The program will produce lines containing the factors of ONLY perfect numbers, W

Question

The program will produce lines containing the factors
of ONLY perfect numbers, WITHOUT DOUBLE-CALCULATING
(each Perfect Number is found and displayed via the same
calculation, so the program doesn't waste time calculating
the same thing twice. In other words, you build the complete
output line, but you only print it one the program has
established you've found the factors of a Perfect Number.
HINT: sprintf
Sample output:
6: 1 2 3 sum: 6
28: 1 2 4 7 14 sum: 28
496: 1 2 4 8 16 31 62 124 248 sum: 496
8128: 1 2 4 8 16 32 64 127 254 508 1016 2032 4064 sum: 8128

this is what i have so far.

#include<iostream>
using namespace std;
int main()
{
for (int i=1 ; ; i++)
{
char buf[1024], *cp;
cp = buf;
cout << i << endl;
int total = 0;
for (int j = 1; j <= i/2; j++)
if (i % j == 0)
total += j;
if (total == i)
{
//cout << i << ": " ;
//for( int x = 1 ; x < i ; x++)
{
cp += sprintf(cp, "%d", i, ": %d" , total);

//
//cout << x << " ";
//}
//cout << "sum: " << i << endl;
}
}
}

Explanation / Answer

#include using namespace std; void perfect(int); int main() { for(int number=2; number

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