題意：給出一個角度，然後求一天中時針、分針、秒針互相的夾角都大於等於這個夾角的概率。。

思路：開始的思路就是列舉，但是由於精度問題，果斷TLE、、、 

膜拜網上各種解法啊，主要就是個數學問題、、但是真心不會啊，這道題學到一些，處理時鐘問題的方法吧，還有區間求交集什麼的。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
inline double Max(double a , double b , double c) {
    double tmp = a > b ? a : b;
    tmp = tmp > c ? tmp : c;
    return tmp;    
}
inline double Min(double a , double b , double c) {
    double tmp = a < b ? a : b;
    tmp = tmp < c ? tmp : c;
    return tmp;    
}
int main() {
    double ss , mm , hh , sm , mh , sh , t_sm , t_mh , t_sh;
    ss = 6.0 , mm = 0.1 , hh = 0.1/12.0;
    sm = 6.0 - 0.1;
    mh = 0.1 - 0.1/12.0;
    sh = 6.0 - 0.1/12.0;//相對速度 
    t_sm = 360.0/sm;
    t_mh = 360.0/mh;
    t_sh = 360.0/sh;//相對週期
    int D;
    double m[3] , n[3] , x[3] , y[3];
    while(~scanf("%d",&D)) {
        if(D == -1) break;
        x[0] = D/sm;
        x[1] = D/mh;
        x[2] = D/sh;
        y[0] = (360.0-D)/sm;
        y[1] = (360.0-D)/mh;
        y[2] = (360.0-D)/sh;
        double st , ed;
        double ans = 0;
        for(m[0] = x[0] , n[0] = y[0] ; m[0]<=43200 ; m[0] += t_sm , n[0] += t_sm) { 
           for(m[1] = x[1] , n[1] = y[1] ; m[1] <= 43200 ; m[1] += t_mh , n[1] += t_mh) {
                if(m[0] > n[1]) continue;
                if(n[0] < m[1]) break;
                for(m[2] = x[2] , n[2] = y[2] ; m[2] <= 43200 ; m[2] += t_sh , n[2] += t_sh) {
                    if(n[2] < m[1] || n[2] < m[0]) continue;
                    if(m[2] > n[0] || m[2] > n[1]) break;
                    st = Max(m[0] , m[1] , m[2]);
                    ed = Min(n[0] , n[1] , n[2]);
                    if(ed > st) ans += ed - st;
                }
            }     
        }         /*取交集 記錄區間的過程*/
        printf("%.3lf\n",ans/432.0);
    } 
}