Problem - 1006_Tick and Tick

Tick and Tick

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10255    Accepted Submission(s): 2879

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

012090-1

 

Sample Output

100.0000.0006.251

 

Author

PAN, Minghao

 

Source

ZJCPC2004

 

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#include<stdio.h>typedef struct{    double left,right;}Section;double d;Section s[3][2];Section Intersection(Section a,Section b){    Section s;    s.left=a.left>b.left?a.left:b.left;    s.right=a.right<b.right?a.right:b.right;    if(s.left>s.right){        s.left=s.right;    }    return s;}Section setSection(double x,double y){    Section s;    /**     *  d<=x*s+y<=360-d(s∈[0,59],s∈N)     *  1.x<0  d-y<=x*s<=360-d-y     *         (360-d-y)/x<=s<=(d-y)/x     *  2.x=0  d<=y<=360-d     *         d-y<=0<=360-d-y     *         d<=180(题设d∈[0,120]【情况可忽略】)     *  3.x>0  d-y<=x*s<=360-d-y     *         (d-y)/x<=s<=(360-d-y)/x     */    if(x<0){        s.left=(360-d-y)/x;        s.right=(d-y)/x;    }    else{        s.left=(d-y)/x;        s.right=(360-d-y)/x;    }    if(s.left<0){        s.left=0;    }    if(s.right>60){        s.right=60;    }    if(s.left>s.right){        s.left=s.right;    }    return s;}double getHappyTime(int x,int y){    double a,b,sum;    Section S;    int i,j,k;    /**     *  d<=|h-m|<=360-d     *  h=30*x+y/2+S/120  m=6*y+S/10     *  1.h-m>=0  => a=S/120-S/10       => b=30.0*x+y/2.0-6.0*y     *                =-11.0/120            =30.0*x-5.5*y     *  2.m-h>=0  => a'=S/10-S/120      => b'=6.0*y-(30.0*x+y/2.0)     *                 =11.0/120=-a          =-30.0*x+5.5*y=-b     */    a=-11.0/120;    b=30.0*x-5.5*y;    s[0][0]=setSection(a,b);    s[0][1]=setSection(-a,-b);    /**     *  d<=|h-s|<=360-d     *  h=30*x+y/2+S/120  s=6*S     *  1.h-s>=0  => a=S/120-6*S        => b=30.0*x+y/2.0     *                =-719.0/120           =30.0*x+0.5*y     *  2.s-h>=0  => a'=6*S-S/120       => b'=-(30.0*x+y/2.0)     *                 =719.0/120=-a         =-30.0*x-0.5*y=-b     */    a=-719.0/120;    b=30.0*x+y/2.0;    s[1][0]=setSection(a,b);    s[1][1]=setSection(-a,-b);    /**     *  d<=|m-s|<=360-d     *  m=6*y+S/10  s=6*S     *  1.m-s>=0  => a=S/10-6*S         => b=6*y     *                =-5.9     *  2.s-m>=0  => a'=6*S-S/10        => b'=-6*y     *                 =5.9=-a               =-b     */    a=-5.9;    b=6*y;    s[2][0]=setSection(a,b);    s[2][1]=setSection(-a,-b);        for(sum=i=0;i<2;i++){                for(j=0;j<2;j++){                        for(k=0;k<2;k++){                S=Intersection(Intersection(s[0][i],s[1][j]),s[2][k]);                sum+=S.right-S.left;            }                    }            }        return sum;}int main(){    double sum;    int hr,min;    while(scanf("%lf",&d)!=EOF&&d!=-1){        for(sum=hr=0;hr<12;hr++){            for(min=0;min<60;min++){                sum+=getHappyTime(hr,min);            }        }        printf("%.3f\n",100.0*sum/(12*60*60));    }    return 0;}