遵循规则的数字序列

Question

我有一个数字序列：1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4,...，我必须猜测序列遵循的规则，并：

• 计算小于n的所有素数之和
• 在序列的元素中查找数字k出现的次数
• 序列中p位置上的数字。输入：
• n，k，p --我已经编写了下一段代码:它达到了它的目的，但我想让您给我一些关于代码风格和性能的提示(而且我考虑的是速度)。# #包括 int素数( int n) { if(n < 2)返回0；if(n == 2)返回1；if(n %2 == 0)返回0；int i；for(i = 3；i <= sqrt(n)；i += 2) { if(n %i == 0)返回0；}== 1；} int main() { int n；int p；int k；scanf( %d，&n )；scanf( %d，&k )；scanf( %d，&p )；int primeSumPerSubSequence = 128；int nAux = n；int primeSum = 0；int subSequences = 0；if( n> 60 ) subSequences =n/ 60；if( n> 60 ){ nAux =n% 60；primeSum = primeSum + primeSumPerSubSequence * subSequences；} int i= 2；int firstNumber = 1；int secondNumber = 2；if( !nAux ) secondNumber = 0；int thirdNumber = 3；int kFound = 0；switch( K){ case 0: kFound =4*子序列；断开；case 1: kFound =8*子序列；例4: kFound =4*子序列；中断；例5: kFound =8*子序列；中断；例6: kFound =4*子序列；中断；案例7: kFound =8*子序列；中断；案例8: kFound =4*子序列；中断；案例9: kFound =8*子序列；中断；}if( kFound == 1)_x( kFound == 2) kFound++；int pNumber = -1；若( !p ) pNumber = 0；若(p% 60 == 0) pNumber = 1；若(p> 60 )p %= 60；而(i <= nAux ){ if( nAux &素数( secondNumber )){ primeSum += secondNumber；} if(( i == p) && ( pNumber =1) pNumber = secondNumber；firstNumber = secondNumber；secondNumber = thirdNumber；thirdNumber = firstNumber + secondNumber；if( thirdNumber >9) thirdNumber %= 10；if( secondNumber == k&(i9) thirdNumber %= 10；if( i == p) pNumber = secondNumber；} if( pNumber == -1) pNumber = 0；printf( "\n%d"，kFound )；printf(“n%d”，pNumber )；返回0；}

我应该编辑这个问题并添加我对算法的想法吗？谢谢，请原谅我英语不好。

Answer 1

这是一种罕见的情况，在这种情况下，评论是绝对必要的，类似于

/* The sequence is Fibonacci numbers modulo 10. It is periodic with a
   period of 60. Each even digits appears for 4 times, and each odd
   digit for 8 times, per period. The sum of primes per period is 128.

           Reviewer's note: are you sure? Looks like you consider 1 as prime.

   To get the desired results, compute the number of entire periods,
   and process the last incomplete part separately. */

• 将subSequences重命名为periods
• 将switch折叠为kFound =(k& 0x01)？8: 4) *周期；
• 我不认为kFound是1或2，因此，如果(( kFound == 1)x( kFound == 2 )) kFound++；似乎没有用。如果我错了，请纠正我。
• 与不完整部分一起工作应分为功能部分。预先计算周期一次，并将其传递给单独的函数(计算素数和、计数次数和在位置返回数字)。