C语言程序设计现代方法第六章课后习题

第一题

#include 
#include
int main(void)
{
	float a, b = 0;
	do
	{
		printf("Enter a number:");
		scanf_s("%f", &a);
		b = (a > b)? a:b;

	} while (a > 0);
	printf("The lagest number entered was %f\n", b);
	system("pause");
	return 0;
}

第二题
方法一
辗转相除法（又名欧几里德法）
设两数为a,b设其中a 做被除数,b做除数，temp为余数
1、大数放a中、小数放b中；
2、求a/b的余数；
3、若temp=0则b为最大公约数；
4、如果temp!=0则把b的值给a、temp的值给a；

#include 
#include
int main(void)
{
	int a, b, temp = 1,c;
	printf("Enter two numbers:");
	scanf_s("%d %d", &a, &b);
	if (b > a)
	{
		c = b;
		b = a;
		a = c;
	}
	while (temp != 0)
	{ 
	temp = a%b;
	if (temp == 0)
		printf("GREATEST COMMON DIVISOR:%d\n", b);
	else
	{
		a = b;
		b = temp;
	}
	}
	system("pause");
	return 0;
}

方法2 穷举法
从1到两数中的较小值，找到所有两数的约数，每次计算较大约数会覆盖上次较小的约数，这样最后输出的就是最大公约数

#include 
#include
int main(void)
{
	int a, b, i,c;
	printf("Enter two numbers:");
	scanf_s("%d %d", &a, &b);
	if (b > a)
	{
		c = b;
		b = a;
		a = c;
	}
	for ( i = 1; i <= b; i++)
	{
		if (a%i == 0)
		{
			if (b%i == 0)
				c = i;
		}
	}
	printf("Greatesy common divisor:%d", c);
	system("pause");
	return 0;
}

还有好多方法可以参考这个https://blog.csdn.net/Brilliance_panpan/article/details/88372432
第三题

#include 
#include
int main(void)
{
	int a, b, i,c,aa,bb;
	printf("Enter a franction:");
	scanf_s("%d/%d", &a, &b);
	aa = a, bb = b;
	if (b > a)
	{
		c = b;
		b = a;
		a = c;
	}
	for (i = 1; i <= b; i++)
	{
		if (a%i == 0)
		{
			if (b%i == 0)
				c = i;
		}
	}
	printf("The lowest terms: %d/%d",(aa/c),(bb/c));
	system("pause");
	return 0;
}

第四题（在5…3的基础上做的）

#include 
#include)
int main(void)
{
	float commission, value = 1, number, price, commissionA;
	while (value != 0)
	{
		printf("Enter number of trade: ");
		scanf_s("%f", &number);
		printf("Enter price of trade: ");
		scanf_s("%f", &price);
		value = price * number;
		if (number < 2000.00f)
			commissionA = (number * 33 + 3) / 100;
		else
			commissionA = (number * 33 + 2) / 100;
		if (value < 2500.f)
			commission = 30.00f + .017f * value;
		else if (value < 6250.00f)
			commission = 56.00f + .0066f * value;
		else if (value < 20000.00f)
			commission = 76.00f + .0034f * value;
		else if (value < 50000.00f)
			commission = 100.00f + .0022f * value;
		else if (value < 500000.00f)
			commission = 155.00f + .0011f * value;
		else
			commission = 255.00f + .0009f * value;
		if (commission < 39)
			commission = 39.00f;
		printf("Commission: $%.2f\n", commission);
		printf("CommissionA: $%.2f\n", commissionA);
	}
	system("pause");
	return 0;
}

第五题

#include 
#include)
int main(void)
{
	int a,b,c;
	printf("Enre the number:");
	scanf_s("%d",&a);
	do
	{
		b = a % 10;
		a = a / 10;
		printf("%d", b);
	} while (a > 0);
	printf("\n");
	system("pause");
	return 0;
}

第六题

#include 
#include
#include 
int main(void)
{
	int a,b;
	printf("Enter a number:");
	scanf_s("%d",&a);
	b = sqrt(a);
	for (int i = 2; i <= b;)
	{
		printf("%d\n", i*i);
		i += 2;
	}
	system("pause");
	return 0;
}

第七题
好像是改下for循环的控制命令就可以了
第八题

#include 
#include
#include 
int main(void)
{
	int a, b, i,j;
	printf("Enter the number of days in mounth:");
	scanf_s("%d", &a);
	printf("Enter the staring day of the week:");
	scanf_s("%d", &b);
	for (i = 0; i < ((b % 7)-1); i++)
		printf("\t");
	for (j = 1; j <= a; j++)
	{
		if ((i + j) % 7 != 0)
			printf("%d\t", j);
		else
			printf("%d\n", j);
	}
	printf("\n");
	system("pause");
	return 0;
}

第九题
第二章的没做。所以。。。
第十题

#include 
#include
int main(void)
{
	int y, m, d ;
	int y1 = 100, m1 = 100, d1 = 100, i = 1;
	printf("Enter a data(mm/dd/yy)：");
	scanf_s("%d/%d/%d", &m, &d, &y);
	while (y + m + d != 0)
	{
		if ((y * 365 + m * 30 + d) < (y1 * 365 + m1 * 30 + d1))
		{
			y1 = y; 
			m1 = m;
			d1 = d;
		}
		printf("Enter a data(mm/dd/yy)：");
		scanf_s("%d/%d/%d", &m, &d, &y);
	}

	printf("%d/%d/%d is the earlest date\n",m1,d1,y1);
	system("pause");
	return 0;
}

第十一题

#include 
#include
int main(void)
{
	int n,i;
	float e=1, j = 1;
	printf("Enter a number:");
	scanf_s("%d", &n);
	for (i = 1; i <= n; i++)
	{   
		j = j * i;
		e = e + 1 / j;
	}
	printf("The E is %7.6f", e);
	system("pause");
	return 0;
}

第十二题

#include 
#include
#include
int main(void)
{
	int i;
	float e,n,num=1, j = 1;
	printf("Enter the number:");
	scanf_s("%d", &n);
	e = exp(1);
	for (i = 1; abs(num-e)>n; i++)
	{   
		j = j * i;
		num = num + 1 / j;
	}
	printf("The E is %.10f\n", num);
	system("pause");
	return 0;
}

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