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
Sample Output
/**************************************************************************** **秒钟的速度s=6°/s，分针是1/10°/s，时针是1/120°/s **所以相对速度s_m=59/10°/s，s_h=719/120°/s，m_h=120/11°/s **所以相差一度所需要的时间sm=10/59 s/°，sh=120/719 s/°，mh=120/11 s/° **他们差360°的周期为tsm=3600/59 s，tsh=43200/719 s，tmh=43200/11 s **需要相差的角度为n。 **rsm>n → n*sm+k1*tsm < t < tsm-n*sm+k1*tsm; **rsh>n → n*sh+k2*tsh < t < tsh-n*sh+k2*tsh; **rmh>n → n*mh+k3*tmh < t < tmh-n*mh+k3*tmh; **三个条件都满足所占的总时间即为时针、分针、秒针相差角度大于n的总时间 ** ****************************************************************************/#include<stdio.h>void main(){ int t; double n,sum,ft1,ft2,ft3,et1,et2,et3,max,min; double sm,sh,mh,tsm,tsh,tmh,fsm,fsh,fmh,esm,esh,emh; sm=10./59.; sh=120./719.; mh=120./11.; tsm=360*sm; tsh=360*sh; tmh=360*mh; while(scanf("%lf",&n)) { if(n<0) break; sum=0; fsm=sm*n; fsh=sh*n; fmh=mh*n; esm=tsm-fsm; esh=tsh-fsh; emh=tmh-fmh; for(ft3=fmh,et3=emh;et3<=43200;et3+=tmh,ft3+=tmh) { for(ft2=fsh,et2=esh;et2<=43200;et2+=tsh,ft2+=tsh) { if(et2<ft3) continue; if(ft2>et3) break; for(t=0,ft1=fsm,et1=esm;et1<=43200;t=t+1,et1=esm+t*tsm,ft1=fsm+t*tsm) { if(et1<ft3 || et1<ft2) continue; if(ft1>et3 || ft1>et2) break; max=ft1; if(ft2>max) max=ft2; if(ft3>max) max=ft3; min=et1; if(et2<min) min=et2; if(et3<min) min=et3; sum+=min-max; } } } sum/=432.; printf("%.3lfn",sum); }}
