Tick and Tick

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20120 Accepted Submission(s): 5262

Problem Description

The three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. You are to calculate how much time in a day that all the hands are happy.

Input

The input contains many test cases. Each of them has a single line with a real number D between 0 and 120, inclusively. The input is terminated with a D of -1.

Output

For each D, print in a single line the percentage of time in a day that all of the hands are happy, accurate up to 3 decimal places.

Sample Input

0
120
90
-1

Sample Output

100.000
0.000
6.251

Author

PAN, Minghao

Source

ZJCPC2004

分析：

三个指针走的角速度：

秒针速度S = 6°/s，分针速度M = (1/10)°/s，时针速度H = (1/120)°/s

这三个指针两两之间的相对速度差为：

秒时相差S_H = (719/120)°/s，秒分相差S_M = (59/10)°/s，分时相差M_H = (120/11)°/s

相差一度需要的时间为

秒时相差SH = (120/719)s/度，秒分相差SM = (10/59)s/度，分时相差MH = (120/11)s/度

相差360°需要的时间为

秒时相差tSH = 43200.0/719，秒分相差tSM = 3600.0/59，分时相差tMH = 43200.0/11

算出两两指针在43200s（12小时）内满足条件时间的区间的交集。

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>

using namespace std;
///秒针角速度 6度/s, 分针0.1度/s,时针1/120度/s
const double sh = 719.0/120 , sm = 59.0/10, mh = 11.0/120;///三个指针两两相对角速度。
const double tsh = 43200.0 / 719, tsm = 3600.0 / 59, tmh = 43200.0 / 11;///三个指针两两相差360度所需的时间。
///将相对角速度变成周期。（即两针间需要多久出现夹角的循环）
/// 同样可求得三个周期的最小公倍数为 43200 秒，即12小时，

double max1(double a, double b, double c)
{
return max(a, max(b, c));
}
double min1(double a, double b, double c)
{
return min(a, min(b, c));
}
int main(void)
{
int D;

while(scanf("%d", &D), D != -1)
{
double bsh, bsm, bmh, esh, esm, emh, total, beginn, endd;
///第一次满足条件的时间。(开始时间）
bsh = D / sh;
bsm = D / sm;
bmh = D / mh;
///第一次出现不满足条件的时间。（结束时间）
esh = (360 - D) / sh;
esm = (360 - D) / sm;
emh = (360 - D) / mh;

total = 0;

for(double bt1 = bsh, et1 = esh; et1<= 43200.000001; bt1 += tsh, et1 += tsh)
{
for(double bt2 = bsm, et2 = esm; et2 <= 43200.000001; bt2 += tsm, et2 += tsm)
{
///判断是否有交集。
if(bt2 > et1)
break;

if(et2 < bt1)
continue;

for(double bt3 = bmh, et3 = emh; et3 <= 43200.000001; bt3 += tmh, et3 += tmh)
{
if(bt3 > et1 || bt3 > et2)
break;

if(et3 < bt1 || et3 < bt2)
continue;

beginn = max1(bt1, bt2, bt3);
endd = min1(et1, et2, et3);

total += (endd - beginn);
}

}
}

printf("%.3f\n", total / 432);

}
return 0;
}

巴特西

HDU 1006 模拟

Tick and Tick

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