Description

Golden ratio base (GRB) is a non-integer positional numeral system that uses the golden ratio (the irrational number (1+√5)/2 ≈ 1.61803399 symbolized by the Greek letter φ) as its base. It is sometimes referred to as base-φ, golden mean base, phi-base, or, phi-nary.       
Any non-negative real number can be represented as a base-φ numeral using only the digits 0 and 1, and avoiding the digit sequence "11" � this is called a standard form. A base-φ numeral that includes the digit sequence "11" can always be rewritten in standard form, using the algebraic properties of the base φ ― most notably that φ + 1 = φ 2 . For instance, 11(φ) = 100(φ). Despite using an irrational number base, when using standard form, all on-negative integers have a unique representation as a terminating (finite) base-φ expansion. The set of numbers which possess a finite base-φ representation is the ring Z[1 + √5/2]; it plays the same role in this numeral systems as dyadic rationals play in binary numbers, providing a possibility to multiply.       
Other numbers have standard representations in base-φ, with rational numbers having recurring representations. These representations are unique, except that numbers (mentioned above) with a terminating expansion also have a non-terminating expansion, as they do in base-10; for example, 1=0.99999….       
Coach MMM, an Computer Science Professor who is also addicted to Mathematics, is extremely interested in GRB and now ask you for help to write a converter which, given an integer N in base-10, outputs its corresponding form in base-φ.      
              

Input

There are multiple test cases. Each line of the input consists of one positive integer which is not larger than 10^9. The number of test cases is less than 10000. Input is terminated by end-of-file.      
              

Output

For each test case, output the required answer in a single line. Note that trailing 0s after the decimal point should be wiped. Please see the samples for more details.      
              

Sample Input

1
2
3
6
10
              

Sample Output

1
10.01
100.01
1010.0001
10100.0101

Hint

 
 
由于φ + 1 = φ 2,两边同乘φ k,得到φ k+1+φ k=φ k+2,说明只有有两位是1,就往前进一位。此外由φ + 1 = φ 2推到的2φ 2=φ 3+1,同理可知:φ k+3+φ k=2φ k+2,说明每一位的2都可以,由它前一位和它的后两位的1构成,这样就能将所有大于2的数降成1.再配合之前的,反复模拟便可得。由于当场没有估算这个数的长度,所以采用两个数组分别存了整数部分和小数部分。整体效率不是非常高,但是在短时间内做出来还是很高兴的。
 
 
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define esp 1e-10
#define N 100 using namespace std; int z[N], x[N], lenz, lenx; bool judge ()
{
if(z[0] && x[0])
return 0;
for (int i = 0; i < lenx; ++i)
if (x[i] > 1 || (x[i] && x[i+1]))
return 0; for (int i = 0; i < lenz; ++i)
if (z[i] > 1 ||(z[i] ==1 && z[i+1] == 1))
return 0; return 1;
} void doz (int i)
{
if (i == lenz-1)
lenz++;
int up = z[i] / 2;
z[i] = z[i] & 1;
z[i+1] += up;
if (i >= 2)
z[i-2] += up;
else
{
if (lenx < 3 - i)
lenx = 3 - i;
x[1-i] += up;
}
} void dox (int i)
{
if (i+3 > lenx)
lenx = i + 3;
int up = x[i] / 2;
x[i] = x[i] & 1;
x[i+2] += up;
if (i == 0)
z[0] += up;
else
x[i-1] += up;
} void qt (int n)
{
memset (z, 0, sizeof(z));
memset (x, 0, sizeof(x));
lenz = 1;
lenx = 0;
z[0] = n;
while (!judge ())
{
for (int i = lenx-1; i >= 0; --i)
{ if (i == 0 && x[i] > 0 && x[i+1] > 0)
{
int up = min (x[i], x[i+1]);
z[0] += up;
x[0] -= up;
x[1] -= up;
continue;
}
else if (x[i] > 0 && x[i+1] > 0)
{
int up = min (x[i], x[i+1]);
x[i-1] += up;
x[i+1] -= up;
x[i] -= up;
continue;
}
if (x[i] > 1)
{
dox (i);
continue;
} }
while(x[lenx-1] == 0)
lenx--;
for (int i = 0; i < lenz; ++i)
{ if (i == 0 && z[i] > 0 && x[0] > 0)
{
if (i == lenz-1)
lenz++;
int up = min (z[i], x[0]);
z[1] += up;
z[0] -= up;
x[0] -= up;
continue;
}
else if (z[i] > 0 && z[i+1] > 0)
{
if (i+3 > lenz)
lenz = i + 3;
int up = min (z[i], z[i+1]);
z[i+2] += up;
z[i+1] -= up;
z[i] -= up;
continue;
}
if (z[i] > 1)
{
doz(i);
continue;
}
}
}
while(x[lenx-1] == 0)
lenx--;
} int main()
{
//freopen ("test.txt", "r", stdin);
int n;
while (scanf ("%d", &n) != EOF)
{
qt (n);
for (int i = lenz - 1; i >= 0; --i)
printf ("%d", z[i]);
if (lenx > 0)
printf (".");
for (int i = 0; i < lenx; ++i)
printf ("%d", x[i]);
printf ("\n");
}
return 0;
}

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