freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-n-step-number-seq...

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id title challengeType videoUrl
598eef80ba501f1268170e1e 斐波那契n步数序列 5

--description--

编写一个函数来生成Fibonacci n步数序列和Lucas序列。第一个参数是n。第二个参数是要返回的元素数。第三个参数将指定是输出Fibonacci序列还是Lucas序列。如果参数为“f”则返回Fibonacci序列如果为“l”则返回Lucas序列。序列必须作为数组返回。更多细节如下

这些数字序列是普通斐波纳契数列的扩展,其中:

对于$ n = 2 $我们有Fibonacci序列;初始值$ \[1,1] $和$ F_k ^ 2 = F\_ {k-1} ^ 2 + F\_ {k-2} ^ 2 $对于$ n = 3 $我们有tribonacci序列;初始值$ \[1,1,2] $和$ F_k ^ 3 = F\_ {k-1} ^ 3 + F\_ {k-2} ^ 3 + F\_ {k-3} ^ 3 $ $ $ = 4 $我们有tetranacci序列;初始值$ \[1,1,2,4] $和$ F_k ^ 4 = F\_ {k-1} ^ 4 + F\_ {k-2} ^ 4 + F\_ {k-3} ^ 4 + F\_ {k -4} ^ 4 $ ...对于一般的$ n> 2 $,我们有斐波那契$ n $ -step序列 - $ F_k ^ n $; $n-1$'斐波那契$ n $ -step序列$ F_k ^ {n-1} $的前$ n $值的初始值;和$ k $'这个$ n $'序列的值是$ F_k ^ n = \\ sum\_ {i = 1} ^ {n} {F\_ {ki} ^ {n}} $

对于$ n $的小值, 希腊数字前缀有时用于单独命名每个系列。

{| style =“text-alignleft;” border =“4”cellpadding =“2”cellspacing =“2”

| + Fibonacci $ n $ -step序列

| - style =“background-colorrgb255,204,255;”

$ n $ !!系列名称!!值

| -

| 2 ||斐波那契|| 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ......

| -

| 3 || tribonacci || 1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ......

| -

| 4 || tetranacci || 1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ......

| -

| 5 || pentanacci || 1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ......

| -

| 6 || hexanacci || 1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ......

| -

| 7 || heptanacci || 1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ......

| -

| 8 || octonacci || 1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...

| -

| 9 || nonanacci || 1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ......

| -

| 10 || decanacci || 1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...

|}

可以在更改初始值的位置生成联合序列:

Lucas系列将两个前面的值相加,例如$ n = 2 $的斐波那契数列,但使用$ [2,1] $作为其初始值。

--hints--

fib_luc是一个功能。

assert(typeof fib_luc === 'function');

fib_luc(2,10,"f")应返回[1,1,2,3,5,8,13,21,34,55]

assert.deepEqual(fib_luc(2, 10, 'f'), ans[0]);

fib_luc(3,15,"f")应返回[1,1,2,4,7,13,24,44,81,149,274,504,927,1705,3136]

assert.deepEqual(fib_luc(3, 15, 'f'), ans[1]);

fib_luc(4,15,"f")应返回[1,1,2,4,8,15,29,56,108,208,401,773,1490,2872,5536]

assert.deepEqual(fib_luc(4, 15, 'f'), ans[2]);

fib_luc(2,10,"l")应返回[ 2, 1, 3, 4, 7, 11, 18, 29, 47, 76] fib_luc(2,10,"l") [ 2, 1, 3, 4, 7, 11, 18, 29, 47, 76]

assert.deepEqual(fib_luc(2, 10, 'l'), ans[3]);

fib_luc(3,15,"l")应返回[ 2, 1, 3, 6, 10, 19, 35, 64, 118, 217, 399, 734, 1350, 2483, 4567 ] fib_luc(3,15,"l") [ 2, 1, 3, 6, 10, 19, 35, 64, 118, 217, 399, 734, 1350, 2483, 4567 ]

assert.deepEqual(fib_luc(3, 15, 'l'), ans[4]);

fib_luc(4,15,"l")应该返回[ 2, 1, 3, 6, 12, 22, 43, 83, 160, 308, 594, 1145, 2207, 4254, 8200 ] fib_luc(4,15,"l") [ 2, 1, 3, 6, 12, 22, 43, 83, 160, 308, 594, 1145, 2207, 4254, 8200 ]

assert.deepEqual(fib_luc(4, 15, 'l'), ans[5]);

fib_luc(5,15,"l")应该返回[ 2, 1, 3, 6, 12, 24, 46, 91, 179, 352, 692, 1360, 2674, 5257, 10335 ]

assert.deepEqual(fib_luc(5, 15, 'l'), ans[6]);

--solutions--