight = 10;
else freight = 8 + (artichoke + beet + carrot) * 0.1;
printf("freight: %.2f\n",freight);
sum = sum - discount + freight;
printf("sum: %.2f\n",sum);
return 0;
}
