一道C语言编程比赛题（时钟追赶问题）

问题描述：

大概的意思是：任意给一个角度，求出时钟上的秒分时指针之间的角度大于等于这个角度的概率。

解决的方案如下：

/*************************************************************************
	> File Name: clock.c
	> Author: Baniel Gao
	> Mail: createchance@163.com 
	> Blog: blog.csdn.net/createchance 
	> Created Time: Tue 08 Apr 2014 05:23:27 PM CST
 ************************************************************************/
#include <stdio.h>

#define HALF_DAY	43200
#define SEC_RATE	6
#define MIN_RATE	(1.0/10)
#define HOUR_RATE	(1.0/120)

void happy_times(double degrees);
static inline double fabs(double val);
static inline double degree_of_circle(double degree);

int main(void)
{
	double degrees;

	printf("Please input a degree: ");
	scanf("%lf", °rees);
	if (degrees > 120 || degrees < 0) {
		printf("The degree must be a number between 0 and 120! \n");
		return -1;
	}
	happy_times(degrees);

	return 0;
}

void happy_times(double degrees)
{
	int sec = 0;
	double gaps[3];
	int happy_times = 0;

	while (sec <= HALF_DAY) {
		if (fabs((SEC_RATE * sec) % 360 - degree_of_circle(MIN_RATE * sec)) <= 180)
			gaps[0] = fabs((SEC_RATE * sec) % 360 - degree_of_circle(MIN_RATE * sec));
		else
			gaps[0] = 360 - fabs((SEC_RATE * sec) % 360 - degree_of_circle(MIN_RATE * sec));

		if (fabs(degree_of_circle(MIN_RATE * sec) - HOUR_RATE * sec) <= 180)
			gaps[1] = fabs(degree_of_circle(MIN_RATE * sec) - HOUR_RATE * sec);
		else
			gaps[1] = 360 - fabs(degree_of_circle(MIN_RATE * sec) - HOUR_RATE * sec);

		if (fabs((SEC_RATE * sec) % 360 - HOUR_RATE * sec) <= 180)
			gaps[2] = fabs((SEC_RATE * sec) % 360 - HOUR_RATE * sec);
		else
			gaps[2] = 360 - fabs((SEC_RATE * sec) % 360 - HOUR_RATE * sec);

		if (gaps[0] >= degrees && gaps[1] >= degrees && gaps[2] >= degrees)
			happy_times++;
		sec++;
	}
	printf("Happy times: %%%.3f \n", happy_times * 100.0 / (sec - 1));
}

static inline double fabs(double val)
{
	if (val > 0.0)
		return val;
	else
		return (0 - val);
}

static inline double degree_of_circle(double degree)
{
	if (degree < 360.0)
		return degree;
	else
		return (degree - (int)(degree / 360.0) * 360);
}

定义一个秒的变量，用来从0自增到43200（12小时，因为每12小时三个指针合并一次），判断每次是否符合问题描述的happy times，符合的话happy times加1，否则不操作。最后用happy times除以总秒数。

一道C语言编程比赛题（时钟追赶问题）,布布扣,bubuko.com

一道C语言编程比赛题（时钟追赶问题）

原文：http://blog.csdn.net/createchance/article/details/23686113