Part a. Make a diagram of memory for point one when n == 2. int expo(int x, int n) { int r, t; if (n == r = 1; else if (n == 1) r = x; else { 0) 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 somewhat 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. Elements a[lo] ... a[hi -1] exist. // PROMISES: Return value is the largest sum that can be made by adding // two elements among a[lo] ... a[hi-1] with adjacent indezes. Hint: It makes sense to have two base cases, not just one.

Part a. Make a diagram of memory for point one when n == 2. int expo(int x, int n) { int r, t; if (n == r = 1; else if (n == 1) r = x; else { 0) 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 somewhat 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. Elements a[lo] ... a[hi -1] exist. // PROMISES: Return value is the largest sum that can be made by adding // two elements among a[lo] ... a[hi-1] with adjacent indezes. Hint: It makes sense to have two base cases, not just one.

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter15: Recursion

Section: Chapter Questions

Problem 8SA

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