HDU 1006(时钟指针转角 **)

题意是说求出在一天中时针、分针、秒针之间距离均在 D 度以上的时间占比。

由于三针始终都在转动，所以要分别求出各个针之间的相对角速度，分别求出三针满足角度差的首次时间，再分别求出不满足角度差的最终时间，取这三个时间段的交集，也就是首次时间的最大值和最终时间的最小值之间的部分，要注意剪枝，去掉多余的部分，否则会超时的，

本人的代码也是借鉴了别人写出来的。代码如下：

#include <bits/stdc++.h>
using namespace std;
const double hm = 11.0/120;//时针分针相对角速度hm = m - h = 360/3600 - 360/(12*3600) = 11/120
const double hs = 719.0/120;//时针秒针相对角速度hs = s - h = 360/60 - 360/(12*3600) = 719/120
const double ms = 59.0/10;//分针秒针相对角速度ms = s - m = 360/60 - 360/3600 = 59/10
const double thm = 43200.0/11;//thm = 360/hm = 43200/11
const double ths = 43200.0/719;//ths = 360/hs = 43200/719
const double tms = 3600.0/59;//tms = 360/ms = 3600/59
const double eps = 1e-6;
double max(double a,double b,double c)
{
    if(a>b) return a>c?a:c;
    return b>c?b:c;
}
double min(double a,double b,double c)
{
    if(a<b) return a<c?a:c;
    return b<c?b:c;
}
int main()
{
    int deg;
    double from1,from2,from3,to1,to2,to3,x1,x2,x3,y1,y2,y3,st,ed,ans;
    while(scanf("%d",&deg))
    {
        if(deg==-1) break;
        from1 = deg/hm;
        from2 = deg/hs;
        from3 = deg/ms;
        to1 = (360-deg)/hm;
        to2 = (360-deg)/hs;
        to3 = (360-deg)/ms;
        ans = 0.0;
        for(x1 = from1, y1 = to1; y1 <= 43200+eps; x1+=thm, y1+=thm)
            for(x2 = from2, y2 = to2; y2 <= 43200+eps; x2+=ths, y2+=ths)
            {
                if(x2>y1) break;
                for(x3 = from3, y3 = to3; y3 <= 43200+eps; x3+=tms, y3+=tms)
                {
                    if(x3>y1 || x3>y2) break;
                    st = max(x1,x2,x3);
                    ed = min(y1,y2,y3);
                    if(st<ed) ans+=ed-st;
                }
            }
        printf("%.3lf
",ans*100.0/43200);//求出时间和的占比
    }
    return 0;
}