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Modulus
October 09, 2002 07:33 AM
How can I calculate the modulus of two numbers using only +, -, *, /, sqrt(), sin(), cos(), tan(), log(), and ln()?
October 09, 2002 07:36 AM
How about:
modulus = a - (a / b) * b
Has to be integer division in order to work...
modulus = a - (a / b) * b
Has to be integer division in order to work...
148
October 09, 2002 08:14 AM
modulo simply is the last part you can not divide any more.
so
will do it.
so
modulo(int a, int b){ while (a >= b) { a-=b; } return a; } will do it.
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1,047
October 09, 2002 08:43 AM
unsigned int divide(unsigned int divisor,unsigned int divident,unsigned int* modulus) { unsigned int division = 0; for(int i=31;i>=0;++i) { if(divisor>>i >= divident) { division |= 1<<i; division -= divident<<i; } } if(modulus) { *modulus = divisor; }} does this work actually? just created it from my freaky brain. i know it doesn''t help you much, but i''m interested:D
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384
October 10, 2002 10:13 PM
no, because
1) it''s an infinate loop
2) it doesnt return anything
3) the algo''s wrong (alright it may not be, but hey - you messed up 1 and 2
)
1) it''s an infinate loop
2) it doesnt return anything
3) the algo''s wrong (alright it may not be, but hey - you messed up 1 and 2
)
314
October 10, 2002 10:21 PM
quote:
Original post by Anonymous Poster
How about:
modulus = a - (a / b) * b
Has to be integer division in order to work...
I used this method all of the time in calculator programs...
(modulus of a divided by b) = a - int(a/b)*b
This works on any number.
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1,047
October 11, 2002 03:08 AM
well, its not infinite, just nearly (int does overflow at 2billions:D)
it returned the modulus yet:D but yes, the division i forgot to return..
the division should not get divident<
that should work now.. :D
"take a look around" - limp bizkit
www.google.com
it returned the modulus yet:D but yes, the division i forgot to return..
the division should not get divident<
unsigned int divide(unsigned int divisor,unsigned int divident,unsigned int* modulus) { unsigned int division = 0; for(int i=31;i>=0;--i) { if(divisor>>i >= divident) { division |= 1<<i; divisor -= divident<<i; } } if(modulus) { *modulus = divisor; } return division;} that should work now.. :D
"take a look around" - limp bizkit
www.google.com
If that's not the help you're after then you're going to have to explain the problem better than what you have. - joanusdmentia
My Page davepermen.net | My Music on Bandcamp and on Soundcloud
712
October 11, 2002 03:41 AM
Can we use inverse trig functions?
acos(cos(pi*numerator/denominator))*denominator/pi
acos(cos(pi*numerator/denominator))*denominator/pi
October 11, 2002 06:43 AM
Yes Beerhunter, I'll try that... Anyway, I don't have int() function
.
About Beerhunter's function, it works for the numerator of 2 and 1 but not for 3.
[edited by - Puzzler183 on October 11, 2002 7:49:05 AM]
.About Beerhunter's function, it works for the numerator of 2 and 1 but not for 3
.[edited by - Puzzler183 on October 11, 2002 7:49:05 AM]
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