//1000 A + B Problem
//1001 Sum Problem
//1002 A + B Problem II
//2000 ASCII码排序
//2001 计算两点间的距离
//C++中精度控制
//1003 Max Sum
//1004 Let the Balloon Rise
//1005 Number Sequence
//1006 Tick and Tick
//
//
//1006 Tick and Tick//BUG版本
#include <iostream>
#include <algorithm>
#include<iomanip>
using namespace std;
int main()
{
double D = 0;
while(cin>>D,D!=-1)
{
double d = D/360.0;
int count = 0;
int hour = 0 , minute = 0 , second = 0;
double h = 0,m = 0,s = 0;
double a,b,c;
int i = 0;
for(i=0; i<3600; ++i)
{
s = second/60.0 ;
m = minute/60.0 + s / 60.0 ;
h = h/60.0 + m/60.0;
second++;
if(second == 60)
{
second = 0;
minute++;
if(minute == 60)
{
minute = 0;
hour++;
}
}
a = s ;
b = m ;
c = h ;
if(a > b)
swap(a,b);
if(b > c)
swap(b,c);
if(a > b)
swap(a,b);
double d1 = b - a;
double d2 = c - b;
double d3 = 1 - (d1 + d2);
if(i%100 == 0)
cout<<fixed<<setprecision(4)<<d<<" "<<d1<<" "<<d2<<" "<<d3<<" "<<count<<" "<<a<<" "<<b<<" "<<c<<endl;
if(d1 >= d && d2 >= d && d3 >= d)
count++;
//if(i%30 == 0)
//{
// cout<<a<<" "<<b<<" "<<c<<" "<<endl;
// cout<<h*360<<" "<<m*360<<" "<<s*360<<endl;
// cout<<hour<<" "<<minute<<" "<<second<<endl<<endl;
//}
}
cout<<count<<endl;
double result = count/3600.0 * 100;
cout<<fixed<<setprecision(10)<<result<<endl;
}
return 0;
}
/*
//网上正确的算法
#include<stdio.h>
#include<stdlib.h>
#include<string>
double num[14];
int hash[13];
int cmp(const void *a,const void *b)
{
return *(double *)a > *(double *)b ? 1 : -1;
}
void hh(double a ,double b)
{
int i;
double start,end;
start = 0;
end = a;
for(i=0;i<13;i++)
if(start <= num[i] && num[i+1] <= end)
hash[i] ++;
start = b;
end = 60;
for(i=0;i<13;i++)
if(start <= num[i] && num[i+1] <= end)
hash[i] ++;
}
void hhh(double a,double b)
{
int i;
if(b<0 || a>60)
return ;
for(i=0;i<13;i++)
if(a <= num[i] && num[i+1] <= b)
hash[i] ++;
}
double happytime(double ms,double hs,double a)
{
//a <= | 6*t-(t+ms)/10 | <= 360-a
//a <= | 6*t-(t+hs)/120 | <= 360-a
//a <= | (t+ms)/10-(t+hs)/120 | <= 360-a
//计算同时满足上边三个条件的t
double aa,ab,ac,ad,ba,bb,bc,bd,ca,cb,cc,cd,sum;
aa = (ms-a*10)/59;
ab = (ms+a*10)/59;
ac = (ms+10*a-3600)/59;
ad = (ms+3600-10*a)/59;
ba = (hs-12*ms-a*120)/11;
bb = (hs-12*ms+a*120)/11;
bc = (hs-12*ms+120*a-43200)/11;
bd = (hs-12*ms+43200-120*a)/11;
ca = (hs-a*120)/719;
cb = (hs+a*120)/719;
cc = (hs+120*a-43200)/719;
cd = (hs+43200-120*a)/719;
num[0] = 0;
num[1] = 60;
num[2] = aa;
num[3] = ab;
num[4] = ac;
num[5] = ad;
num[6] = ba;
num[7] = bb;
num[8] = bc;
num[9] = bd;
num[10] = ca;
num[11] = cb;
num[12] = cc;
num[13] = cd;
qsort(num,14,sizeof(num[0]),cmp);
memset(hash,0,sizeof(hash));
sum = 0;
hh(aa,ab);
hh(ba,bb);
hh(ca,cb);
hhh(ac,ad);
hhh(bc,bd);
hhh(cc,cd);
for(int i=0;i<13;i++)
if(hash[i]>=6)
sum += num[i+1] - num[i];
return sum;
}
int main()
{
int i,start;
double ans,a;
while(scanf("%lf",&a),a!=-1)
{
ans = 0;
start = 0;
for(i=0;i<720;i++)
{
if(i==109)
i = i;
ans += happytime(start%3600,start,a);
start += 60;
}
printf("%.3lf\n",ans*100/43200);
}
return 0;
}
*/
/*
// 分析：
//1，题中(1 <= A, B <= 1000, 1 <= n <= 100,000,000)，可知用蛮力肯定行不通。
//2， (A * f(n - 1) + B * f(n - 2)) mod 7 =(A%7*f(n-1)+B%7*f(n-2))%7
//3，因f(i)和f(i+1) 只有49种组合，因为(f(i),(i+1)均只有7种选择,就是只能是0，1，2，3，4，5，6中的一个。故周期<=49。
//解题：
//1，先求周期，顺便把第一个周期的f(n)求出来。
//2，利用周期，直接求其余f(n)。
//1005 Number Sequence
#include <iostream>
using namespace std;
int main()
{
int a = 0,b = 0;
int num;
int result[100] = {0};
cin>>a>>b>>num;
while(a!=0 || b!=0 || num!=0)
{
memset(result,0,100);
result[1] = 1;
result[2] = 1;
int i = 3;
int loop = 0;
a = a % 7;
b = b % 7;
for( ; i<=99; ++i)
{
result[i] = (a * result[i-1] + b * result[i-2])%7;
//cout<<i<<" "<<result[i]<<endl;
if(result[i] == result[5]&&result[i-1] == result[4]&&result[i-2] == result[3]&& i > 5)
break;
}
loop = i - 5;
//WA的
//cout<<"loop="<<loop<<endl;
//int temp = num % loop;
//cout<<"temp="<<temp<<endl;
//if(temp == 0) temp = loop;
//cout<<result[temp]<<endl;
//AC的
if(num<5) cout<<result[num]<<endl;
else cout<<result[(num-5) % loop + 5]<<endl;
cin>>a>>b>>num;
}
return 0;
}
*/
/*
#include <iostream>
using namespace std;
#define M 52
int main()
{
int f[10000];
int a,b,n;
while(cin>>a>>b>>n&&(a||b||n) )
{
f[1]=1;
f[2]=1;
a=a%7;
b=b%7;
for(int i=3;i<=M;i++)
{
f[i]=(a*f[i-1]+b*f[i-2]) % 7;
if(f[i-1]==f[3]&&f[i]==f[4]&&i>4) break;
}
int t=i-4;
if(n<4) cout<<f[n]<<endl;
else cout<<f[(n-4)%t+4]<<endl;
}
return 0;
}
*/
/*
//Time Limit Exceeded
#include <iostream>
using namespace std;
int main()
{
int a = 0,b = 0;
int num;
int result[4] = {0};
cin>>a>>b>>num;
while(a!=0 && b!=0 && num!=0)
{
result[1] = 1;
result[2] = 1;
int i = 3;
for( ; i<=num; ++i)
{
result[3] = (a * result[2] + b * result[1])%7;
result[1] = result[2];
result[2] = result[3];
}
if(num<=2)
cout<<"1"<<endl;
else
cout<<result[3]<<endl;
cin>>a>>b>>num;
}
return 0;
}
*/
/*
//1004 Let the Balloon Rise
#include <iostream>
#include <string>
using namespace std;
int main()
{
int n;
string s[1000];
int num[1000] = {0};
while(cin>>n,n!=0)
{
memset(num,0,1000);
int iMax = 0;
int iNum = 0;
int i = 0;
for( ; i<n; ++i)
{
cin>>s[i];
int j = 0;
for( ; j<=i; ++j)
{
if(s[j] == s[i])
++num[i];
}
if(iMax < num[i])
{
iMax = num[i];
iNum = i;
}
}
cout<<s[iNum]<<endl;
}
return 0;
}
*/
/*460B
#include <iostream>
using namespace std;int main(){int n;cin>>n;int times = 0;while(n--){++times;int nums = 0;cin>>nums;int posB = 1,posE = 1;int now = 0;int max = 0;int times1 = 0;int tempB = 1;while(nums--){++times1;int temp;cin>>temp;if(now+temp < temp){tempB = times;now = temp;}else{now += temp;}if(now > max){posB = tempB;posE = times1;max = now;}}cout<<"Case "<<times<<":"<<endl;cout<<max<<" "<<posB<<" "<<posE<<endl;if(n!=0)cout<<endl; }return 0;}
*/
/*
//1003 Max Sum 673B
//1.首先，读取第一个数据，令now和max等于第一个数据，初始化pos1,pos2,x位置
//2.然后，读入第二个数据，判断
//①. 若是now+temp<temp，表示当前读入的数据比之前存储的加上当前的还大，说明可以在当前另外开始记录，更新now=temp
//②. 反之，则表示之前的数据和在增大，更新now=now+temp
//3.之后，把now跟max做比较，更新或者不更新max的值，记录起始、末了位置
//4.循环2~3步骤，直至读取数据完毕。
#include <iostream>
using namespace std;
int main()
{
int n;
cin>>n;
int times = 0;
while(n--)
{
++times;
int nums = 0;
int temp;
cin>>nums>>temp;
int posB = 1,posE = 1;
int now = temp;
int max = temp;
int times1 = 1;
--nums;
int tempB = 1;
while(nums--)
{
++times1;
cin>>temp;
if(now+temp < temp)
{
tempB = times1;
now = temp;
}
else
{
now += temp;
}
if(now > max)
{
posB = tempB;
posE = times1;
max = now;
}
}
cout<<"Case "<<times<<":"<<endl;
cout<<max<<" "<<posB<<" "<<posE<<endl;
if(n!=0)
cout<<endl;
}
return 0;
}
/*
/*
//C++中精度控制
#include <iostream>
#include<iomanip>//格式控制函数声明
using namespace std;
int main()
{
double n1=0.123456789,n2=0.123;
cout<<fixed<<setprecision(6)<<n1<<endl;//精确到小数点后6位 //一定要这样写，不然没效果
cout<<fixed<<setprecision(6)<<n2<<endl;//精确到小数点后6位
cout<<fixed<<setprecision(9)<<n1<<endl;//精确到小数点后9位
cout<<fixed<<setprecision(9)<<n2<<endl;//精确到小数点后9位
return 0;
}
*/
/*
//2001 计算两点间的距离
//使用勾股定理
#include <stdio.h>
#include <math.h>
int main()
{
double x1 = 0,y1 = 0,x2 = 0,y2 = 0;
while(scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2)!=EOF)
{
double x = x1 - x2;
double y = y1 - y2;
double result = sqrt(x*x + y*y);
printf("%.2lf\n",result);
}
return 0;
}
*/
/*
//2000 ASCII码排序
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
string s;
while(cin>>s)
{
if(s[0]>=s[1]) swap(s[0],s[1]);
if(s[1]>=s[2]) swap(s[1],s[2]);
if(s[0]>=s[1]) swap(s[0],s[1]);
cout<<s[0]<<" "<<s[1]<<" "<<s[2]<<endl;
}
return 0;
}
/*
/*
//1002 A + B Problem II
//大数求和，主要思想：使用字符串存储大数，从2位数最低位开始相加，
//并注意如果>=10就要产生进位，注意的地方：2个数位数不同而带来的错位
//和最后结果可能比最大数位数大的情况（如 ：99 + 2 = 101 ）
#include <iostream>
#include <string>
using namespace std;
int main()
{
int n = 0,times = 1;
string s1,s2,result;
cin>>n;
while(n--)
{
cin>>s1>>s2;
int l1 = s1.length();
int l2 = s2.length();
int nR = 0;
while(l1 >0 || l2 >0)
{
int n1,n2;
if(l1!=0)
n1 = (char)s1[--l1] - 48;
else
n1 = 0;
if(l2!=0)
n2 = (char)s2[--l2] - 48;
else
n2 = 0;
nR = nR + n1 + n2;
if(nR>=10)
{
int iTemp = nR - 10;
nR = 1;
char c = 48 + iTemp;
result = c + result;
}
else
{
char c = 48 + nR;
nR = 0;
result = c + result;
}
}
if(nR) result = '1' +result;
cout<<"Case "<<times++<<":"<<endl;
cout<<s1<<" + "<<s2<<" = "<<result<<endl;
if(n!=0)
cout<<endl;
result = "";
s1 = "";
s1 = "";
}
return 0;
}
*/
/*
//1001 Sum Problem
#include <stdio.h>
int main()
{
int num = 0;
int i = 0,result = 0;
while(scanf("%d",&num)!=EOF)
{
for(i=1; i<=num; ++i)
result +=i;
printf("%d\n\n",result);
result = 0;
}
return 0;
}
*/
//1000 A + B Problem
/*
#include <iostream>
using namespace std;
int main()
{
int a,b;
while(cin>>a>>b)
{
cout<<a+b<<endl;
}
return 0;
}
*/
/*
#include <stdio.h>
int main()
{
int a,b;
while(scanf("%d%d",&a,&b)!=EOF)
{
printf("%d\n",a+b);
}
return 0;
}
*/