//Programming Exercise 1
/*把用分钟表示的时间转换成用小时和分钟表示的时间，
使用#define创建一个表示60的常量或const常量。
通过while loop让用户重复输入值，直到用户输入小于或等于0的值才停止循环
*/
#include <stdio.h>
#define MIN_P_HOUR 60 //一小时的分钟数
int main(void)
{
int sec;
int min;
int hour;
printf("Enter the time in minute (<=0 to quit): \n");
scanf("%d", sec);
while (sec > 0)
{
}
hour = sec / MIN_P_HOUR;
min = sec % MIN_P_HOUR;
printf("the time equals to: %d hour %d minutes. \n");
}
//Programming Exercise 2
/*提示用户输入一个整数，然后打印从该数到比该数大10的所有整数
例如输入5，则打印5~15的所有整数，包括5和15，
要求打印的各值之间要有一个空格、制表符或换行符分开
*/
#include <stdio.h>
int main(void)
{
int num;
printf("Please enter a number: \n");
scanf("%d", &num);
while (num <= num+10)
{
printf("\t%d", num);
}
return 0;
}
//Programming Exercise 3
/*提示用户输入天数，然后将其转换成周数和天数
例如用户输入18，则转换成2周4天。以下面的格式显示结果
18 days are 2 weeks, 4 days.
通过while loop使用户重复输入天数
，直到用户输入一个非正值时，循环结束
*/
#include <stdio.h>
#define DAYS_P_WEEK 7 //一周的天数
int main(void)
{
unsigned int days;
unsigned int weeks;
unsigned int left;
printf("Please enter a number of days(<=0 to quit): \n");
scanf("%d", &days);
while (days > 0)
{
weeks = days / DAYS_P_WEEK;
left = days % DAYS_P_WEEK;
printf("%u days are %u weeks, %u days \n",
days, weeks, left);
printf("Please enter a number of days(<=0 to qiut): \n");
scanf("%u", days);
}
return 0;
}
//Programming Exercise 4
/*提示用户输入一个身高(单位：厘米)
并分别以厘米和英寸为单位显示该值,允许有小数部分
程序应该能让用户重复输入身高，直到用户输入一个非正值。
其输出示例如下：
Enter a height in centimeters: 182
182.0 cm = 5 feet, 11.7 inches
Enter a height in centimeters (<=0 to quit): 168.7
168.0 cm = 5 feet, 6.4 inches
Enter a height in centimeters (<=0 to qiut): 0
bye
*/
#include <stdio.h>
#define INCH_P_CM 0.3937008 //一厘米是0.3937008英寸
#define FEET_P_CM 0.0328084 //一厘米是0.032808英尺
#define INCH_P_FEET 12 //一英尺(feet)是12英寸(inch)
int main(void)
{
float height;
int feet;
float inch;
printf("Enter a height in centimeters: ");
scanf("%f", &height);
while(height > 0)
{
feet = (int)(height * FEET_P_CM) ;
inch = (cm * INCH_P_CM) - (feet * INCH_P_FEET);
printf("%.1f cm = %d feet, %.1 inches \n");
printf("Enter a height in centimeters (<=0 to quit): ");
scanf("%f", &height);
}
printf("bye \n");
return 0;
}
//Programming Exercise 5
/*
修改Listing5.13，使其可以和用户进行交互，根据用户输入的数进行计算
即，用一个读入的变量来代替20
*/
#include <stdio.h>
int main(void)
{
int count, sum;
int end; //存放用户输入的数据
count = 0;
sum = 0;
printf("Enter a number to add for one to it: \n");
scanf("%d", &end);
while (count++ < end)
sum = sum + count;
printf("sum = %d \n", sum);
return 0;
}
//Programming Exercise 6
/*
修改编程练习5的程序，使其能计算整数的平方和
（可以认为，第一天赚$1,第二天赚$4,第三天赚$9以此类推）
修改程序，使其可以跟用户进行交互，即根据用户的输入来计算
（即，用用户输入的第一个值来代替20）
*/
#include <stdio.h>
int main(void)
{
int count;
int sum;
int end;
count = 0;
sum = 0;
printf("Enter a number and i will return from\
1 to sum of all its cubes: \n");
while (count++ < end)
{
sum = count * count * count + sum;
}
printf("sum of all its cube: %d \n", sum);
return 0;
}
//Programming Exercise 7
/*
提示用户输入一个double类型的数据，并打印该数的立方
自己设计一个函数，并打印立方值
main函数要把用户输入的值传递给该函数
*/
#include <stdio.h>
int main(void)
{
double num;
double cube;
printf("Please input a number and ");
printf("I will give you its cube: \n");
scanf("%lf", &num);
cube = cube(num);
printf("The cube is : %f: ", cube);
return 0;
}
double cube(double num)
{
return num * num * num;
}
//Programming Exercise 8
/*
编写一个程序，显示求模运算的结果
把用户输入的第一个整数作为求模运算符的第二个运算对象
该数在运算过程中保持不变。
后用户输入的数作为第一个运算对象。
当用户输入一个非正值时，循环结束
其输出示例如下：
This program computes moduli.
Enter an integer to serve as the second operand : 256
Now enter the first operand : 438
438 % 256 is 182
Enter next number for first operand (<= 0 to quit) :
1234567 % 256 is 135
Enter next number for first operand (<= 0 to quit) : 0
Done
*/
#include <stdio.h>
int main(void)
{
int first_op;
int sec_op;
int result ;
printf("This program computes moduli. \n");
printf("Enter an integer to serve as the second operand : ");
scanf("%d", &sec_op);
printf("Now enter the first operand : ");
scanf("%d", &sec_op);
while (first_op > 0)
{
result = first_op % sec_op;
printf("%d %% %d is %d \n", first_op, sec_op, result);
printf("Enter next number for first operand (<= 0 to quit) : ");
scanf("%d", &first_op);
}
return 0;
}