Answer:
The answer of this question is given below into explanation section
Explanation:
answer (a)
I visited the carrerbuilder dot com and search for data entry job. The link of the posting is given below
https://www.careerbuilder.com/jobs?utf8=%E2%9C%93&keywords=data+entry&location=
answer(B)-Requirements of the the job
- Previous office experience (data entry experience a plus)
- Proficient with a computer and computer software (Excel knowledge required)
- Excellent verbal and written communication skills
- The ability to multi-task and work in a team-oriented environment
- High School Diploma / G.E.D.
- Ability to meet background check and drug screening requirements
answer(C)-Tasks of the job
- Open, sort, and scan documents
- Track all incoming supplies and samples
- Data entry of samples that come in
- Assist with documentation and maintaining of data
- Prepare and label information for processing
- Review and correct any data entry error or missing information
answer (d)
I have 3 years of experience in organization administration where I managed the organization data, generated reports and communicated verbally and written within the organization efficiently.
Answer:
#include <stdio.h>
int fib(int n) {
if (n <= 0) {
return 0;
}
if (n <= 2) {
return 1;
}
return fib(n-1) + fib(n-2);
}
int main(void) {
for(int nr=0; nr<=20; nr++)
printf("Fibonacci %d is %d\n", nr, fib(nr) );
return 0;
}
Explanation:
The code is a literal translation of the definition using a recursive function.
The recursive function is not per se a very efficient one.
Answer:
Each time you insert a new node, call the function to adjust the sum.
This method has to be called each time we insert new node to the tree since the sum at all the
parent nodes from the newly inserted node changes when we insert the node.
// toSumTree method will convert the tree into sum tree.
int toSumTree(struct node *node)
{
if(node == NULL)
return 0;
// Store the old value
int old_val = node->data;
// Recursively call for left and right subtrees and store the sum as new value of this node
node->data = toSumTree(node->left) + toSumTree(node->right);
// Return the sum of values of nodes in left and right subtrees and
// old_value of this node
return node->data + old_val;
}
This has the complexity of O(n).
Explanation:
