acm--1006

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

//用角速度考虑解决问题

#include <cstdio>
#include <math.h>
#include <iostream>
#include <algorithm>

#define V_SEC 6.0 //秒针角速度
#define V_MIN 0.1 //分针角速度
#define V_HOU 1.0/120 //时针角速度

#define A_SEC s*6 //秒针角度
#define A_MIN m*6+s*0.1 //分针角度
#define A_HOU h*30+m*0.5+s/120.0 //时针角度

using namespace std;
struct interval { //区间
double l; //left
double r; //right
};
double Angle; //角度
int s=0; //秒数

interval solve(double v,double a) { //解方程
//Angle<=v*t+a<=360-Angle;,并且和[0,60]取交集
interval p;
if(v>0) {
p.l=(Angle-a)/v;
p.r=(360-Angle-a)/v;
} else {
p.l=(360-Angle-a)/v;
p.r=(Angle-a)/v;
}
if(p.l< 0)p.l= 0;
if(p.r>60)p.r=60;
if(p.l>=p.r)p.l=p.r=0;
return p;
}

interval jiao(interval a,interval b) {
interval p;
p.l=max(a.l,b.l);
p.r=min(a.r,b.r);
if(p.l>=p.r)p.l=p.r=0;
return p;
}

double happytime(int h,int m) { //计算h时m分 满足题意的秒数
double v_diff;//速度差
double a_diff;//角度差
interval s0[3][2];
interval s1;
/*解方程 Angle<=|hh-mm|<=360-Angle*/
v_diff=V_HOU-V_MIN;
a_diff=A_HOU-A_MIN;//时针分针夹角
s0[0][0]=solve( v_diff, a_diff);
s0[0][1]=solve(-v_diff,-a_diff);
/*解方程 Angle<=|hh-ss|<=360-Angle*/
v_diff=V_HOU-V_SEC;
a_diff=A_HOU-A_SEC;//时针秒针夹角
s0[1][0]=solve( v_diff, a_diff);
s0[1][1]=solve(-v_diff,-a_diff);
/*解方程 Angle<=|mm-ss|<=360-Angle*/
v_diff=V_MIN-V_SEC;
a_diff=A_MIN-A_SEC;//分针秒针夹角
s0[2][0]=solve( v_diff, a_diff);
s0[2][1]=solve(-v_diff,-a_diff);
/*
六个区间，选三个取交集
因为绝对值的式子得到的两个区间要并，而三个不同表达式
的区间要交，故这样做
*/
double res=0;
for(int i=0; i<2; i++)
for(int j=0; j<2; j++)
for(int k=0; k<2; k++) {
s1=jiao(jiao(s0[0][i],s0[1][j]),s0[2][k]);
res+=s1.r-s1.l;
}
return res;
}

int main() {
int h,m;
while(scanf("%lf",&Angle)) {
if(Angle==-1)break;
double res=0;
for(h=0; h<12; h++)
for(m=0; m<60; m++)
res+=happytime(h,m);
printf("%.3lf\n",res*100.0/43200);
}
}

//没有找到恰当的方法解决问题，思考角度不够