]]>
]]>
]]>
]]>
]]>
]]>
#include <stdio.h>
void to_binary(unsigned long n);
int main(void)
{
unsigned long number;
printf("Enter an integer (q to quit): \n");
while(scanf("%lu", &number) == 1)
{
printf("Binary equivalent: ");
to_binary(number);
putchar('\n');
printf("Enter an integer (q to quit): \n");
}
printf("Done!\n");
return 0;
}
void to_binary(unsigned long n)
{
int r;
r = n % 2;
if(n >= 2)
to_binary(n / 2);
putchar('0' + r);
return;
}
复习(fn)?/span>void to_binary(unsigned long n);
int main(void)
{
unsigned long number;
printf("Enter an integer (q to quit): \n");
while(scanf("%lu", &number) == 1)
{
printf("Binary equivalent: ");
to_binary(number);
putchar('\n');
printf("Enter an integer (q to quit): \n");
}
printf("Done!\n");
return 0;
}
void to_binary(unsigned long n)
{
int r;
r = n % 2;
if(n >= 2)
to_binary(n / 2);
putchar('0' + r);
return;
}
1、实际参数和形式参量有何不同Q?br />{：(x)
形式参量Q也被称为Ş式参敎ͼ是一个变量，它在被调函数中进行定义。实际参数是在函数调用中出现的|它被赋值给形式参量。可以把实际参数认ؓ(f)是在函数被调用时用来初始化Ş式参量的倹{?br />2、写Z面所描述的各个函数的ANSI函数头。注意：(x)只写出函数头卛_Q不需要实现?br />a.donut()接受一个intcd的参敎ͼ然后输出若干?Q输?的数目等于参数的倹{?br />b.gear()接受两个intcd的参数ƈq回intcd的倹{?br />c.stuff_it()的参数包括一个doublecd的g及一个doublecd变量的地址Q功能是把第一个数值存攑ֈ指定的地址中?br />{：(x)
a.void donut(int n)
b.int gear(int n, int m)
c.void stuff_it(double n, double * d)
3、只写出下列函数的ANSI C函数_(d)不需要实现函数?br />a.n_to_char()接受一个intcd的参数ƈq回一个charcd的倹{?br />b.digits()接受的参数是一个doublecd的数值和一个intcd的数|q回值类型是int?br />c.random()不接受参敎ͼq回intcd的数倹{?br />{：(x)b.int gear(int n, int m)
c.void stuff_it(double n, double * d)
a.char n_to_char(int n)
b.int digits(double n, int m)
c.int random(void)
4、设计一个实C整数相加q将l果q回的函数?br />{：(x)b.int digits(double n, int m)
c.int random(void)
int plus(int n, int m)
{
return n + m;
}
5、假如问?中的函数实现两个doublecd的数值相加，那么应该如何修改原函敎ͼ{
return n + m;
}
{：(x)
double plus(double n, double m)
{
return n + m;
}
6、设计函数alter()Q其输入参数是两个intcd的变量x和yQ功能是分别这两个变量的数值改为它们的和以及它们的差?br />{：(x)Q?span style="background-color: yellow;">注意Q下面这U写法是错误的！Q！Q?/span>{
return n + m;
}
void alter(int x, int y)
{
x = x + y;
y = x - y;
}
正确的写法如下：(x){
x = x + y;
y = x - y;
}
void alter(int * u, int * v)
{
int temp;
temp = *u;
*u = *u + *v;
*v = temp - *v;
}
7、判断下面的函数定义是否正确?br />{
int temp;
temp = *u;
*u = *u + *v;
*v = temp - *v;
}
void salami(num)
{
int num, count;
for(count = 1; count <= num; num++)
printf("O salami mio!\n");
}
{：(x){
int num, count;
for(count = 1; count <= num; num++)
printf("O salami mio!\n");
}
有错误。num应该在salami()的参数列表中而不是在花括号之后声明，而且应该是count++而不是num++?br />8、编写一个函敎ͼ使其q回3个整数参C的最大倹{?br />{：(x)
int max(int x, int y, int z)
{
int max;
if(x > y)
if(x > z)
max = x;
else
max = z;
else
if(y > z)
max = y;
else
max = z;
return max;
}
or Q?span style="background-color: yellow;">更简z一点）{
int max;
if(x > y)
if(x > z)
max = x;
else
max = z;
else
if(y > z)
max = y;
else
max = z;
return max;
}
int max(int x, int y, int z)
{
int max = x;
if(y > max)
max = y;
if(z > max)
max = z;
return max;
}
9、给定下面的输出Q?br />Please choose one of the following:{
int max = x;
if(y > max)
max = y;
if(z > max)
max = z;
return max;
}
1)copy files 2)move files
3)remove files 4)quit
Enter the number of your choice:
a.用一个函数实现菜单的昄Q且该菜单有4个用数字~号的选项q要求你选择其中之一Q输出应该如题中所C）?br />b.~写一个函敎ͼ该函数接受两个intcd的参敎ͼ(x)一个上界和一个下界。在函数中，首先从输入终端读取一个整敎ͼ如果该整C在上下界规定的范围内Q则函数重新昄菜单Q用本题目a(chn)部分中的函数Q以再次提醒用户输入新倹{如果输入数值在规定的范围内Q那么函数应该将数D回给调用函数?br />c.使用本题目a(chn)和b部分中的函数~写一个最的E序。最的意思是该程序不需要实现菜单中所描述的功能；它只需要显C些选项q能获取正确的响应即可?br />{：(x)Q参考课后答案）
#include <stdio.h>
void menu(void);
int get_input(int, int);
int main(void)
{
int res;
menu();
while((res = get_input(1, 4)) != 4)
printf("I like choice %d.\n", res);
printf("Bye!\n");
return 0;
}
void menu(void)
{
printf("Please choose one of the following: \n");
printf("1)copy files 2)move files\n");
printf("3)remove files 4)quit\n");
printf("Enter the number of your choice: \n");
}
int get_input(int min, int max)
{
int number;
scanf("%d", &number);
while(number < min || number > max)
{
printf("%d is not a valid choice; try again.\n", number);
menu();
scanf("%d", &number);
}
return number;
}
~程l习(fn)void menu(void);
int get_input(int, int);
int main(void)
{
int res;
menu();
while((res = get_input(1, 4)) != 4)
printf("I like choice %d.\n", res);
printf("Bye!\n");
return 0;
}
void menu(void)
{
printf("Please choose one of the following: \n");
printf("1)copy files 2)move files\n");
printf("3)remove files 4)quit\n");
printf("Enter the number of your choice: \n");
}
int get_input(int min, int max)
{
int number;
scanf("%d", &number);
while(number < min || number > max)
{
printf("%d is not a valid choice; try again.\n", number);
menu();
scanf("%d", &number);
}
return number;
}
1?br />
#include <stdio.h>
double min(double, double);
int main(void)
{
printf("One of the smaller of the two numbers is %.2f", min(23.34, 12.11));
return 0;
}
double min(double x, double y)
{
return x < y ? x : y;
}
2?br />double min(double, double);
int main(void)
{
printf("One of the smaller of the two numbers is %.2f", min(23.34, 12.11));
return 0;
}
double min(double x, double y)
{
return x < y ? x : y;
}
#include <stdio.h>
void chline(char ch, int i, int j);
int main(void)
{
chline('$', 3, 5);
return 0;
}
void chline(char ch, int i, int j)
{
int index;
for(index = 1; index < i; index++)
putchar(' ');
for(index = 1; index <= j - i + 1; index++)
putchar(ch);
}
3?br />void chline(char ch, int i, int j);
int main(void)
{
chline('$', 3, 5);
return 0;
}
void chline(char ch, int i, int j)
{
int index;
for(index = 1; index < i; index++)
putchar(' ');
for(index = 1; index <= j - i + 1; index++)
putchar(ch);
}
#include <stdio.h>
void chline(char ch, int col, int row);
int main(void)
{
chline('$', 3, 5);
return 0;
}
void chline(char ch, int col, int row)
{
int i, j;
for(i = 0; i < row; i++)
{
for(j = 0; j < col; j++)
putchar(ch);
putchar('\n');
}
}
4?br />void chline(char ch, int col, int row);
int main(void)
{
chline('$', 3, 5);
return 0;
}
void chline(char ch, int col, int row)
{
int i, j;
for(i = 0; i < row; i++)
{
for(j = 0; j < col; j++)
putchar(ch);
putchar('\n');
}
}
#include <stdio.h>
double computer(double a, double b);
int main(void)
{
printf("%.2f?.2f的谐均值是Q?.3f\n", 0.3, 0.5, computer(0.3, 0.5));
return 0;
}
double computer(double a, double b)
{
double result;
result = 1 / ((1/a + 1/b) / 2);
return result;
}
5?br />double computer(double a, double b);
int main(void)
{
printf("%.2f?.2f的谐均值是Q?.3f\n", 0.3, 0.5, computer(0.3, 0.5));
return 0;
}
double computer(double a, double b)
{
double result;
result = 1 / ((1/a + 1/b) / 2);
return result;
}
#include <stdio.h>
void larger_of(double *, double *);
int main(void)
{
double x = 23.3;
double y = 34.4;
printf("Originally x = %.1f; y = %.1f\n", x, y);
larger_of(&x, &y);
printf("Now x = %.1f; y = %.1f\n", x, y);
return 0;
}
void larger_of(double * u, double * v)
{
double temp;
temp = *u > *v ? *u : *v;
*u = temp;
*v = temp;
}
6、（W一ơ码的程序读取到换行W的时候也?x)打印出来，会(x)给人看不明白的感觉Q烦性按[Enter]键的时候就退出@环，不要dEOFQ?br />void larger_of(double *, double *);
int main(void)
{
double x = 23.3;
double y = 34.4;
printf("Originally x = %.1f; y = %.1f\n", x, y);
larger_of(&x, &y);
printf("Now x = %.1f; y = %.1f\n", x, y);
return 0;
}
void larger_of(double * u, double * v)
{
double temp;
temp = *u > *v ? *u : *v;
*u = temp;
*v = temp;
}
#include <stdio.h>
#include <ctype.h>
void printchar(char ch);
int main(void)
{
char ch;
printf("误入要分析的东西：(x)\n");
while((ch = getchar()) != EOF)
{
printchar(ch);
}
return 0;
}
void printchar(char ch)
{
if(isalpha(ch))
{
printf("%c %d\n", ch, toupper(ch) % 'A' + 1);
}
}
修改之后Q程序如下：(x)#include <ctype.h>
void printchar(char ch);
int main(void)
{
char ch;
printf("误入要分析的东西：(x)\n");
while((ch = getchar()) != EOF)
{
printchar(ch);
}
return 0;
}
void printchar(char ch)
{
if(isalpha(ch))
{
printf("%c %d\n", ch, toupper(ch) % 'A' + 1);
}
}
#include <stdio.h>
#include <ctype.h>
int show_c_location(char ch);
int main(void)
{
char ch;
printf("Please enter some characters: \n");
while((ch = getchar()) != '\n')
printf("%c-%d ", ch, show_c_location(ch));
return 0;
}
int show_c_location(char ch)
{
int result;
if(isalpha(ch))
result = toupper(ch) - 'A' + 1;
else
result = -1;
return result;
}
7?br />#include <ctype.h>
int show_c_location(char ch);
int main(void)
{
char ch;
printf("Please enter some characters: \n");
while((ch = getchar()) != '\n')
printf("%c-%d ", ch, show_c_location(ch));
return 0;
}
int show_c_location(char ch)
{
int result;
if(isalpha(ch))
result = toupper(ch) - 'A' + 1;
else
result = -1;
return result;
}
#include <stdio.h>
double power(double n, int p);
int main(void)
{
double x, xpow;
int exp;
printf("Enter a number and the positive integer power");
printf(" to which\nthe number will be raised. Enter q");
printf(" to quit.\n");
while(scanf("%lf%d", &x, &exp) == 2)
{
xpow = power(x, exp);
printf("%.3g to power %d is %.5g\n", x, exp, xpow);
printf("Enter next pair of numbers or q to quit.\n");
}
printf("Hope you enjoyed this power trip -- bye!\n");
return 0;
}
double power(double n, int p)
{
int i;
double result = 1;
if(n != 0)
{
if(p > 0)
{
for(i = 1; i <= p; i++)
result *= n;
}
else if(p < 0)
{
for(i = 1; i <= -p; i++)
result *= (1 / n);
}
else
result = 1;
}
else
{
if(p == 0)
result = 1;// 0?ơ方是一个有争议的数Q本题认Z(x)得到1
else
result = 0;
}
return result;
}
8?br />double power(double n, int p);
int main(void)
{
double x, xpow;
int exp;
printf("Enter a number and the positive integer power");
printf(" to which\nthe number will be raised. Enter q");
printf(" to quit.\n");
while(scanf("%lf%d", &x, &exp) == 2)
{
xpow = power(x, exp);
printf("%.3g to power %d is %.5g\n", x, exp, xpow);
printf("Enter next pair of numbers or q to quit.\n");
}
printf("Hope you enjoyed this power trip -- bye!\n");
return 0;
}
double power(double n, int p)
{
int i;
double result = 1;
if(n != 0)
{
if(p > 0)
{
for(i = 1; i <= p; i++)
result *= n;
}
else if(p < 0)
{
for(i = 1; i <= -p; i++)
result *= (1 / n);
}
else
result = 1;
}
else
{
if(p == 0)
result = 1;// 0?ơ方是一个有争议的数Q本题认Z(x)得到1
else
result = 0;
}
return result;
}
#include <stdio.h>
double power(double n, int p);
int main(void)
{
double x, xpow;
int exp;
printf("Enter a number and the positive integer power");
printf(" to which\nthe number will be raised. Enter q");
printf(" to quit.\n");
while(scanf("%lf%d", &x, &exp) == 2)
{
xpow = power(x, exp);
printf("%.3g to power %d is %.5g\n", x, exp, xpow);
printf("Enter next pair of numbers or q to quit.\n");
}
printf("Hope you enjoyed this power trip -- bye!\n");
return 0;
}
double power(double n, int p)
{
double result = 1;
if(n != 0)
{
if(p > 0)
result = n * power(n, p-1);
else if(p < 0)
result = (1/n) * power(n, p+1);
else
result = 1;
}
else
{
if(p == 0)
result = 1;// 0?ơ方是一个有争议的数Q本题认Z(x)得到1
else
result = 0;
}
return result;
}
9?br />double power(double n, int p);
int main(void)
{
double x, xpow;
int exp;
printf("Enter a number and the positive integer power");
printf(" to which\nthe number will be raised. Enter q");
printf(" to quit.\n");
while(scanf("%lf%d", &x, &exp) == 2)
{
xpow = power(x, exp);
printf("%.3g to power %d is %.5g\n", x, exp, xpow);
printf("Enter next pair of numbers or q to quit.\n");
}
printf("Hope you enjoyed this power trip -- bye!\n");
return 0;
}
double power(double n, int p)
{
double result = 1;
if(n != 0)
{
if(p > 0)
result = n * power(n, p-1);
else if(p < 0)
result = (1/n) * power(n, p+1);
else
result = 1;
}
else
{
if(p == 0)
result = 1;// 0?ơ方是一个有争议的数Q本题认Z(x)得到1
else
result = 0;
}
return result;
}
#include <stdio.h>
void to_base_n(unsigned long n, int range);
int main(void)
{
unsigned long number;
int range;
printf("误入要转换的无W号整数和所规定的进制数Q?nbsp;\n");
while(scanf("%lu %d", &number, &range) == 2)
{
if(range >= 2 && range <= 10)
{
printf("无符h?lu转换?dq制Cؓ(f)Q?nbsp;", number, range);
to_base_n(number, range);
putchar('\n');
printf("误入要转换的无W号整数和所规定的进制数Q?nbsp;\n");
}
else
printf("所规定的进制数的范围是2~10Q请输入正确的数字\n");
}
printf("Done!\n");
return 0;
}
void to_base_n(unsigned long n, int range)
{
int r;
r = n % range;
if(n >= range)
to_base_n(n / range, range);
putchar('0' + r);
return;
}
10、（题意理解不清楚，借鉴CSDN——vs9841原作者的做法Q脑子太W，实在想不出来Q?br />void to_base_n(unsigned long n, int range);
int main(void)
{
unsigned long number;
int range;
printf("误入要转换的无W号整数和所规定的进制数Q?nbsp;\n");
while(scanf("%lu %d", &number, &range) == 2)
{
if(range >= 2 && range <= 10)
{
printf("无符h?lu转换?dq制Cؓ(f)Q?nbsp;", number, range);
to_base_n(number, range);
putchar('\n');
printf("误入要转换的无W号整数和所规定的进制数Q?nbsp;\n");
}
else
printf("所规定的进制数的范围是2~10Q请输入正确的数字\n");
}
printf("Done!\n");
return 0;
}
void to_base_n(unsigned long n, int range)
{
int r;
r = n % range;
if(n >= range)
to_base_n(n / range, range);
putchar('0' + r);
return;
}
#include <stdio.h>
int Fibonacci(int n);
int main(void)
{
int n = 9;
printf("当n?dӞ斐LU_数gؓ(f)%d", n, Fibonacci(9));
return 0;
}
int Fibonacci(int n)
{
int a, b, i;
a = 0;
b = 1;
int sum;
if(n == 0)
return 0;
if(n == 1)
return 1;
else
{
for(i = 2; i <= n; i++)
{
sum = a + b;
a = b;
b = sum;
}
return sum;
}
}
ȝQM来说~程l习(fn)相对以往来说要简单了Q但W?0题没明白什么意思，所以只能借鉴别h的了Q真是天下文章一大抄Q?img src ="http://www.aygfsteel.com/BeautifulMan/aggbug/428322.html" width = "1" height = "1" />int Fibonacci(int n);
int main(void)
{
int n = 9;
printf("当n?dӞ斐LU_数gؓ(f)%d", n, Fibonacci(9));
return 0;
}
int Fibonacci(int n)
{
int a, b, i;
a = 0;
b = 1;
int sum;
if(n == 0)
return 0;
if(n == 1)
return 1;
else
{
for(i = 2; i <= n; i++)
{
sum = a + b;
a = b;
b = sum;
}
return sum;
}
}
]]>
1、putchar(getchar())是一个有效的表达式，它实C么功能？getchar(putchar())也有效吗Q?br />{：(x)
语句putchar(getchar())使程序读取下一个输入字Wƈ打印它，getchar()的返回g为putchar()的参数。getchar(putchar())则不是合法的Q因为getchar()不需要参数而putchar()需要一个参数?br />2、下面的每个语句实现什么功能？
a.putchar('H');
b.putchar('\007');
c.putchar('\n');
d.putchar('\b');
{：(x)b.putchar('\007');
c.putchar('\n');
d.putchar('\b');
a. 昄字符H
b.如果pȝ使用ASCII字符~码Q则发出一声警?br />c.把光标移动到下一行的开?br />d.退后一?br />3、假设?zhn)有一个程序countQ该E序对输入的字符q行l计。用countE序设计一个命令行命o(h)Q对文gessay中的字符q行计数q将l果保存在名为essayct的文件中?br />{：(x)
count < essay > essayct
4、给定问?中的E序和文Ӟ下面哪个命o(h)是正的Q?br />{：(x)
a.essayct <essay
b.count essay
c.essay >count
{：(x)
c是正的?br />5、EOF是什么？
{：(x)
它是由getchar()和scanf()q回的信P一个特定的|Q用来表明已l到达了文g的结?br />6、对l出的输入，下面每个E序D늚输出是什么（假定ch是intcd的，q且输入是缓冲的Q？
a. 输入如下所C：(x)
If you quit, I will.[enter]
E序D如下所C：(x)
while ((ch = getchar()) != 'i')
putchar(ch);
b. 输入如下所C：(x)
Harhar[enter]
E序D如下所C：(x)
while ((ch = getchar()) != '\n')
{
putchar(ch++);
putchar(++ch);
}
{：(x)
a.If you qu
b.HJacrthjacrt
7、C如何处理h不同文g和换行约定的不同计算机系l？
{：(x)
C的标准I/O库把不同的文件Ş式映ؓ(f)l一的流Q这样就可以按相同的方式对它们进行处理?br />8、在~冲pȝ中把数D入与字符输入相合时Q?zhn)所面(f)的潜在问题是什么？
{：(x)
数字输入跌I格和换行符Q但是字W输入ƈ不是q样。假设?zhn)~写了这L(fng)代码Q?br />
int score;
char grade;
printf("Enter the score.\n");
scanf("%d", &score);
printf("Enter the letter grade.\n");
grade = getchar();
假设(zhn)输入分?8Q然后按下回车键来把分数发送给E序Q?zhn)同时也发送了一个换行符Q它?x)成Z一个输入字W被d到grade中作为等U的倹{如果在字符输入之前q行了数字输入，应该添加代码以在获取字W输入之前剔除换行字W?br />~程l习(fn)char grade;
printf("Enter the score.\n");
scanf("%d", &score);
printf("Enter the letter grade.\n");
grade = getchar();
1?br />
#include <stdio.h>
int main(void)
{
int ch;
int count = 0;
while((ch = getchar()) != EOF) // 包括换行W?/span>
count++;
printf("The number of characters is %d\n", count);
return 0;
}
2、（觉得q题难的！Q！看了一些他人写的例子，直胡说八道！Q！不过q是完美解决了）int main(void)
{
int ch;
int count = 0;
while((ch = getchar()) != EOF) // 包括换行W?/span>
count++;
printf("The number of characters is %d\n", count);
return 0;
}
#include <stdio.h>
int main(void)
{
int ch;
int i = 0;
while((ch = getchar()) != EOF)
{
if(ch >= 32) // 可打印字W?/span>
{
putchar(ch);
printf("/%d ", ch);
i++;
}
else if(ch == '\n') // 打印换行W?/span>
{
printf("\\n");
printf("/%d ", ch);
putchar(ch); // 清除输入~冲区里面的换行W?/span>
i = 0; // i|ؓ(f)0重新开始计敎ͼ因ؓ(f)题目要求每次遇到一个换行符时就要开始打C个新?/span>
}
else if(ch == '\t') // 打印制表W?/span>
{
printf("\\t");
printf("/%d ", ch);
i++;
}
else // 打印控制字符
{
putchar('^');
putchar(ch + 64);
printf("/%d ", ch);
}
if(i == 10)
{
putchar('\n');
i = 0;
}
}
return 0;
}
q行l果如下Q?br />int main(void)
{
int ch;
int i = 0;
while((ch = getchar()) != EOF)
{
if(ch >= 32) // 可打印字W?/span>
{
putchar(ch);
printf("/%d ", ch);
i++;
}
else if(ch == '\n') // 打印换行W?/span>
{
printf("\\n");
printf("/%d ", ch);
putchar(ch); // 清除输入~冲区里面的换行W?/span>
i = 0; // i|ؓ(f)0重新开始计敎ͼ因ؓ(f)题目要求每次遇到一个换行符时就要开始打C个新?/span>
}
else if(ch == '\t') // 打印制表W?/span>
{
printf("\\t");
printf("/%d ", ch);
i++;
}
else // 打印控制字符
{
putchar('^');
putchar(ch + 64);
printf("/%d ", ch);
}
if(i == 10)
{
putchar('\n');
i = 0;
}
}
return 0;
}
I love you!
I/73 /32 l/108 o/111 v/118 e/101 /32 y/121 o/111 u/117Q每行打?0个|
!/33 \n/10Q每ơ遇C个换行符时就开始一个新行）
My hello world^A
M/77 y/121 /32 h/104 e/101 l/108 l/108 o/111 /32 w/119Q每行打?0个|
o/111 r/114 l/108 d/100 ^A/1 \n/10Q每ơ遇C个换行符时就开始一个新行）
^Z
3?br />I/73 /32 l/108 o/111 v/118 e/101 /32 y/121 o/111 u/117Q每行打?0个|
!/33 \n/10Q每ơ遇C个换行符时就开始一个新行）
My hello world^A
M/77 y/121 /32 h/104 e/101 l/108 l/108 o/111 /32 w/119Q每行打?0个|
o/111 r/114 l/108 d/100 ^A/1 \n/10Q每ơ遇C个换行符时就开始一个新行）
^Z
#include <stdio.h>
#include <ctype.h>
int main(void)
{
int ch;
int low_count = 0, up_count = 0;
while((ch = getchar()) != EOF)
{
if(islower(ch))
low_count++;
if(isupper(ch))
up_count++;
}
printf("A number of capital letters: %d\n", up_count);
printf("A number of lower case letters: %d\n", low_count);
return 0;
}
4?br />#include <ctype.h>
int main(void)
{
int ch;
int low_count = 0, up_count = 0;
while((ch = getchar()) != EOF)
{
if(islower(ch))
low_count++;
if(isupper(ch))
up_count++;
}
printf("A number of capital letters: %d\n", up_count);
printf("A number of lower case letters: %d\n", low_count);
return 0;
}
#include <stdio.h>
#include <ctype.h>
#include <stdbool.h>
int main(void)
{
char ch;
long chars = 0L; // l计单词的字W数
int words= 0; // 单词?/span>
bool inword = false; // 如果ch在一个单词中Q则inword为true
printf("Enter text to be analyzed: \n");
while((ch = getchar()) != EOF)
{
if(!isspace(ch) && !ispunct(ch))
chars++;
if(!isspace(ch) && !inword)
{
inword = true;
words++;
}
if(isspace(ch) && inword)
inword = false;
}
printf("The average number of words per word: %ld\n", chars / words);
return 0;
}
5、（二分搜烦法W一ơ碰见，搞了大半天了Q借鉴的是CSDN-----vs9841作者的做法Q不q稍微加了下工）#include <ctype.h>
#include <stdbool.h>
int main(void)
{
char ch;
long chars = 0L; // l计单词的字W数
int words= 0; // 单词?/span>
bool inword = false; // 如果ch在一个单词中Q则inword为true
printf("Enter text to be analyzed: \n");
while((ch = getchar()) != EOF)
{
if(!isspace(ch) && !ispunct(ch))
chars++;
if(!isspace(ch) && !inword)
{
inword = true;
words++;
}
if(isspace(ch) && inword)
inword = false;
}
printf("The average number of words per word: %ld\n", chars / words);
return 0;
}
#include <stdio.h>
char get_choice(void);
char get_first(void);
int main(void)
{
int low = 1, high = 100, guess = 50;
char ch;
printf("Pick an integer from 1 to 100. I will try to guess it\n");
printf("Un
is your number %d?\n", guess);
while((ch = get_choice()) != 'q')
{
if(ch == 'a')
{
printf("I knew I could do it!\n");
break;
}
else if(ch == 'b')
{
printf("It is too small!\n");
low = guess + 1;
}
else if(ch == 'c')
{
printf("It is too big!\n");
high = guess - 1;
}
guess = (low + high) / 2;
printf("Unis your number %d?\n", guess);
}
printf("Done!\n");
return 0;
}
char get_choice(void)
{
int ch;
printf("Enter the letter of your choice: \n");
printf("a. right b. too small\n");
printf("c. too big q. quit\n");
ch = get_first();
while((ch < 'a' || ch > 'c') && ch != 'q')
{
printf("Please respond with a, b, c, or q.\n");
ch = get_first();
}
return ch;
}
char get_first(void)
{
int ch;
ch = getchar();
while(getchar() != '\n')
continue;
return ch;
}
6?br />char get_choice(void);
char get_first(void);
int main(void)
{
int low = 1, high = 100, guess = 50;
char ch;
printf("Pick an integer from 1 to 100. I will try to guess it\n");
printf("Un
is your number %d?\n", guess);while((ch = get_choice()) != 'q')
{
if(ch == 'a')
{
printf("I knew I could do it!\n");
break;
}
else if(ch == 'b')
{
printf("It is too small!\n");
low = guess + 1;
}
else if(ch == 'c')
{
printf("It is too big!\n");
high = guess - 1;
}
guess = (low + high) / 2;
printf("Un
is your number %d?\n", guess);}
printf("Done!\n");
return 0;
}
char get_choice(void)
{
int ch;
printf("Enter the letter of your choice: \n");
printf("a. right b. too small\n");
printf("c. too big q. quit\n");
ch = get_first();
while((ch < 'a' || ch > 'c') && ch != 'q')
{
printf("Please respond with a, b, c, or q.\n");
ch = get_first();
}
return ch;
}
char get_first(void)
{
int ch;
ch = getchar();
while(getchar() != '\n')
continue;
return ch;
}
char get_first(void)
{
int ch;
while((ch = getchar()) == '\n')
continue;
while(getchar() != '\n')
continue;
return ch;
}
7?br />{
int ch;
while((ch = getchar()) == '\n')
continue;
while(getchar() != '\n')
continue;
return ch;
}
#include <stdio.h>
#define WORK_OVERTIME 40
#define MULTIPLE 1.5
#define RATE1 0.15
#define RATE2 0.20
#define RATE3 0.25
#define BREAK1 300
#define BREAK2 450
#define BASE1 (BREAK1 * RATE1)
#define BASE2 (BASE1 + (BREAK2 - BREAK1) * RATE2)
char get_choice(void);
char get_first(void);
int main(void)
{
int hour, choise;
double total, tax, net_pay;
double base_pay; // 基本工资{不能?define来定义了Q因为它要随着E序而改变了Q书上真是胡说八?/span>
while((choise = get_choice()) != 'q')
{
switch(choise)
{
case 'a':
base_pay = 8.15;
break; // break只是DE序qswitch语句Q蟩到switch之后的下一条语句！Q！
case 'b':
base_pay = 9.33;
break;
case 'c':
base_pay = 10.00;
break;
case 'd':
base_pay = 11.20;
break;
default:
printf("Program error!\n");
break;
}
printf("Please enter the hour used: ");
scanf("%d", &hour); // 获取每周工作时数时没有像书上那样判断，我偷懒了Q！Q?/span>
if(hour <= WORK_OVERTIME)
{
total = hour * base_pay;
if (total <= BREAK1)
{
tax = total * RATE1;
net_pay = total - tax;
}
else
{
tax = BASE1 + (total - BREAK1) * RATE2;
net_pay = total - tax;
}
}
else
{
total = base_pay * WORK_OVERTIME + (hour - WORK_OVERTIME) * MULTIPLE * base_pay;
if(total <= BREAK2)
{
tax = BASE1 + (total - BREAK1) * RATE2;
net_pay = total - tax;
}
else
{
tax = BASE2 + (total - BREAK2) * RATE3;
net_pay = total - tax;
}
}
printf("The total pay: %.2f; tax: %.2f; net pay: %.2f\n", total, tax, net_pay);
}
printf("Bye!\n");
return 0;
}
char get_choice(void)
{
int ch;
printf("*****************************************************************\n");
printf("Enter number corresponding to the desired pay rate or action:\n");
printf("a) $8.75/hr\tb) $9.33/hr\n");
printf("c) $10.00/hr\td) $11.20/hr\n");
printf("q) quit\n");
printf("*****************************************************************\n");
printf("Please enter your choise: ");
ch = get_first();
while((ch < 'a' || ch > 'd') && ch != 'q')
{
printf("Please respond with a, b, c, d, or q.\n");
ch = get_first();
}
return ch;
}
char get_first(void)
{
int ch;
while((ch = getchar()) == '\n')
continue;
while(getchar() != '\n')
continue;
return ch;
}
8?br />#define WORK_OVERTIME 40
#define MULTIPLE 1.5
#define RATE1 0.15
#define RATE2 0.20
#define RATE3 0.25
#define BREAK1 300
#define BREAK2 450
#define BASE1 (BREAK1 * RATE1)
#define BASE2 (BASE1 + (BREAK2 - BREAK1) * RATE2)
char get_choice(void);
char get_first(void);
int main(void)
{
int hour, choise;
double total, tax, net_pay;
double base_pay; // 基本工资{不能?define来定义了Q因为它要随着E序而改变了Q书上真是胡说八?/span>
while((choise = get_choice()) != 'q')
{
switch(choise)
{
case 'a':
base_pay = 8.15;
break; // break只是DE序qswitch语句Q蟩到switch之后的下一条语句！Q！
case 'b':
base_pay = 9.33;
break;
case 'c':
base_pay = 10.00;
break;
case 'd':
base_pay = 11.20;
break;
default:
printf("Program error!\n");
break;
}
printf("Please enter the hour used: ");
scanf("%d", &hour); // 获取每周工作时数时没有像书上那样判断，我偷懒了Q！Q?/span>
if(hour <= WORK_OVERTIME)
{
total = hour * base_pay;
if (total <= BREAK1)
{
tax = total * RATE1;
net_pay = total - tax;
}
else
{
tax = BASE1 + (total - BREAK1) * RATE2;
net_pay = total - tax;
}
}
else
{
total = base_pay * WORK_OVERTIME + (hour - WORK_OVERTIME) * MULTIPLE * base_pay;
if(total <= BREAK2)
{
tax = BASE1 + (total - BREAK1) * RATE2;
net_pay = total - tax;
}
else
{
tax = BASE2 + (total - BREAK2) * RATE3;
net_pay = total - tax;
}
}
printf("The total pay: %.2f; tax: %.2f; net pay: %.2f\n", total, tax, net_pay);
}
printf("Bye!\n");
return 0;
}
char get_choice(void)
{
int ch;
printf("*****************************************************************\n");
printf("Enter number corresponding to the desired pay rate or action:\n");
printf("a) $8.75/hr\tb) $9.33/hr\n");
printf("c) $10.00/hr\td) $11.20/hr\n");
printf("q) quit\n");
printf("*****************************************************************\n");
printf("Please enter your choise: ");
ch = get_first();
while((ch < 'a' || ch > 'd') && ch != 'q')
{
printf("Please respond with a, b, c, d, or q.\n");
ch = get_first();
}
return ch;
}
char get_first(void)
{
int ch;
while((ch = getchar()) == '\n')
continue;
while(getchar() != '\n')
continue;
return ch;
}
#include <stdio.h>
char get_choice(void);
char get_first(void);
float get_float(void);
int main(void)
{
char choise;
float first_number, second_number;
while((choise = get_choice()) != 'q')
{
printf("Enter first number: ");
first_number = get_float();
printf("Enter second number: ");
second_number = get_float();
switch(choise)
{
case 'a':
printf("%.1f + %.1f = %.1f\n", first_number, second_number, first_number + second_number);
break;
case 's':
printf("%.1f - %.1f = %.1f\n", first_number, second_number, first_number - second_number);
break;
case 'm':
printf("%.1f * %.1f = %.1f\n", first_number, second_number, first_number * second_number);
break;
case 'd':
if(second_number == 0)
{
printf("Enter a number other than 0: ");
second_number = get_float();
printf("%.1f / %.1f = %.1f\n", first_number, second_number, first_number / second_number);
}
else
printf("%.1f / %.1f = %.1f\n", first_number, second_number, first_number / second_number);
break;
default:
printf("Program error!\n");
break;
}
}
printf("Bye.\n");
return 0;
}
char get_choice(void)
{
int ch;
printf("Enter the operation of your choice: \n");
printf("a. add\ts. subtract\n");
printf("m. multiply\td. divide\n");
printf("q. quit\n");
ch = get_first();
while(ch != 'a' && ch != 's' && ch != 'm' && ch != 'd' && ch != 'q')
{
printf("Please respond with a, s, m, d, or q.\n");
ch = get_first();
}
return ch;
}
char get_first(void)
{
int ch;
while((ch = getchar()) == '\n')
continue;
while(getchar() != '\n')
continue;
return ch;
}
float get_float(void)
{
float input;
char ch;
while((scanf("%f", &input)) != 1)
{
while((ch = getchar()) != '\n')
putchar(ch);
printf(" is not a number.\nPlease enter a ");
printf("number, such as 2.5, -1.78E8, or 3: ");
}
return input;
}
char get_choice(void);
char get_first(void);
float get_float(void);
int main(void)
{
char choise;
float first_number, second_number;
while((choise = get_choice()) != 'q')
{
printf("Enter first number: ");
first_number = get_float();
printf("Enter second number: ");
second_number = get_float();
switch(choise)
{
case 'a':
printf("%.1f + %.1f = %.1f\n", first_number, second_number, first_number + second_number);
break;
case 's':
printf("%.1f - %.1f = %.1f\n", first_number, second_number, first_number - second_number);
break;
case 'm':
printf("%.1f * %.1f = %.1f\n", first_number, second_number, first_number * second_number);
break;
case 'd':
if(second_number == 0)
{
printf("Enter a number other than 0: ");
second_number = get_float();
printf("%.1f / %.1f = %.1f\n", first_number, second_number, first_number / second_number);
}
else
printf("%.1f / %.1f = %.1f\n", first_number, second_number, first_number / second_number);
break;
default:
printf("Program error!\n");
break;
}
}
printf("Bye.\n");
return 0;
}
char get_choice(void)
{
int ch;
printf("Enter the operation of your choice: \n");
printf("a. add\ts. subtract\n");
printf("m. multiply\td. divide\n");
printf("q. quit\n");
ch = get_first();
while(ch != 'a' && ch != 's' && ch != 'm' && ch != 'd' && ch != 'q')
{
printf("Please respond with a, s, m, d, or q.\n");
ch = get_first();
}
return ch;
}
char get_first(void)
{
int ch;
while((ch = getchar()) == '\n')
continue;
while(getchar() != '\n')
continue;
return ch;
}
float get_float(void)
{
float input;
char ch;
while((scanf("%f", &input)) != 1)
{
while((ch = getchar()) != '\n')
putchar(ch);
printf(" is not a number.\nPlease enter a ");
printf("number, such as 2.5, -1.78E8, or 3: ");
}
return input;
}
]]>
]]>