杭电ACM HDU 1006 Tick and Tick
Tick and Tick
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6204 Accepted Submission(s): 1687
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
Recommend
JGShining
首先想到的是一秒一秒的试。
/**
* 采用离散的点，精度无法满足
*/
#include <cstdio>
int is(double h,double m,double s,int d){
double hs,hm,ms;
hs=h>s?h-s:s-h;
hm=h>m?h-m:m-h;
ms=m>s?m-s:s-m;
if(hs>180)hs=360-hs;
if(hm>180)hm=360-hm;
if(ms>180)ms=360-ms;
if(ms>=d&&hs>=d&&hm>=d)return 1;
return 0;
}
int main(){
int d,c,t;
double h,m,s;
while(scanf("%d",&d),~d){
t=c=h=m=s=0;
for(;t<3600*12;t++){
s+=6;
m+=1.0/10;
h+=1.0/120;
if(s>=360)s-=360;
if(m>=360)m-=360;
if(h>=360)h-=360;
if(is(h,m,s,d))c++;
}
printf("%.3f\n",c*2/864.00);
}
}
做了好久无果，参考大神的代码，原来用角速度，然后暴力搜索720min。
/**
* @date 2013-2
* @name Tick and Tick
* 用角速度
*/
#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;
}
/*
hh=30*h+m/2+s/120
mm=6*m+s*0.1
ss=6*s
Angle<=|hh-mm|<=360-Angle
Angle<=|hh-ss|<=360-Angle
Angle<=|mm-ss|<=360-Angle
*/
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);
}
}