C程序设计(第五版 谭浩强著)再学习第五章

while的框架如下图所示。
总结一句话就是先判断，再执行。

例题5-1

#include 
//1+2+3+...100
//本题要求使用while的方法。
int i = 1;
int sum = 0;
int main(void)
{
	while (i <= 100)
	{
		sum = sum + i;
		i = i + 1;
	}
	printf("1+2+3+4+5...+100 = %d", sum);
	return 0;
}

例题5-2
小知识点

#include 
int i = 1;
int sum = 0;
int main(void)
{
	do
	{
		sum = sum + i;
		i = i + 1;
	} while(i<=100);
	printf("1+2+3+...100 = %d", sum);
	return 0;
}

例题5-3

#include 
int main()
{
	int i, sum = 0;
	printf("please enter i,i=?");
	scanf("%d", &i);
	while (i <= 10)
	{
		sum = sum + i;
		i++;
	}
	printf("sum=%d\n", sum);
	return 0;
}

#include 
int main()
{
	int i, sum = 0;
	printf("please enter i,i=?");
	scanf("%d", &i);
	do
	{
		sum = sum + i;
		i++;
	} while (i <= 10);
	printf("sum=%d\n", sum);
	return 0;
}

注意:for循环基本结构

例题5-4

#include 
#define total_amount 10000
int sum = 0;
int amount = 0;
int i = 0;
double average = 0.0;
int main(void)
{
	for (i = 1; i <= 1000; i++)
	{
		printf("please enter amount:");
		scanf("%d", &amount);
		sum = sum + amount;
		if (sum >= 10000)
		{
			break;
		}
	}
	printf("sum = %d, i = %d, average = %f", sum, i, (double)sum/ (double)i);
	return 0;
}

例题5-5

#include 
int i = 0;
int main(void)
{
	for (i = 100; i <= 200; i++)
	{
		if (i % 3 == 0)
			continue;
		printf("%d  ", i);
	}
	return 0;
}

例题5-6

思路：这里可以用for循环的嵌套来实现。

#include 
int i = 0, j = 0;
int main(void)
{
	for (i = 1; i <= 4; i++)
	{
		for (j = 1; j <= 5; j++)
			printf("%d\t", i * j);
		printf("\n");
	}
	return 0;
}

例题5-7

#include 
double sum = 0.0;
int i = 1;
int sign = 1;
double pi = 0.0;
int main(void)
{
	while (1.0 / i >= 1e-6)
	{
		sum = sum + sign * (1.0 / i);
		sign = -sign;
		i = i + 2;
	}
	pi = sum * 4;
	printf("pi=%10.8f", pi);
	return 0;
}

例题5-8

#include 
int n = 0;
int sum = 0;
int f1 = 1,f2 = 1,f3 = 0;
//1 1 2 3 5 8
int main(void)
{
	printf("%9d\n%9d\n", f1, f2);
	for (n = 1; n <= 38; n++)
	{
		sum = f1 + f2;
		printf("%9d\n", sum);
		f1 = f2;
		f2 = sum;
	}
	return 0;
}

可以看出打印的过长，可以再将程序优化一下。

#include 
int n = 0;
int sum = 0;
int f1 = 1, f2 = 1, f3 = 0;
//1 1 2 3 5 8
int main(void)
{
	for (n = 1; n <= 20; n++)
	{
		printf("%12d%12d", f1,f2);
		f1 = f1 + f2;
		f2 = f1 + f2;
		if (n % 2 == 0)
			printf("\n");
	}
	return 0;
}

例题5-9

#include 
#include 
int n = 0, i = 0;
int main(void)
{
	printf("please type a number\n");
	scanf("%d", &n);
	for (i = 2; i <= n - 1; i++)
	{
		if (n % i == 0)
		{
			printf("this number is not prime");
			break;
		}
	}
	if(i == n)
		printf("this number is prime");
	return 0;
}

可以把判断进一步的缩小，这样程序可以进一步的优化。

#include 
#include 
int n = 0, i = 0;
int main(void)
{
	printf("please type a number\n");
	scanf("%d", &n);
	for (i = 2; i <= (int)sqrt(n); i++)
	{
		if (n % i == 0)
		{
			printf("this number is not prime");
			break;
		}
	}
	if (i == (int)sqrt(n) + 1)
		printf("this number is prime");
	return 0;
}

例题5-10

#include 
#include 
int n = 0, i = 0;
int number = 0;
int main(void)
{
	for (number = 100; number <= 200; number++)
	{
		for (i = 2; i <= (int)sqrt(number); i++)
		{
			if (number % i == 0)
			{
				//printf("this number is not prime\n");
				break;
			}
		}
	if (i == (int)sqrt(number) + 1)
		printf("%d is prime\n",number);
	}
	return 0;
}

例题5-11

#include 
char passward = 0;
int main(void)
{
	printf("please type passward\n");
	while ((passward = getchar())!= '\n')
	{
		if (('a' <= passward && passward <= 'z') || ('A' <= passward && passward <= 'Z'))
		{ 
			if(('a' <= passward && passward <= 'v')|| ('A' <= passward && passward <= 'V'))
				passward = passward + 4; 
			else                          
				passward = passward - 22;
		}
		else
		{ }
		printf("%c", passward);
	}
	return 0;
}

习题5-2

#include 
double sum = 0.0;
int i = 1;
int sign = 1;
double pi = 0.0;
int count = 0;
int main(void)
{
	while (1.0 / i >= 1e-6)
	{
		sum = sum + sign * (1.0 / i);
		sign = -sign;
		i = i + 2;
		count = count + 1;
	}
	pi = sum * 4;
	printf("pi=%10.8f\n", pi);
	printf("count = %d\n", count);
	return 0;
}

#include 
double sum = 0.0;
int i = 1;
int sign = 1;
double pi = 0.0;
int count = 0;
int main(void)
{
	while (1.0 / i >= 1e-8)
	{
		sum = sum + sign * (1.0 / i);
		sign = -sign;
		i = i + 2;
		count = count + 1;
	}
	pi = sum * 4;
	printf("pi=%10.8f\n", pi);
	printf("count = %d\n", count);
	return 0;
}

习题5-3

#include 
unsigned int m = 0, n = 0;
unsigned int max = 0, min = 0;
unsigned int i = 0;
unsigned int a1 = 0 , a2 = 0;
int main(void)
{
	printf("please type m value and n value\n");
	scanf("%d%d", &m, &n);
	//compare m value and n value
	if (m > n)
	{
		max = m;
		min = n;
	}
	else
	{
		max = n;
		min = m;
	}
	for (i = 2; i <= min; i++)
	{
		if (m % i == 0 && n % i == 0)
		{
			a1 = i; 
		}
	}
	for (i = max; i <= min*max; i++)
	{
		if (i % m == 0&& i % n == 0)
		{
			a2 = i;
			break;
		}
	}
	printf("a1=%d\n", a1);
	printf("a2=%d\n", a2);
	return 0;
}

最大公约数和最小公倍数计算器

借鉴答案

#include 
int p = 0, r = 0, m = 0, n = 0, temp = 0;
int main(void)
{
	printf("please type m value and n value:");
	scanf("%d%d", &n, &m);  
	if (n < m);//make sure n value > m value
	{
		temp = n;
		n = m;
		m = temp;
	}
	p = n * m;
	while (m != 0)
	{
		r= n % m ;
		n = m;
		m = r;
	}
	printf("最大公约数为:%d\n",n);
	printf("最小公倍数为:%d\n",p/n);
	return 0;
}

习题5-4

#include 
//count1 英语字符个数
//count2 空格字符个数
//count3 数字字符个数
//count4 其他字符个数
char a = 0;
int count1 = 0;
int count2 = 0;
int count3 = 0;
int count4 = 0;
int main(void)
{
	printf("please type your message:");
	while ((a= getchar()) != '\n')
	{
		if (('A' <= a && a <= 'Z') || ('a' <= a && a <= 'z'))
			count1 = count1 + 1;
		else if (a == ' ')
			count2 = count2 + 1;
		else if (('0' <= a && a <= '9'))
			count3 = count3 + 1;
		else 
			count4 = count4 + 1;
	}
	printf("count1 = %d,count2 = %d,count3 = %d,count4 = %d\n", count1, count2, count3, count4);
	return 0;
}

习题5-5

#include 
#include 
int n = 0;
int sum1 = 0;
int sum = 0;
int a = 0;

int main(void)
{
	printf("please type a value and n value:");
	scanf("%d%d",&a,&n);
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= i; j++)
		{
			sum1 = sum1 + a * (int)pow(10,j-1);
		}
		sum = sum + sum1;
		sum1 = 0;
	}
	printf("sum = %d", sum);
	return 0;
}

习题5-6

#include 
double sum1 = 1;
double sum = 0;
int a = 0;

int main(void)
{
	for (int i = 1; i <= 20; i++)
	{
		for (int j = 1; j <= i; j++)
		{
			sum1 = sum1*j;
		}
		sum = sum + sum1;
		sum1 = 1;
	}
	printf("sum = %22.15e\n", sum);
	return 0;
}

习题5-7

#include 
double sum1 = 0;
double sum2 = 0;
double sum3 = 0;
int main(void)
{
	for (int i = 1; i <= 100; i++)
	{
		if(1 <= i && i <= 100)
			sum1 = sum1 + i;
		if (1 <= i && i <= 50)
			sum2 = sum2 + i*i;
		if (1 <= i && i <= 10)
			sum3 = sum3 + 1.0/i;
	}
	printf("sum = %f", sum1 + sum2 + sum3);
	return 0;
}

习题5-8

#include 
#include 
int sum = 0;
int main(void)
{
	for (int i = 100; i <= 999; i++)
	{
		sum = (int)pow(i / 100, 3) + (int)pow(i / 10 % 10, 3) + (int)pow(i % 10, 3);
		if (sum == i)
			printf("i = %d\n", i);
	}
	return 0;
}

思来想去，使用下面的代码会更加稳妥。

#include 
#include 
int sum = 0;
int i = 0, j = 0, k = 0;
int main(void)
{
	for (int count = 100; count <= 999; count++)
	{
		i = count / 100;
		j = count / 10 % 10;
		k = count % 10;
		sum = i * i * i + j * j * j + k * k * k;
		if (sum == count)
			printf("count = %d\n", count);
	}
	return 0;
}

习题5-9

#include 
int sum = 0;
int i = 0, j = 0;
int main(void)
{
	for (i = 1; i <= 1000; i++)
	{
		sum = 0;
		for (j = 1; j < i; j++)
		{
			if (i % j == 0)
				sum = sum + j;
		}
		if (sum == i)
			printf("i = %d\n", i);
	}
	return 0;
}

习题5-10

#include 
double sum = 0;
double a1 = 1, b1 = 2;
double bottle = 0;
int main(void)
{
	for (int i = 1; i <= 20; i++)
	{
		sum = sum + b1 / a1;
		bottle = b1;
		b1 = a1 + b1;
		a1 = bottle;
	}
	printf("sum = %16.10f", sum);
	return 0;
}

习题5-11

#include 
double height = 100;
double sum = 0;
int main(void)
{
	for (int i = 1; i <= 10; i++)
	{
		if(i==1)
			sum = sum + height;
		else 
			sum = sum + 2*height;
		height = height * 0.5;
	}
	printf("sum = %f,height = %f\n", sum, height);
	return 0;
}

#include 
int sum = 1;
int main(void)
{
	for (int i = 1; i <= 9; i++)
	{
		sum = (sum + 1) * 2;
	}
	printf("sum = %d\n", sum);
	return 0;
}

习题5-13

#include 
#include 
double x0 = 0, x1 = 0;
double a = 0;
int main(void)
{
	printf("please type a value:");
	scanf("%lf", &a);
	x0 = a / 2;
	x1 = 0.5 * (x0 + a / x0);
	while (fabs(x1 - x0) >= 1e-5)
	{
		x0 = x1;
		x1 = 0.5 * (x0 + a / x0);
	}
	printf("the square root of %f is %f ", a, x1);
	return 0;
}

习题5-14
下面是牛顿迭代法的公式。注意，本题做出的前提是理解牛顿迭代法。

#include 
#include 
double x1=0, x0 = 0, f = 0, f1 = 0;
int main(void)
{
	x1 = 1.5;
	do
	{
		x0 = x1;
		f = 2 * x0 * x0 * x0 - 4 * x0 * x0 + 3 * x0 - 6;
		f1 = 6 * x0 * x0 - 8 * x0 + 3;
		x1 = x0 - f / f1;
	} while (fabs(x1 - x0) >= 1e-5);
	printf("the root of equation is %f\n", x1);
	return 0;
}

习题5-15 略~~

习题5-16

#include 
int n = 0;
int len = 0;
int main(void)
{
	printf("-------------------------\n");
	printf("\
   *   \n\
  ***  \n\
 ***** \n\
*******\n\
 ***** \n\
  ***  \n\
   *   \n");
	printf("-------------------------\n");
	printf("please type n value:"); //for example n = 7  len = 3
	scanf("%d", &n);                //for example n = 9  len = 4
	len = n / 2;
	for (int i = 1; i <= 2*len+1; i++)
	{
		if (i <= len)
		{
			for (int j = 1; j <= len - i + 1; j++)
				putchar(' ');
			for (int j = 1; j <= n - (len - i + 1) * 2; j++)
				putchar('*');
			for (int j = 1; j <= len - i + 1; j++)
				putchar(' ');
			putchar('\n');
		}
		else if (i == len + 1)
		{
			for (int j = 1; j <= n; j++)
				putchar('*');
			putchar('\n');
		}    
		else 
		{           
			for (int j = 1; j <= i-len-1; j++)
				putchar(' ');
			for (int j = 1; j <= n - (i - len - 1) * 2; j++)
				putchar('*');
			for (int j = 1; j <= i - len - 1; j++)
				putchar(' ');
			putchar('\n');
		}
	}
	return 0;
}

习题5-17 略~~