如何减少此代码的执行时间？

Question

Problem

Consider the sequence D of the last decimal digits of the first N Fibonacci numbers, ie D = (F0%10,F1%10,…,FN−1%10) .

Now, you should perform the following process:

Let D=(D1,D2,…,Dl)

If l=1, the process ends.

Create a new sequence

E=(D2,D4,…,D2⌊l/2⌋)

In other words, E is the sequence created by removing all odd-indexed elements from D

Change D to E

When this process terminates, the sequence D contains only one number. You have to find this number.

Input

The first line of the input contains a single integer T denoting the number of test cases.

The description of T test cases follows. The first and only line of each test case contains a single integer N

Output

For each test case, print a single line containing one integer ― the last remaining number.

Code

#include <stdio.h>
#include <stdlib.h>

int test(int *arr, int n);

int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    int *arr;
    arr = (int *)malloc((n + 1) * sizeof(int));
    arr[1] = 0;
    arr[2] = 1;

    for (int i = 3; i <= n; i++) {
      arr[i] = arr[i - 1] + arr[i - 2];
    }

    /*
    for(int k=1;k<=n;k++){
      printf("%d ",arr[k] );
    }
    printf("\n");
    */

    printf("%d\n", (test(arr, n)) % 10);
  }
}

int test(int *arr, int n) {
  if (n == 1) {
    return arr[1];
  } else {
    for (int i = 1; i <= (n / 2); i++) {
      arr[i] = arr[2 * i];
    }
    return test(arr, n / 2);
  }
}

Answer 1

Using the algorithm from https://math.stackexchange.com/questions/681674/recursively-deleting-every-second-element-in-a-list ,

1. Find the largest integer A , such that 2^A < N .
2. Find Fibonnaci(2^A - 1) % 10

Answer 2

Adding to Bill Lynch 's answer , which is itself based on this other answer by happymath :

You will always end up getting 2 ⁿ − 1 where n is maximum integer such that 2 ⁿ < K

I'd like to point out another useful mathematical property.

In number theory, the nth Pisano period, written π(n), is the period with which the sequence of Fibonacci numbers taken modulo n repeats. ( https://en.wikipedia.org/wiki/Pisano_period )

Here we need to consider the case where n = 10, π(10) = 60 and the last decimal digits correspond to the OEIS sequence A003893 :

0, 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1

So that there's no need to calculate the actual Fibonacci number, nor to generate all the sequence up to N.