求斐波那契单词的第n个字符
定义【摘抄自Wiki】
Let 
be "0" and 
be "01". Now 
(the concatenation of the previous sequence and the one before that).
The infinite Fibonacci word is the limit
We have:
0
01
010
01001
01001010
0100101001001
...
The first few elements of the infinite Fibonacci word are:
0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, ...
Problem:
求斐波那契单词的第n个字符。即写如下函数：
char f(index)
思路：
初始条件：f(0)=0, f(1)=1
斐波那契单词分为两部分，Sn-1和Sn-2。
假设index >= length(Sn-1)，那么f(index) = f(index-length(Sn-1))
。
不断递减index直到index <= 1为止，那么即可得出f(index)的值。
C++代码如下：
#include
#include
#include
char
getCharInFibonacciWord(
int
index)
{
using
namespace
std;
assert(index >=0);
// The 48th fibonacci number is already out of range of int.
static
int
fibs[48] = {1,2,0};
while
( index > 1)
{
int
i=0;
for
(
int
i=1; i<_countof(fibs); ++i)
{
if
(fibs[i] == 0)
fibs[i] = fibs[i-1] + fibs[i-2];
if
(fibs[i] > index)
{
index -= fibs[i-1];
break
;
}
}
}
return
(index == 0) ?
'0'
:
'1'
;
}
int
_tmain(
int
argc, _TCHAR* argv[])
{
for
(
int
i=0; i<100; ++i)
std::cout << getCharInFibonacciWord(i);
std::cout << std::endl;
return
0;
}