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

本人实在太菜：

参考自该博主，顺便附上链接膜一波：https://blog.csdn.net/yuyanggo/article/details/45765811

解析：这道题，其实想通了也不难，甚至可以说太简单了。

           时钟的时针、分针、秒针一天内的三次重合点：0：00  12:00   24:00

           所以计算这道题的是时候，只需要考虑0:00到12:00即可。

设：ws     表示秒针的角速度

                   wm    表示分针的角速度

                   wh     表示时针的角速度

                

                  wsm=ws-wm   表示秒针相对于分针的角速度 

                  wsh=ws-wh     表示秒针相对于时针的角速度

                  wmh=wm-wh   表示分针相对于时针的角速度

                  

            那么就有：(t1，t2，t3<=12*60*60)

360*(i-1)+d<=wsm*t1<=360*i-d

                  360*(j-1)+d<=wsh*t2<=360*j-d

                  360*(k-1)+d<=wmh*t3<=360*k-d

                

            t1、t2、t3可取时间的交集即为答案。

           

           具体实现如下：

              用 sum 来记录happy time，以秒为单位。

1.枚举 i ，得到 t1  与 i 对应的左右界：L、R

                若 L>=12*60*60 退出循环。

2.枚举 j ，得到 t2  与 j 对应的左右界：L2、R2。

                此时应满足[L2，R2]∩[L，R] ≠ 空集。

3.枚举 k ，得到t3 与 k 对应的左右界：L3、R3

                 此时应满足[L3，R3]∩[L2，R2] ≠ 空集，则 [L3，R3]能使上述三个不等式均成立，为happy time。

                 sum+=R3-L3。

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;

double sum,d,wsm,wsh,wmh;

int main() {
    wsm=59/10.0,wsh=719/120.0,wmh=11/120.0;
    double l,r,l2,r2,l3,r3;
    int i,j,k;
    while(scanf("%lf",&d),d>-0.5) {
        for(sum=0,i=1;; i++) {
            l=(360*i-360+d)/wsm;
            r=(360*i-d)/wsm;
            if(l>12*60*60)break;

            j=(int)((l*wsh+d)/360);
            if((360*j-d)/wsh<=l)j++;
            for(;; j++) {
                l2=(360*j-360+d)/wsh;
                r2=(360*j-d)/wsh;
                if(l2<l)l2=l;
                if(r2>r)r2=r;
                if(l2>=r2)break;
                k=(int)((l2*wmh+d)/360);
                if((360*k-d)/wmh<=l2)k++;
                for(;; k++) {
                    l3=(360*k-360+d)/wmh;
                    r3=(360*k-d)/wmh;
                    if(l3<l2)l3=l2;
                    if(r3>r2)r3=r2;
                    if(l3>=r3)break;
                    sum+=r3-l3;
                }
            }
        }
        printf("%.3lf\n",sum/(12*60*60)*100);
    }
    return 0;
}