#1、编写一个程序读取输入，读到#字符停止，然后报告读取空格数，换行符数目以及所有的其它字符数目。
#include<stdio.h>
int main(void)
{
int space = 0, line_break = 0, other = 0;
char ch;
while((ch = getchar()) != '#')
{
if(' ' == ch)
{
space++;
}
else if('\n' == ch)
{
line_break++;
}
else
{
other++;
}
}
printf("space = %d\nline_break = %d\nother = %d\n", space, line_break, other);
return 0;
}
2．编写一个程序读取输入，读到#字符停止。程序要打印每个输入的字符以及对应的ASCII码（十进制）。一行打印8个字符。建议：使用字符计数和求模运算符(%)在每8个循环周期时打印一个换行符。
#include<stdio.h>
int main()
{
char ch;
int count = 0;
printf("Please enter text(# to terminate):\n");
while((ch = getchar()) != '#')
{
printf("%c:%d ", ch, ch);
count++;
if(count%8 == 0 && count != 0)
{
printf("\n");
}
}
return 0;
}
3．编写一个程序，读取整数，直到用户输入0。输入结束后，程序应该报告输入的偶数（不包括0）个数、这些偶数的平均值，输入的奇数个数以及奇数的平均值。
#include<stdio.h>
int main()
{
int number = 1;
int odd_number = 0, even_number = 0;
int count_odd = 0, count_even = 0;
printf("Please enter integer to be analyzed(0 to terminate):\n");
while(scanf("%d", &number) == 1 && number != 0)
{
if(number%2 != 0)
{
odd_number += number;
count_odd++;
}
if(number%2 == 0)
{
even_number += number;
count_even++;
}
}
odd_number /= count_odd ;
even_number /= count_even;
printf("奇数有%d个，平均值 = %d\n", count_odd, odd_number);
printf("偶数有%d个，平均值 = %d\n", count_even, even_number);
return 0;
}
4．使用if else语句编写一个程序读取输入，读到#停止。用感叹号代替句号，用两个感叹号代替原来的感叹号，最后报告进行了多少次替代。
#include<stdio.h>
int main()
{
char ch = 0;
int count1 = 0, count2 = 0;
printf("Please enter text to be analyzed:(# to terminate):\n");
while ((ch = getchar()) != '#')
{
if ('.' == ch)
{
putchar('!');
count1++;
}
else if ('!' == ch)
{
putchar('!');
putchar('!');
count2++;
}
else
{
putchar(ch);
}
}
printf("\n!代替.一共%d次, !!代替!一共%d次\n", count1, count2);
return 0;
}
5．用switch重写练习4。
#include<stdio.h>
int main()
{
char ch = 0;
int count1 = 0, count2 = 0;
printf("Please enter text to be analyzed:(# to terminate):\n");
while ((ch = getchar()) != '#')
{
switch(ch)
{
case '.': putchar('!'); count1++; break;
case '!': putchar('!'); putchar('!'); count2++; break;
default: putchar(ch); break;
}
}
printf("\n!代替.一共%d次, !!代替!一共%d次\n", count1, count2);
return 0;
}
6.编写一个程序读取输入，直接#，并报告序列ei出现的次数。
说明： 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()
{
char ch = ' ', pre_ch = ' ';
int count = 0;
while((ch = getchar()) != '#')
{
if(ch == 'e')
{
pre_ch = 'e';
}
if(pre_ch == 'e' && ch == 'i')
{
count++;
pre_ch = ' ';
}
}
printf("ei has appeared %d", 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/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>
#define RATE1 0.15
#define RATE2 0.2
#define RATE3 0.25
int main()
{
double work_time;
double sum_sanary, tax, after_tax_sanary;
scanf("%lf", &work_time);
if(work_time > 40)
{
work_time += (work_time - 40)*1.5;
}
sum_sanary = work_time * 10;
if(sum_sanary < 300)
{
tax = sum_sanary*RATE1;
}
else if(sum_sanary <= 450)
{
tax = 300*RATE1+ (sum_sanary - 300)*RATE2;
}
else
{
tax = 300*RATE1 + 150*RATE2 + (sum_sanary -450)*RATE3;
}
after_tax_sanary = sum_sanary - tax;
printf("sum_sanary = %.2lf\ntax = %.2lf\nafter_tax_sanary"
"= %.2lf\n", sum_sanary, tax, after_tax_sanary);
return 0;
}
8. 修改练习7中的假设a，使程序提供一个选择等级的菜单。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 desired pay rate or action:
1）$8.75/hr 2）$9.33/hr
3）$10.00/hr 4）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，直接输入5。如果输入1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>
#define sanary1 8.75
#define sanary2 9.33
#define sanary3 10.00
#define sanary4 11.20
#define RATE1 0.15
#define RATE2 0.2
#define RATE3 0.25
int main()
{
double work_time;
double sum_sanary, tax, after_tax_sanary;
int number;
printf("*******************************************************************************\n");
printf("Enter the number corresponding to 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");
while(scanf("%d", &number) && number != 5)
{
if(number < 1 && number > 5)
{
printf("Please input correct command!\n");
}
printf("Please input your work_time:\n");
scanf("%lf", &work_time);
if(work_time > 40)
{
work_time += (work_time - 40)*1.5;
}
switch(number)
{
case 1: sum_sanary = work_time * sanary1; break;
case 2: sum_sanary = work_time * sanary2; break;
case 3: sum_sanary = work_time * sanary3; break;
case 4: sum_sanary = work_time * sanary4; break;
}
if(sum_sanary < 300)
{
tax = sum_sanary*RATE1;
}
else if(sum_sanary <= 450)
{
tax = 300*RATE1+ (sum_sanary - 300)*RATE2;
}
else
{
tax = 300*RATE1 + 150*RATE2 + (sum_sanary -450)*RATE3;
}
after_tax_sanary = sum_sanary - tax;
printf("sum_sanary = %.2lf\ntax = %.2lf\nafter_tax_sanary"
"= %.2lf\n", sum_sanary, tax, after_tax_sanary);
}
printf("quit!\n");
return 0;
}
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>
#include<math.h>
int main()
{
int i;
int m;
printf("please input the upper number:");
scanf("%d", &m);
if(m == 0)
printf("0不是素数!\n");
if(m==1)
printf("1不是素数!\n");
while(m > 1)
{
for(i=2; i <= sqrt(m); i++)//有等号
{
if( m%i ==0)
break;
}
if(i > sqrt(m))
printf("%d\n", m);//是素数
m--;
}
return 0;
}
10.
#include<stdio.h>
#define RATE1 0.15
#define RATE2 0.28
int main()
{
double sum_sanary, tax;
int number = 0;
printf("种类\t\t税金\n");
printf("单身\t\t前17850美金按15%%，超出部分按28%%\t请按1\n");
printf("户主\t\t前23900美金按15%%，超出部分按28%%\t请按2\n");
printf("已婚，共有\t前29750美金按15%%，超出部分按28%%\t请按3\n");
printf("已婚，离异\t前14875美金按15%%，超出部分按28%%\t请按4\n");
printf("退出请按5\n\n");
while(number != 5)
{
printf("请选择税金种类:\n");
scanf("%d", &number);
printf("Please input your sum_sanary:\n");
scanf("%lf", &sum_sanary);
switch(number)
{
case 1:
if(sum_sanary <= 17850)
tax = sum_sanary*RATE1;
else
tax = 17850*RATE1 + (sum_sanary-17850)*RATE2;
break;
case 2:
if(sum_sanary <= 23900)
tax = sum_sanary*RATE1;
else
tax = 23900*RATE1 + (sum_sanary-23900)*RATE2;
break;
case 3:
if(sum_sanary <= 29750)
tax = sum_sanary*RATE1;
else
tax = 29750*RATE1 + (sum_sanary-29750)*RATE2;
break;
case 4:
if(sum_sanary <= 14875)
tax = sum_sanary*RATE1;
else
tax = 14875*RATE1 + (sum_sanary-14875)*RATE2;
break;
}
printf("tax = %.2lf\n\n", tax);
}
return 0;
}
11.
#include<stdio.h>
int main()
{
double n_artichoke = 0.0;
double n_beet = 0.0;
double n_carrot = 0.0;
double freight = 0.0;
double n_pound = 0.0;
char choice = 0;
double total_cost = 0.0;
double discount = 0.0;
double total_weight = 0.0;
while ('q' != choice)
{
printf("Please choose the item you want to buy:\n");
printf("%-20s%-20s\n%-20s%-20s\n", "a) Artichoke", "b) Beet", "c) Carrot", "q) Quit");
printf("Now enter you choose:");
choice = getchar();
switch (choice)
{
case 'a':
printf("How many pounds of Artichoke do you want to buy:");
scanf("%lf", &n_pound);
n_artichoke += n_pound;
break;
case 'b':
printf("How many pounds of Beet do you want to buy:");
scanf("%lf", &n_pound);
n_beet += n_pound;
break;
case 'c':
printf("How many pounds of Carrot do you want to buy:");
scanf("%lf", &n_pound);
n_carrot += n_pound;
break;
case 'q':
continue;
default:
printf("Your choice is invalid! Please choose again.\n");
break;
}
while (getchar() != '\n');
}
printf("%-20s%-20s%-20s%-20s\n", "Category", "Price", "Pounds", "Total price");
printf("%-20s%-20s%-20.2lf%-20.2lf\n", "Artichoke", "$2.05/pound", n_artichoke, (n_artichoke * 2.05));
printf("%-20s%-20s%-20.2lf%-20.2lf\n", "Beet", "$1.15/pound", n_beet, (n_beet * 1.15));
printf("%-20s%-20s%-20.2lf%-20.2lf\n", "Carrot", "$1.09/pound", n_carrot, (n_carrot * 1.09));
total_cost = n_artichoke * 2.05 + n_beet * 1.15 + n_carrot * 1.09;
printf("Total_cost:%.2lf\n", total_cost);
if (total_cost > 100)
{
discount = total_cost * 0.05;
printf(" Discount:%2lf\n", discount);
total_cost -= discount;
}
total_weight = n_artichoke + n_beet + n_carrot;
if (0 < total_weight && total_weight <= 5.0)
{
freight = 6.5;
}
else if (5.0 < total_weight && total_weight <= 20)
{
freight = 14.0;
}
else if (20 < total_weight )
{
freight = 14 + (total_weight - 20) * 0.5;
}
printf(" Total_freight:%.2lf\n", freight);
printf(" Final cost:%.2lf\n", (total_cost + freight));
return 0;
}
来源：oschina
链接：https://my.oschina.net/u/4741747/blog/4870799