The sum of squared digits

This is a discussion on The sum of squared digits within the C++ Programming forums, part of the General Programming Boards category; I currently want to use the 7 digits: 1234567 to make the sum of squared digits so 1^2+2^2+3^2+4^2+5^2+6^2+7^2=140 I have ...

  1. #1
    T-Mac wiz23's Avatar
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    The sum of squared digits

    I currently want to use the 7 digits: 1234567 to make the sum of squared digits so 1^2+2^2+3^2+4^2+5^2+6^2+7^2=140

    I have already done this code as follows:
    b = (id/1000000)^2 + ((id%1000000)/100000)^2 +
    ((id%100000)/10000)^2 + ((id%10000)/1000)^2 +
    ((id%1000)/100)^2 + ((id%100)/10)^2 + (id%10)^2;

    b should equal to 140, but I only get the sum of 2. Can somebody please help me.

  2. #2
    Registered User Micko's Avatar
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    a^b is not power in C/C++, it's a bitwise operation XOR
    Read this:
    bitwise opearation

    You should use function pow() declared in cmath header
    Last edited by Micko; 04-10-2005 at 03:17 AM.
    Gotta love the "please fix this for me, but I'm not going to tell you which functions we're allowed to use" posts.
    It's like teaching people to walk by first breaking their legs - muppet teachers! - Salem

  3. #3
    T-Mac wiz23's Avatar
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    Is this what you mean?
    #include <cmath>

    I have done this as follows:
    b = (id/1000000)pow(1,2) + ((id%1000000)/100000)pow(1,2) + ((id%100000)/10000)pow(1,2) + ((id%10000)/1000)pow(1,2) + ((id%1000)/100)pow(1,2) + ((id%100)/10)pow(1,2) + (id%10)pow(1,2);

    I think is wrong, so can you please give me an example. I am new to c++ so I need to look at some example so I can get a better undstanding. Thanx.

  4. #4
    Chief Code Coloniser!
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    OI! Ease up on the double posting

    (other thread)

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    & the hat of GPL slaying Thantos's Avatar
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    Locked due to double post
    Th power function

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