int f (int n)
{ if (n == 0) then return n;
else
return n + f(n-2);
}
The function $f(n)$ is defined with two cases:
We need to calculate the value of $f(100)$. Let's trace the execution:
$f(100) = 100 + f(98)$
$f(98) = 98 + f(96)$
$f(96) = 96 + f(94)$
$...$
$f(4) = 4 + f(2)$
$f(2) = 2 + f(0)$
$f(0) = 0$
By substituting the results back up the chain, we see that:
$f(100) = 100 + 98 + 96 + ... + 4 + 2 + 0$
The value $f(100)$ represents the sum of even numbers from 2 up to 100.
The series is $ S = 2 + 4 + 6 + \dots + 98 + 100 $.
This is an arithmetic progression.
First, find the number of terms ($k$). Using the formula $ l = a + (k-1)d $: $ 100 = 2 + (k-1)2 $ $ 98 = (k-1)2 $ $ 49 = k-1 $ $ k = 50 $ So, there are 50 terms in the series.
Next, calculate the sum ($S_k$) using the formula $ S_k = \frac{k}{2}(a + l) $: $ S_{50} = \frac{50}{2}(2 + 100) $ $ S_{50} = 25 \times 102 $ $ S_{50} = 2550 $
Thus, the function $f$ returns 2550 when $n = 100$.
The process of removing recursion involves replacing recursive function calls with:
#include <stdio.h>
int foo(int S[], int size){
if(size == 0) return 0;
if(size == 1) return 1;
if(S[0] != S[1]) return 1 + foo(S + 1, size - 1);
return foo(S + 1, size - 1);
}
int main(){
int A[] = {0, 1, 2, 2, 2, 0, 0, 1, 1};
printf("%d", foo(A, 9));
return 0;
}
The value printed by the given C program is _______ . (Answer in integer)
int bar(int n){
if (n == 1) return 0;
else return 1 + bar(n/2);
}
int foo(int n){
if (n == 1) return 1;
else return 1 + foo(bar(n));
}int func(int start, int end){
int length=end+1-start;
if((length < 1)||(start < 0)||(end < 0)){ return(0); }
if(length%3==0){
return(func(start+1, end));
} else if(length%3==1){
return(1+func(start, end-1));
} else {
return(func(start+2, end));
}
}What is the output of the given C language code snippet?
#include<stdio.h>
void f(int n)
{
if(n==0)
return;
printf("%d ",n);
f(n-1);
}
int main()
{
f(3);
return 0;
}