Free Practice / PYQ
What is the output of the following C function when called with fun(5)?
int fun(int n) {
if (n <= 1) return 1;
return n + fun(n - 1);
}fun(5) = 5 + fun(4) = 5 + 4 + fun(3) = 5 + 4 + 3 + fun(2) = 5 + 4 + 3 + 2 + fun(1) = 5 + 4 + 3 + 2 + 1 = 15.
Consider the following recursive C function:
void solve(int n) {
if (n <= 0) return;
printf("%d ", n % 2);
solve(n / 2);
}What is printed when solve(13) is called?
solve(13): prints 13%2 = 1, calls solve(6).
solve(6): prints 6%2 = 0, calls solve(3).
solve(3): prints 3%2 = 1, calls solve(1).
solve(1): prints 1%2 = 1, calls solve(0).
solve(0): returns.
Output: 1 0 1 1.
What does the following recursive C function compute for a non-negative integer n?
int foo(int n) {
if (n == 0) return 0;
return (n % 10) + foo(n / 10);
}(n % 10) extracts the last digit, and foo(n / 10) recursively processes the remaining digits, summing them up.
Consider the following recursive C function:
int my_func(int a, int b) {
if (b == 0) return 0;
if (b % 2 == 0) return my_func(a + a, b / 2);
return my_func(a + a, b / 2) + a;
}What does my_func(a, b) calculate for positive integers a and b?
This implements Russian Peasant Multiplication, recursively computing a * b in O(log b) steps.
What is the output of the following C program?
#include <stdio.h>
void print_val(int n) {
if (n == 0) return;
print_val(n - 1);
printf("%d ", n);
}
int main(void) {
print_val(4);
return 0;
}The printf statement is after the recursive call, so numbers are printed during unwinding: 1 2 3 4.