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Artyom0805 [142]
3 years ago
9

Assuming P != NP, which of the following is true ?

Computers and Technology
2 answers:
7nadin3 [17]3 years ago
8 0

Answer:

B. NP-complete \cap P = \Phi

Explanation:

The is because no NP-Complete problem can be solved in polynomial time.

Assuming one NP-Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time. If that is the case, then NP and P set become same which contradicts the given condition.

White raven [17]3 years ago
7 0

Answer:

Answer is B, see explanations.

Explanation:

Answer is (B) NP−complete∩P=ϕ

Since, P≠NP, there is at least one problem in NP, which is harder than all P problems. Lets take the hardest such problem, say X. Since, P≠NP,X∉P .

Now, by definition, NP−complete problems are the hardest problems in NP and so X problem is in NP−complete. And being in NP, X can be reduced to all problems in NP−complete, making any other NP−complete problem as hard as X. So, since X∉P, none of the other NP−complete problems also cannot be in P.

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#include <iostream>

#include <stdlib.h>

using namespace std;

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

int a, b, c, d;

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cout<<"Nickles? ";

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int quarters = coins/25;

 

coins = coins%25;

 

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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)

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}

else

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cout<<"Not correct"<<endl;

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

 

}

 

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

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

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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.
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