The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:
by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that 
returns indeed the sum of the first n-1 odd numbers, and we have to proof that 
returns the sum of the first n odd numbers. By the recursive logic, we have
and by induction, is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So, is the sum of the first n odd numbers, as required:
Here is a Python program:
tmp = input().split(' ')
c = tmp[0]; s = tmp[1]
ans=0
for i in range(len(s)):
if s[i] == c: ans+=1
# the ans variable stores the number of occurrences
print(ans)
Answer: hello the options related to your question is missing attached below is the complete question and options
answer : Read the status register ( A )
Explanation:
To determine if the device is busy or ready to accept a new command the line of action is to write a code that will Read the status register. A status register is a hardware register which contains the information about the processor of the device ( i.e. whether the processor is busy or ready )
Answer:
Fullmetal Alchemist: Brotherhood (160,975)
Explanation:
Fullmetal Alchemist: Brotherhood (160,975)
Answer:
Computer engineering is a branch of engineering that integrates several fields of computer science and electronic engineering required to develop computer hardware and software.
Explanation:
