Question

If n is odd, then f(n) = 2 * f(n/2) + [n/2]. if n is even, let f(n) = f(n-1) + s(n), where s(n) is the number of ones in binary representation of n. The base case is f(0) = 0. Find the value f(200) +f(210)

Write code in java languge.

Answer 1

To find the value of f(200) + f(210), we implement a Java program with two functions: one recursive function defined by the given rules and another function to count the number of ones in the binary representation of a number. The provided code snippet demonstrates how to calculate these values.

Step-by-step explanation:

The question relates to a recursive function definition in the context of computer programming, specifically Java language. To solve the problem, we must implement this function f(n) in Java, adhering to the rules defined for odd and even values of n. The base case for the recursion is given as f(0) = 0. For even n, f(n) is the sum of f(n-1) and the number of ones (s(n)) in the binary representation of n. When n is odd, the function is defined as 2 * f(n/2) + [n/2].

To find the value of f(200) + f(210), we would write a Java program that calculates these values using the given recursive function. We would also need a helper function to count the number of ones in the binary representation of a number (s(n)).

Here's a simplified code snippet illustrating the implementation:

int f(int n) {
if (n == 0) return 0;
if (n % 2 == 0) return f(n-1) + s(n);
else return 2 * f(n/2) + (n/2);
}

int s(int n) {
return Integer.bitCount(n);
}

public static void main(String[] args) {
int result = f(200) + f(210);
System.out.println("The value of f(200) + f(210) is " + result);
}