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Old 2010-12-08, 00:20   #10
3.14159
 
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Quote:
Originally Posted by CRGreathouse View Post
Yes. But it's sometimes useful to reduce only partially: (a - Ax)(b - Bx) mod x = a * b mod x. This is used in some efficient algorithms where fully reducing at each step would be more costly.
List a few examples.
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