Part a. Make a diagram of memory for point one when n == 2. int expo(int x, int n) { int r, t; if (n == 0) r = 1; else if (n == 1) r = x; else { t = expo (x, n / 2); r = t * t; if (n % 2 == 1) r *= t:...

Extracted text: Part a. Make a diagram of memory for point one when n == 2. int expo(int x, int n) { int r, t; if (n == 0) r = 1; else if (n == 1) r = x; else { t = expo (x, n / 2); r = t * t; if (n % 2 == 1) r *= t: } //-- -point one return r; int main(void) { int i; i = expo(4, 5); return 0; Part b. Using divide-and-conquer recursion to solve a problem similar to Prob- lem 3 is somewaiät tricky but a good test of programming skill. Write a function definition that provides a divide-and-conquer implementation of the following interface: double max_adj_sum(const double *a, int lo, int hi); // REQUIRES: hi - lo >= 2. // PROMISES: Return value is the largest sum that can be made by adding // Elements a[lo] ... a[hi -1] exist. two elements among a[lo] ... a[hi-1] with adjacent inderes. Hint: It makes sense to have two base cases, not just one.