Consider the following ANSI C program.

The output of the program upon execution is \(\underline{\hspace{2cm}}\).
Step 1: Understand base conditions.
The function returns \( q \) when both \( x \le 0 \) and \( y \le 0 \).
Step 2: Recursive behavior.
If one variable is non-positive, recursion continues by decreasing the other by \( q \).
If both are positive, the function branches into two recursive calls.
Step 3: Evaluate \( \text{foo}(15,15,10) \).
\[
\text{foo}(15,15,10) = \text{foo}(15,5,10) + \text{foo}(5,15,10)
\]
Each of these further expands until base cases are reached.
This forms a binary recursion tree where each leaf contributes a value of \( 10 \).
Step 4: Count base case occurrences.
The recursion generates exactly 6 base case hits.
Step 5: Compute final result.
\[
\text{Result} = 6 \times 10 = 60
\]
% Final Answer
Final Answer: \[ \boxed{60} \]
def f(a, b):
if (a == 0):
return b
if (a % 2 == 1):
return 2 * f((a - 1) / 2, b)
return b + f(a - 1, b)
print(f(15, 10))
The value printed by the code snippet is 160 (Answer in integer).
Consider the following Python code snippet.
def f(a, b):
if (a == 0):
return b
if (a % 2 == 1):
return 2 * f((a - 1) / 2, b)
return b + f(a - 1, b)
print(f(15, 10))
The value printed by the code snippet is 160 (Answer in integer).
Consider the following ANSI C function:
int SomeFunction(int x, int y)
{
if ((x == 1) || (y == 1)) return 1;
if (x == y) return x;
if (x > y) return SomeFunction(x - y, y);
if (y > x) return SomeFunction(x, y - x);
} The value returned by SomeFunction(15, 255) is \(\underline{\hspace{2cm}}\).