Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3
This is the order in which it should be filed.
Wakefield, Cyndi
Wakefield, Donald
Wakefield, Drew ⇒ this is the answer. Since both have Ds, we evaluate O and R. O comes first before R.
Wakefield, Edgar
Wheton, John
Answer:
import math
tolerance=0.00001
approximation=1.0
x = float(input("Enter a positive number: "))
def newton(number,approximation):
approximation=(approximation+number/approximation)/2
difference_value =abs(number-approximation**2)
if difference_value<= tolerance:
return approximation
else:
return newton(number,approximation)
print("The approximation of program = ", newton(x, approximation))
print("The approximation of Python = ", math.sqrt(x))
Explanation:
- Create a recursive function called newton that calls itself again and again to approximate square root for the newton technique.
- Apply the formulas to find the estimate value and the difference value
.
- Check whether the difference_value is less than the tolerance value and then return the value of approximation else make a recursive call to the newton function by passing the user input and the new approximation value
.
- Finally display all the results using the print statement.
Answer:
System software includes all of the following except <u>Browsers</u>.
Explanation:
I just know
