Calculation Problem

Example: 28 = 1 + 2 + 4 + 7 + 14 ==> if one fully paid equal to the sum of its divisor is

my c-code this:

Code:

# Include <stdio.h> 
# Include <conio.h> 

int main () 
( 
 long z, i, sum = 0; 

 for(z = 1, z <= 1000000; z + +) 
 ( 
  for(i = 1; i <z; i + +) 
  ( 
   if(z% i == 0) sum + = i; 
  ) 
  if(sum == x) printf ("% \ n ld", z); 
  sum = 0; 
 ) 

 return 0; 
)

now I have the problem that it takes very long (on my athlon 1800 + I got after 5 min terminated because he was only at about 80 000)

can this be done efficiently?

 

Code:

for (i = 1; i <z; i + +) 
if (z% i == 0) sum + = i; 

looks like for Maich

Code:

for (i = 1, i * i <z; i + +) 
if (z% i == 0) (sum + = i; sum + = z / i;) 
if (z% i == 0) sum + = i;

 

is the last line as aware of?

 

Code:

for (z = 1, z <= 1000000; z + +) 
 ( 
  for (i = 1; i <z; i + +) 
  ( 
   if (z% i == 0) sum + = i; 
  ) 
  if (sum == x) printf ("% ld \ n", z); 
  sum = 0; 
 )

You could be in the "i loop" Check whether the sum is too large and, if yes cancel.

Code:

for (i = 1; i <z; i + +) 
  ( 
   if (z% i == 0) 
     (Sum + = i; 
       if (sum> z) = z i / / or break, (I do not break :-)) 
     ) 
  )

I too have to believe that the i-loop search only to z / 2:
for (i = 1, (i * 2) <= z; i + +)

and I would look for from big to small
for (i = (z +1) / 2; i> 0, - i) / / the z +1 to be sure,
/ / is that even the half of tested

But then I must terminate with i = 0 to change

 

Code:

# Include <stdio.h> 
# Include <math.h> 
int main () 
( 
  long z, i, sum = 1, / / there is always a divisor 

  for (z = 1, z <= 1000000; z + +) 
    ( 
      for (i = sqrt (z), i> 1, - i) / / without a splitter to the root 
        (If (z% i == 0) 
             (Sum + = i; 
               sum + = z / i / / splitter, each has its "co-parter" 
               if (sum> z) i = 0; / / demolition as the sum too large 
             ) 
        ) 
       if (sum == x) printf ("% ld \ n", z); 
       sum = 1, / / there is always a divisor 
    ) 

  return 0; 
)

NEN on Pentium 450s in about two minutes the whole Mille calculated through
but only when the figures found

if I do not know all of the

 

That's all. But since you'll have to come up, many time-saving things, so you can find as many. The fifth is 33,550,336, and already the sixth 8589869056 ...

 

It may interest one or the other too:

Perfect numbers are always a sum of consecutive numbers, eg:
6 = 1 +2 +3,
28 = 1 +2 +3 +4 +5 +6 +7,
496 = 1 +2 +3 +4 +5 +...+ 30 +31
8128 = 1 +2 +3 +4 +5 +...+ 126 +127

Euclid proved using the geometric series:

n for some numbers p = 1 +2 +4 +8 +...+ 2 ^ n = 2 ^ (n +1) -1 is a prime number.
In each such case, 2 ^ n · p is perfect.
6 = 2 * (2 ² -1)
28 = 2 ² * (2 ³ -1)

It was later proved by Euler, that this rule all even, perfect numbers provides - if you find a suitable prime. Therefore, virtually every new large prime Fund is associated with a further big perfect number.

The question of whether it is also odd perfect numbers, is still unclear. In case of existence of these would need to be larger than 10 ^ 100 and have at least 11 different prime divisors.

 

Yes.
Top I add the paired dividers.
So for 36 I'll do as
2 18
3 12
4 9
and down again I'll do the 6
and the one I've forgotten.

yes, if every perfect number ne triangular number must be, because then you have still NEN fine accelerator.

 

f I want to study up to z / 2 why I did then as a termination condition not> or <z / 2 z and square root but i?

 

because sqrt (z) is already sufficient.
z / 2 would be a waste of time.

dividers for each i with z% i == 0, the opposites meet g = z and z% i% g == 0. so you only need to go to root, as would i == g. Otherwise, is always a small and a larger than the root.

 

creative, Make sure you understand the difference between CODE & QUOTE tag. Refer http://www.cfanatic.com/misc.php?do=bbcode if you have any doubts.

 

but why?
As I see, all tags are correct in my post...

 

I did that for you.