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sashaice [31]
3 years ago
15

Im trying to call the keys in a dictionary I have called "planet_dict". in order for them to be included in the for loop, the va

lues of the keys have to be within a range of
273 <= x <= 373
I understand how to do this with certain values, but idk how to call them from a dictionary.

Computers and Technology
1 answer:
zavuch27 [327]3 years ago
4 0

You can do something like this. My code iterates through the dictionary keys and then we use that key to get a value. We check if the value is between 273 and 373 and if it is, it's a water planet. My code is just a general idea of what to do. Instead of printing, you could add the key to a list and then print the contents of the list.

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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)

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

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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:

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  • 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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