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miskamm [114]
2 years ago
8

int) You are the head of a division of a big Silicon Valley company and have assigned one of your engineers, Jim, the job of dev

ising an algorithm to sort through an English text of n words and convert it into an Esperanto document. Jim comes up with an algorithm which takes 2n2+2n bit operations to handle an input text with n words. Suppose the computers in your business can handle one bit operation every nanosecond (1 nanosecond =10−9 seconds). How many nanoseconds would it take Jim's algorithm to convert a text with 10 wor
Computers and Technology
1 answer:
sergeinik [125]2 years ago
4 0

Answer:

Correct option is E

Explanation:

a) 2n^2+2^n operations are required for a text with n words

Thus, number of operations for a text with n=10 words is 2\cdot 10^2+2^{10}=1224 operation

Each operation takes one nanosecond, so we need 1224 nanoseconds for Jim's algorithm

b) If n=50, number of operations required is 2\cdot 50^2+2^{50}\approx 1.12589990681\times 10^{15}

To amount of times required is 1.12589990681\times 10^{15} nanoseconds which is

1125899.90685 seconds (we divided by 10^{9}

As 1$day$=24$hours$=24\times 60$minutes$=24\times 60\times 60$seconds$

The time in seconds, our algortihm runs is \frac{1125899.90685}{24\cdot 60\cdot 60}=13.0312 days

Number of days is {\color{Red} 13.0312}

c) In this case, computing order of number of years is more important than number of years itself

We note that n=100 so that 2(100)^2+2^{100}\approx 1.267650600210\times 10^{30} operation (=time in nanosecond)

Which is 1.267650600210\times 10^{21} seconds

So that the time required is 1.4671881947\times 10^{16} days

Each year comprises of 365 days so the number of years it takes is

\frac{1.4671881947\times 10^{16}}{365}=4.0197\times 10^{13} years

That is, 40.197\times 10^{12}=$Slightly more than $40$ trillion years$

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A school has 100 lockers and 100 students. All lockers are closed on the first day of school. As the students enter, the first s
kati45 [8]

Solution :

public class NewMain {  

   public_static void main_(String[] args) {  

       boolean[] _locker = new boolean_[101];  

       // to set all the locks to a false NOTE:  first locker is the number 0. Last locker is the 99.

       for (int_i=1;i<locker_length; i++)

       {

       locker[i] = false;

       }

       // first student opens all lockers.

       for (int i=1;i<locker.length; i++)        {

       locker[i] = true;

       }

       for(int S=2; S<locker.length; S++)

       {

          for(int k=S; k<locker.length; k=k+S)

          {

              if(locker[k]==false) locker[k] = true;

              else locker[k] = false;      

          }            

       }

       for(int S=1; S<locker.length; S++)

       {

           if (locker[S] == true) {

               System.out.println("Locker " + S + " Open");

           }

         /* else {

               System.out.println("Locker " + S + " Close");

           } */

       }

   }

}

5 0
1 year ago
Early Intel processors (e.g., the 8086) did not provide hardware support for dual-mode operation (i.e., support for a separate u
Firlakuza [10]

Answer:

Uneven use of resources

Explanation:

Potential problem associated with supporting multi - user operation without hardware support is:

Uneven use of resources: In a situation where we assign a set of resources to user 1 and if a new user comes, then it would be difficult to allocate new resources to him. The processor would get confused between the two users. And the tasks would not be completed. This can affect task processing.

8 0
2 years ago
#A year is considered a leap year if it abides by the #following rules: # # - Every 4th year IS a leap year, EXCEPT... # - Every
lara [203]

Answer:

To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.

Similarly check for 400th year and multiple 0f 4. The following C program describes the function.

#include<stdio.h>

#include<stdbool.h>

bool is_leap_year(int year);

void main()

{

int y;

bool b;

 

printf("Enter the year in yyyy format: e.g. 1999 \n");

scanf("%d", &y);     // taking the input year in yyyy format.

 

b= is_leap_year(y);  //calling the function and returning the output to b

if(b==true)

{

 printf("Thae given year is a leap year \n");

}

else

{

 printf("The given year is not a leap year \n");

}

}

bool is_leap_year(int year)

{

if(year%100==0)   //every 100th year

{

 if(year%400==0)   //every 400th year

 {

  return true;

 }

 else

 {

  return false;

 }

}

if(year%4==0)  //is a multiple of 4

{

 return true;

}

else

{

 return false;

}

}

Explanation:

Output is given as image

5 0
2 years ago
If the object instance is created in a user program, then the object instance can access both the public and private members of
rewona [7]

Answer:

False

Explanation:

The private member of a class is not accessible by using the Dot notation ,however the private member are those which are not accessible inside the class they are accessible outside the class  .The public member are accessible inside the class so they are accessible by using the dot operator .

<u>Following are the example is given below in C++ Language </u>

#include<iostream>   // header file

using namespace std;  

class Rectangle

{    

   private:  

       double r; // private member  

   public:      

       double  area()  

       {     return 3.14*r*r;  

       }        

};  

int main()  

{    

  Rectangle r1;// creating the object  

   r1.r = 3.5;  

 double t= r1.area(); // calling

cout<<" Area is:"<<t;  

   return 0;  

}  

Output:

compile time error is generated

<u>The correct program to access the private member of class is given below </u>

#include<iostream>   // header file

using namespace std;  

class Rectangle

{    

   private:  

       double r; // private member  

   public:      

       double  area()  

       {    

r1=r;

double t2=3.14*r2*r2;

return(t2); // return the value  

       }        

};  

int main()  

{    

  Rectangle r1;// creating the object  

   r1.r = 1.5;  

 double t= r1.area(); // calling

cout<<" Area is:"<<t;  

   return 0;  

}  

Therefore the given statement is False

5 0
2 years ago
When this program is compiled and executed on an x86-64 Linux system, it prints the string 0x48\n and terminates normally, even
Alexxandr [17]

Answer:

I get 0x55 and this the linking address of the main function.

use this function to see changes:

/* bar6.c */

#include <stdio.h>

char main1;

void p2()

{

printf("0x%X\n", main1);

}

Output is probably 0x0

you can use your original bar6.c with updaated foo.c

char main;

int main() // error because main is already declared

{

  p2();

   //printf("Main address is 0x%x\n",main);

  return 0;

}

Will give u an error

again

int main()

{

  char ch = main;

  p2(); //some value

  printf("Main address is 0x%x\n",main); //some 8 digit number not what printed in p2()

  printf("Char value is 0x%x\n",ch); //last two digit of previous line output

  return 0;

}

So the pain in P2() gets the linking address of the main function and it is different from address of the function main.

Now char main (uninitialized) in another compilation unit fools the compiler by memory-mapping a function pointer on a char directly, without any conversion: that's undefined behavior. Try char main=12; you'll get a multiply defined symbol main...

Explanation:

5 0
1 year ago
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