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Calculation Problem

Discussion in 'C' started by inspiration, Jun 13, 2010.

  1. inspiration

    inspiration New Member

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    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?
     
  2. techme

    techme New Member

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    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;
     
  3. inspiration

    inspiration New Member

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    is the last line as aware of?
     
  4. creative

    creative New Member

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    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
     
  5. creative

    creative New Member

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    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
     
  6. meyup

    meyup New Member

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    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 ...
     
  7. meyup

    meyup New Member

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    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.
     
  8. techme

    techme New Member

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    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.
     
  9. inspiration

    inspiration New Member

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    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?
     
  10. techme

    techme New Member

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    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.
     
  11. NewsBot

    NewsBot New Member

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  12. creative

    creative New Member

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    but why?
    As I see, all tags are correct in my post...
    :)
     
  13. NewsBot

    NewsBot New Member

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    I did that for you.
     

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