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fiasKO [112]
3 years ago
13

Write a function named “createPurchaseOrder” that accepts the quantity (integer), the cost per item(double), and the description

(string). It will return a newly created PurchaseOrderobject holding that information if they are valid: quantity and cost per item cannot be negative and the description can not be empty or blank(s). When it is invalid, it will return NULL to indicate that it cannot create such PurchaseOrderobject.
Computers and Technology
1 answer:
Leona [35]3 years ago
4 0
#include
Program: using namespace std;
string createPurchaseOrder0;
int main(
{
cout<return 0;
}
string createPurchaseOrder(
{
int qty;
double costPerltem;
string description,info="":
cout<<"Enter Quantity:
cin>>qty;
cout<<"Enter cost per item: "
cin>>costPerltem;
cout<<"Enter Description: "
cin>>description;
if(qty<0 I| costPerltem<0
Idescription.compare(''"')==0)
cout<<'InThe entered data is invalid!":
info="":
else
"
cout<<"'InThe entered data is valid!":
info=info+"'(nQuantity: "+to_string (qty) +" In";
info=info+"Cost per item:
"†to_string (costPerltem)+"In";
info=info+"Description: "description+" In";
return info;

Output:
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4 years ago
. In the select algorithm that finds the median we divide the input elements into groups of 5. Will the algorithm work in linear
8090 [49]

Answer:

we have that it grows more quickly than linear.

Explanation:

It will still work if they are divided into groups of 77, because we will still know that the median of medians is less than at least 44 elements from half of the \lceil n / 7 \rceil⌈n/7⌉ groups, so, it is greater than roughly 4n / 144n/14 of the elements.

Similarly, it is less than roughly 4n / 144n/14 of the elements. So, we are never calling it recursively on more than 10n / 1410n/14 elements. T(n) \le T(n / 7) + T(10n / 14) + O(n)T(n)≤T(n/7)+T(10n/14)+O(n). So, we can show by substitution this is linear.

We guess T(n) < cnT(n)<cn for n < kn<k. Then, for m \ge km≥k,

\begin{aligned} T(m) & \le T(m / 7) + T(10m / 14) + O(m) \\ & \le cm(1 / 7 + 10 / 14) + O(m), \end{aligned}

T(m)

​

 

≤T(m/7)+T(10m/14)+O(m)

≤cm(1/7+10/14)+O(m),

​

therefore, as long as we have that the constant hidden in the big-Oh notation is less than c / 7c/7, we have the desired result.

Suppose now that we use groups of size 33 instead. So, For similar reasons, we have that the recurrence we are able to get is T(n) = T(\lceil n / 3 \rceil) + T(4n / 6) + O(n) \ge T(n / 3) + T(2n / 3) + O(n)T(n)=T(⌈n/3⌉)+T(4n/6)+O(n)≥T(n/3)+T(2n/3)+O(n) So, we will show it is \ge cn \lg n≥cnlgn.

\begin{aligned} T(m) & \ge c(m / 3)\lg (m / 3) + c(2m / 3) \lg (2m / 3) + O(m) \\ & \ge cm\lg m + O(m), \end{aligned}

T(m)

​

 

≥c(m/3)lg(m/3)+c(2m/3)lg(2m/3)+O(m)

≥cmlgm+O(m),

​

therefore, we have that it grows more quickly than linear.

5 0
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julia-pushkina [17]

Answer:

Could you seperate them if they are different answers? I'm lost, sorry!

Explanation:

5 0
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