Question

Write a program which given an integer n, classifies it as 'perfect', 'abundant' or 'deficient'. A number is perfect if the sum of its divisors ( excluding the number itself) is equal to the number, e.g. 6. A number is abundant if the sum of its divisors(excluding the number itself) is greater than the number, e.g. 12. A number is deficient if it is neither perfect nor abundant

Answer 1

What do you need help with in this problem?

Answer 2

Give at least a pseudocode so that I can understand what to do

Answer 3

you should try this program out yourself first, before asking people to do it for you.

Answer 4

#include <iostream>
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using namespace std;
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int main()
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{
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int number,counter,total;
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/* Asks for user input*/
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cout << "Welcome to the program Factors, this program check to see\n"
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<< "if the number you enter is deficent, perfect or abundant\n"
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<< "this program will read in numbers till you enter -1 and or negitive numbers";
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while (number > -1)
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{
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counter = 1;
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cout << endl;
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cout << "Please enter a Integer:\n";
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cin >> number;
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cout << endl;
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if (number % counter == 0)
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{
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total= counter + (number % counter);
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cout << total << "+";
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counter++;
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}
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if (number < total)
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{
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cout << endl;
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cout << number <<" is abundant\n";
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}
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else if (number>total)
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{
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cout << endl;
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cout << number << " is deficient\n";
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cout << endl;
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}
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else
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{
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cout << endl;
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cout << number << " is Perfect\n";
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cout << endl;
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}
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}
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cout<< endl << endl << endl;
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system ("pause");
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return 0;
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}

Answer 5

errrors?

Answer 6

Yeah

Answer 7

I mean, what are they?

Answer 8

its inaccurate

Answer 9

I think something is wrong with my values?

Answer 10

But i cannot put my finger on it...

Answer 11

The thing which is annoying is you have to add up EVERY possible combination. So for 12 it's 3 + 4 + 2 + 6....

Answer 12

its like adding prime factors of 12 excluding itself?

Answer 13

so if it is working on a number that isnt perfect
It will work through the IF statements to see if it is abundant and another to see if it is deficient

Answer 14

I will post the code properly

Answer 15

#include <iostream>
using namespace std;
int main()

{
int number = 0,counter,total;

/* Asks for user input*/

cout << "Welcome to the program Factors, this program check to see\n"

<< "if the number you enter is deficent, perfect or abundant\n"

<< "this program will read in numbers till you enter -1 and or negitive numbers";


while (number > -1)

{

cout << endl;

cout << "Please enter a Integer:\n";

cin >> number;

if (number < 0) { break; }

cout << endl;



counter = 1; total = 0;

while (counter < number) {

if (number % counter == 0) {

cout << total << "+" << counter;

total += counter;

cout << "=" << total << endl;

}

counter++;

}



if (number < total){

cout << endl;

cout << number <<" is abundant\n";

} else if (number>total) {

cout << endl;

cout << number << " is deficient\n";



cout << endl;

} else {

cout << endl;

cout << number << " is Perfect\n";

cout << endl;

}



}



cout<< endl << endl << endl;

system ("pause");

return 0;

}

Answer 16

#include <iostream>

using namespace std;

/*-------------
-------------*/


int main ()
{
int highNum, abunTotal, defTotal, perfTotal;

abunTotal = 0; // sum of divisors more than the number itself
defTotal = 0; // sum of divisors less than the number itself
perfTotal = 0; // sum of divisors equal to the number itself
int sum = 0;

cout<< "Please enter a positive integer:";
cin>> highNum;

for(int i = 1; i <= highNum; i++ )
{
sum = 0;
for(int j = 1; j < i; j++ )
{
if((i % j) == 0)
{
sum += j;
}
}

if ( sum == i) //the number is perfect
{
perfTotal++;
cout << "PERFECT NUMBER: " << i << endl;
}

if ( sum < i) //the number is deficient
{
defTotal++;
}

if ( sum > i) //the number is abundant
{
abunTotal++;
}
}

cout << "The perfTotal is:" << perfTotal << endl;
cout << "The defTotal is:" << defTotal << endl;
cout << "The abunTotal is:"<< abunTotal << endl;
cout << endl;
}