费氏级数(Fibonacci)的学习
费氏级数(Fibonacci)的学习
Written by SkyJacker 2007-05-31
Http://www.cnpack.org
(1)费氏级数数值序列：0,1,1,2,3,5,8,13...
(2)表示式归纳：Fn = Fn-1 + Fn-2
(3)存储空间：用无符号 4 字节整数来表示，共有 48 个数据
(4)意义描述：一对刚生下来的兔子, 两个月之后会繁殖一对兔子，求 N 月之后共有多少对兔子。
(5)参考资料：
http://translate.google.com/translate?hl=zh-CN&sl=zh-TW&u=http://www.stat.nctu.edu.tw/MISG/SUmmer_Course/C_language/Cdatastru/DataFib.htm&sa=X&oi=translate&resnum=1&ct=result&prev=/search%3Fq%3D%25E8%25B4%25B9%25E6%25B0%258F%2B%25E7%25BA%25A7%25E6%2595%25B0%2B%25E5%2585%2594%25E5%25AD%2590%26hl%3Dzh-CN%26lr%3D%26newwindow%3D1%26sa%3DG
http://translate.google.com/translate?hl=zh-CN&sl=zh-TW&u=http://web.nlhs.tyc.edu.tw/~b305/fibonacci&sa=X&oi=translate&resnum=9&ct=result&prev=/search%3Fq%3D%25E8%25B4%25B9%25E6%25B0%258F%2B%25E7%25BA%25A7%25E6%2595%25B0%2B%25E5%2585%2594%25E5%25AD%2590%26start%3D10%26hl%3Dzh-CN%26lr%3D%26newwindow%3D1%26sa%3DN
(6)关键字：繁殖、一对兔子、两个月
(7)计算实例：
有繁殖能力的兔子|兔子总数| 月份
0 1 1
1 1 2
1 2 3
2 3 4
3 5 5
5 8 6
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657
46368
75025
121393
196418
317811
514229
832040
1346269
2178309
3524578
5702887
9227465
14930352
24157817
39088169
63245986
102334155
165580141
267914296
433494437
701408733
1134903170
1836311903
2971215073
}
//(8)函数实现：
// 费氏级数: 有繁殖能力的兔子数
function FeiShiJiShu(N: Integer): DWORD;
var
Fn, Fn1, Fn2: DWORD;
I: Integer;
begin
Fn := 0;
if n <=1 then
Result := Fn
else
begin
Fn2 := 0;
Fn1 := 1;
Log(IntToStr(Fn2));
Log(IntToStr(Fn1));
for I:= 0 to N-3 do
begin
if Fn1 >= (High(Fn1) div 2) then // 防止溢出
Break;
Fn := Fn2 + Fn1;
Log(IntToStr(Fn)); // 只须打印 Fn
Fn2 := Fn1;
Fn1 := Fn
end;
Result := Fn;
end;
end;
{
(9)相关特性: 费氏级数与黄金分割点(fn/fn+1)
性质：
0,1,1,2,3,5,8,13,21,34
fn+2/fn+1 = 1 + fn/fn+1
13/8 = 1 + 5/8
h1 := 433494437 / 701408733 =
h2 := 701408733 / 1134903170 =
h3 := 1134903170 / 1836311903 =
h4 := 1836311903 / 2971215073 =
计算实例：
0
1
0.000000000000000000
1.000000000000000000
0.500000000000000000
0.666666666666666667
0.600000000000000000
0.625000000000000000
0.615384615384615385
0.619047619047619048
0.617647058823529412
0.618181818181818182
0.617977528089887640
0.618055555555555556
0.618025751072961373
0.618037135278514589
0.618032786885245902
0.618034447821681864
0.618033813400125235
0.618034055727554180
0.618033963166706530
0.618033998521803400
0.618033985017357939
0.618033990175597087
0.618033988205325051
0.618033988957902001
0.618033988670443186
0.618033988780242683
0.618033988738303007
0.618033988754322538
0.618033988748203621
0.618033988750540839
0.618033988749648102
0.618033988749989097
0.618033988749858848
0.618033988749908599
0.618033988749889596
0.618033988749896854
0.618033988749894082
0.618033988749895141
0.618033988749894736
0.618033988749894891
0.618033988749894832
0.618033988749894854
0.618033988749894846
0.618033988749894849
0.618033988749894848
0.618033988749894848
}
function FeiShiJiShu_HuangJin(N: Integer): DWORD;
const
cnF18 = '%.18f';
var
Fn, Fn1, Fn2: DWORD;
a1, a2: Double;
I: Integer;
begin
Fn := 0;
if n <=1 then
Result := Fn
else
begin
Fn2 := 0;
Fn1 := 1;
Log(IntToStr(Fn2));
Log(IntToStr(Fn1));
for I:= 0 to N-3 do
begin
if Fn1 >= (High(Fn1) div 2) then // 防止溢出
Break;
Fn := Fn2 + Fn1;
a1 := Fn1;
a2 := Fn2;
Log(Format(cnF18, [a2 / a1]));
Fn2 := Fn1;
Fn1 := Fn
end;
Result := Fn;
end;
end;
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