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garik1379 [7]
3 years ago
5

Consider a variation of Sequential Search that scans a list to return the number of occurrences of a given search key in the lis

t. Does its efficiency differ from the efficiency of classic Sequential Search? Explain.
Computers and Technology
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

Considering the fact that the sequential search key must go through the entire list, The search is no different from a classic sequential search.

Explanation:

Sequential search is a form of search in computer science that searches for an item by comparing sequentially items in an ordered list. In big O notation, it take the sequential search O(n) time, with n being the number of items in the list.

This is the same as searching for the occurrence of an item multiple times in the list, as it must go through the entire list items.

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Write a program that helps a young student learn to make change. Use the random number generator to generate change amounts betw
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Answer:

#include <iostream>

#include <stdlib.h>

using namespace std;

int main()

{

srand(time(0));

 

int coins = rand()%99 + 1;

int temp = coins;

cout<<"Enter the coins needed to make $0."<<coins<<"?"<<endl<<endl;

int a, b, c, d;

cout<<"Quarters? ";

cin>>a;

cout<<"Dimes? ";

cin>>b;

cout<<"Nickles? ";

cin>>c;

cout<<"Pennies? ";

cin>>d;

 

int quarters = coins/25;

 

coins = coins%25;

 

int dimes = coins/10;

 

coins = coins%10;

 

int nickles = coins/5;

coins = coins%5;

 

int pennies = coins/1;

 

if(((a*25) + (b*10) + (c*5) + (d*1))==temp)

{

cout<<"That's correct"<<endl;

 

if(a!=quarters || b!=dimes || c!=nickles || d!=pennies)

{

cout<<"The optimized solution is "<<endl;

}

}

else

{

cout<<"Not correct"<<endl;

cout<<"The solution is "<<endl;

 

}

 

cout<<"\tQuarters: "<<quarters<<endl;

cout<<"\tDimes: "<<dimes<<endl;

cout<<"\tNickles: "<<nickles<<endl;

cout<<"\tPennies: "<<pennies<<endl;

return 0;

}#include <iostream>

#include <stdlib.h>

using namespace std;

int main()

{

srand(time(0));

 

int coins = rand()%99 + 1;

int temp = coins;

cout<<"Enter the coins needed to make $0."<<coins<<"?"<<endl<<endl;

int a, b, c, d;

cout<<"Quarters? ";

cin>>a;

cout<<"Dimes? ";

cin>>b;

cout<<"Nickles? ";

cin>>c;

cout<<"Pennies? ";

cin>>d;

 

int quarters = coins/25;

 

coins = coins%25;

 

int dimes = coins/10;

 

coins = coins%10;

 

int nickles = coins/5;

coins = coins%5;

 

int pennies = coins/1;

 

if(((a*25) + (b*10) + (c*5) + (d*1))==temp)

{

cout<<"That's correct"<<endl;

 

if(a!=quarters || b!=dimes || c!=nickles || d!=pennies)

{

cout<<"The optimized solution is "<<endl;

}

}

else

{

cout<<"Not correct"<<endl;

cout<<"The solution is "<<endl;

 

}

 

cout<<"\tQuarters: "<<quarters<<endl;

cout<<"\tDimes: "<<dimes<<endl;

cout<<"\tNickles: "<<nickles<<endl;

cout<<"\tPennies: "<<pennies<<endl;

return 0;

}

Explanation:

  • Use the built-in rand function to generate a random coin.
  • Get all the required values for quarters, dimes, nickles and pennies and store them in variables.
  • Check if the coins give a value of temp, then answer is correct and the values of user will be added.
  • Optimum solution is displayed  If it does not matches with the optimum solution. Display that the answer is wrong along with solution.
6 0
3 years ago
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