# data structure c++

Budget $30-5000 USD

Note -----Please make the first three programs as 1 program must be commented on every step you do. [url removed, login to view] an algorithm that reads a list of integers from the keyboard,creates a linked list out of them and prints the result. [url removed, login to view] an algorithm that accepts alinked list traverse it,and returns the data in the node with the minimum key value. [url removed, login to view] an algorithm that traverses alinked list and deletes all nodes whose keys are negative. ----------------------------------------------------------------------------------------------------------------------- ¯ [url removed, login to view] a program in c++. calculating in floating point arithmetic 9x^4 - y^4 + 2y^2 =1 (9x^4 - y^4 )+2^2 and (2y^2 - y ^4)+9^4 in the following differnt ways making 10 ways altogether. u^4 = u*u*u*u,(u^2)^2 ,u^4, exp(4 ln u )and 10 ^(4log u) with u = x , y using the values instead x = 10864 y= 18817. the correct value is exactly 1 but you will get very different values instead. this exercise is intended to show you that floating point arithmetic is not true arithmetic.

## Deliverables

Note -----Please make the first three programs as 1 program must be commented on every step you do. [url removed, login to view] an algorithm that reads a list of integers from the keyboard,creates a linked list out of them and prints the result. [url removed, login to view] an algorithm that accepts alinked list traverse it,and returns the data in the node with the minimum key value. [url removed, login to view] an algorithm that traverses alinked list and deletes all nodes whose keys are negative. ----------------------------------------------------------------------------------------------------------------------- [url removed, login to view] a program in c++. calculating in floating point arithmetic 9x^4 - y^4 + 2y^2 =1 (9x^4 - y^4 )+2^2 and (2y^2 - y ^4)+9^4 in the following differnt ways making 10 ways altogether. u^4 = u*u*u*u,(u^2)^2 ,u^4, exp(4 ln u )and 10 ^(4log u) with u = x , y using the values instead x = 10864 y= 18817. the correct value is exactly 1 but you will get very different values instead. this exercise is intended to show you that floating point arithmetic is not true arithmetic

## Platform

os 95 98 2000 melinum

## Deadline information

must be done by due date