Tick and Tick

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22203 Accepted Submission(s): 5877

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

题意：时针，分针和秒针都厌倦了其余两针，只有在与其余两针保持n度以上的距离才能感到高兴。问一天中，三针都高兴的时间占总时间的百分比。注意时间是连续的。

分析：

时间是连续的，不能枚举。。我一开始就去枚举。。。

推荐一个大神的题解：（我的太垃圾）

https://blog.csdn.net/a601025382s/article/details/37814777（看了大神博客就懂了）

不要脸的贴一下代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define max_v 10000
int 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("%.3lf
",sum);
    }
    return 0;
}