清览题库--C语言程序设计第五版编程题解析（1）

#include <stdio.h>
int main()
{
	int h, m, n, p;
	double z;//最后输出要求是浮点数，提前double
	scanf_s("%d:%d", &h, &m);//正常情况下应用scanf,但VS2022本身不承认scanf，只承认scanf_s,据说是安全性的问题
	p = m * 6;//分钟直接用输入的数字×每分钟对应的角度。
	n = h * 5 * 6 + p * 30 / 360;//时针×每个大空隙每个大空隙中的小空隙对应的角度+通过比例求出的角度
	z = 360 - n + p;//求角度差
	printf("%.2lf", z);//输出时要求保留小数点后两位，通过.2实现
	return 0;
}

#include <stdio.h>
#include <math.h>
int main()
{
	int	h, m;
	double	a;
	scanf( "%d:%d", &h, &m );
	if ( h >= 0 && h <= 12 && m >= 0 && m <= 59 )
	{
		double a1, a2; /* 时针和分针以0点为起点走过的角度 */
		a2	= m * 1.0 / 60 * 360;
		a1	= h * 1.0 / 12 * 360 + m * 1.0 / 60 * 30;
		a	= fabs( a2 - a1 );
		a	= a > 180 ? 360 - a : a;
		printf( "%.2lf", a );
	}else
		printf( "input error" );
	return(0);
}

#include <stdio.h>
int main() {
  double r,h;
  double z = 3.14;
  scanf("%lf %lf",&r,&h);
  printf("圆周长:%0.2lf\n",2*z*r);
  printf("圆面积:%0.2lf\n",z*r*r);
  printf("圆球表面积:%0.2lf\n",4*z*r*r);
  printf("圆球体积:%0.2lf\n",(4*z*r*r*r)/3);
  printf("圆柱体积:%0.2lf\n",z*r*r*h);
}

#include <stdio.h>
#define PI 3.14
int main()
{
	double r, h, l, s, ss, qv, v;
	scanf( "%lf%lf", &r, &h );
	if ( r > 0 && h > 0 )
	{
		l	= 2 * PI * r;
		s	= PI * r * r;
		ss	= 4.0 * PI * r * r;
		qv	= 4.0 / 3 * PI * r * r * r;
		v	= PI * r * r * h;
		printf( "圆周长:%.2lf\n圆面积:%.2lf\n圆球表面积:%.2lf\n圆球体积:%.2lf\n圆柱体积:%.2lf\n", l, s, ss, qv, v );
	}else
		printf( "input error" );
	return(0);
}