56 lines
		
	
	
		
			1.7 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
		
		
			
		
	
	
			56 lines
		
	
	
		
			1.7 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
|   | --- | |||
|  | id: 5900f4ed1000cf542c50fffe | |||
|  | challengeType: 5 | |||
|  | title: 'Problem 384: Rudin-Shapiro sequence' | |||
|  | videoUrl: '' | |||
|  | localeTitle: 问题384:Rudin-Shapiro序列 | |||
|  | --- | |||
|  | 
 | |||
|  | ## Description
 | |||
|  | <section id="description">将序列a(n)定义为n的二进制展开(可能重叠)中相邻的1对的数量。例如:a(5)= a(1012)= 0,a(6)= a(1102)= 1,a(7)= a(1112)= 2 <p>定义序列b(n)=( -  1)a(n)。该序列称为Rudin-Shapiro序列。还要考虑b(n)的总和序列:。 </p><p>这些序列的前几个值是:n 0 1 2 3 4 5 6 7 a(n)0 0 0 1 0 0 1 2 b(n)1 1 1 -1 1 1 -1 1 s(n)1 2 3 2 3 4 3 4 </p><p>序列s(n)具有显着特性,即所有元素都是正的,并且每个正整数k恰好出现k次。 </p><p>定义g(t,c),其中1≤c≤t,作为s(n)中的索引,其中t在s(n)中出现第c次。例如:g(3,3)= 6,g(4,2)= 7,g(54321,12345)= 1220847710。 </p><p>设F(n)为由下式定义的斐波那契数:F(0)= F(1)= 1且F(n)= F(n-1)+ F(n-2),n> 1。 </p><p>定义GF(t)= g(F(t),F(t-1))。 </p><p>找到ΣGF(t)为2≤t≤45。 </p></section> | |||
|  | 
 | |||
|  | ## Instructions
 | |||
|  | <section id="instructions"> | |||
|  | </section> | |||
|  | 
 | |||
|  | ## Tests
 | |||
|  | <section id='tests'> | |||
|  | 
 | |||
|  | ```yml | |||
|  | tests: | |||
|  |   - text: <code>euler384()</code>应返回3354706415856333000。 | |||
|  |     testString: 'assert.strictEqual(euler384(), 3354706415856333000, "<code>euler384()</code> should return 3354706415856333000.");' | |||
|  | 
 | |||
|  | ``` | |||
|  | 
 | |||
|  | </section> | |||
|  | 
 | |||
|  | ## Challenge Seed
 | |||
|  | <section id='challengeSeed'> | |||
|  | 
 | |||
|  | <div id='js-seed'> | |||
|  | 
 | |||
|  | ```js | |||
|  | function euler384() { | |||
|  |   // Good luck! | |||
|  |   return true; | |||
|  | } | |||
|  | 
 | |||
|  | euler384(); | |||
|  | 
 | |||
|  | ``` | |||
|  | 
 | |||
|  | </div> | |||
|  | 
 | |||
|  | 
 | |||
|  | 
 | |||
|  | </section> | |||
|  | 
 | |||
|  | ## Solution
 | |||
|  | <section id='solution'> | |||
|  | 
 | |||
|  | ```js | |||
|  | // solution required | |||
|  | ``` | |||
|  | </section> |