(3) There is a function void fnE(int i, int num) that runs in 1000-i steps, regardless of what num is. Consider the following code snippet: void fnA(int S[]) { int n = s.length; for (int i=0;i

Implement in C Programming 6.11.2: Modify an array parameter. Write a function SwapArrayEnds() that swaps the first and last elements of the function's array parameter. Ex: sortArray = {10, 20, 30, 40} becomes {40, 20, 30, 10}. #include <stdio.h> /* Your solution goes here  */ int main(void) {   const int SORT_ARR_SIZE = 4;   int sortArray[SORT_ARR_SIZE];   int i;   int userNum;    for (i = 0; i < SORT_ARR_SIZE; ++i) {      scanf("%d", &sortArray[i]);   }    SwapArrayEnds(sortArray, SORT_ARR_SIZE);    for (i = 0; i < SORT_ARR_SIZE; ++i) {      printf("%d ", sortArray[i]);   }   printf("\n");    return 0;}

Write a parallel program for the following code   /* PROGRAM Sieve */ #include <stdlib.h> #include <math.h> #define n 100 boolean Prime[n+1]; int i,num,loc; main( ) { for (i = 1; i <= n; i++) Prime[i] = TRUE; for (num = 2; num <= Sqrt(n); num++) if (Prime[num]) { loc = num + num; while (loc <= n) { Prime[loc] = False; } } loc = loc + num; }

For each of the following function, indicate the class (n) the function belongs (use the simplest g (n) possible in your answer. a) (n² + 5)1⁰ c) 5n+3 + 7+1 b) 4nlog2(n+2)² + (n+2)³log2(n/2) d) 2n + 3n+ n²+ 4n