Number Theory Summary 2012-09-05

Relatively dull, a little bit number theory took a long time to figure out, a small summary.

① Greatest Common divisor (Euclidean method)

Function gcd (a,b:longint): Longint;

Begin

If B=0 then Gcd:=a

else GCD:=GCD (b,a mod b);

End

② least common multiple

LCM (A, B) *gcd (A, b) =a*b

LCM (A, B) =a*b/gcd (A, b);

③ primes table

Sieve method

④ Prime number Test

Miller-rabbin Test:

If present and n coprime positive integer A satisfies a^ (n-1) =1 (mod n) (Fermat theorem is pushed), then N is a pseudo prime number based on a. The base B (s), which is not more than n-1, is continuously selected to calculate whether b^ (n-1) =1 (mod n) is available each time.    If each time is established, it is almost possible to conclude that N is prime. (from the Contest of Algorithmic Arts and Informatics)

⑤ Extended Euclid

If D=GCD (A, B) then there must be X, y to make ax+by=d.

Function extended_gcd (A,b:longint;var x,y:longint): Longint;

var k:longint;

Begin

If B=0 then BEGIN

Extended_gcd:=a;

x:=1;y:=0;

End

ELSE begin

EXTENDED_GCD:=EXTENDED_GCD (b,a mod b,x,y);

K:=x;

X:=y;

y:=k-(a div b) *y;

End

End

⑥ Fast Power (non-recursive notation)

Function Pow (k,q:longint): Longint;

Begin

Pow:=1;

While q<>0 do

Begin

If q mod 2=1 then pow:=pow*k;

K:=k*k;

Q:=q Div 2;

End

End

⑦ linear modal equation

Solution Ax=c (mod b)

Ax+by=c

Set D=GCD (A, b);

The necessary and sufficient condition for an integer solution to the problem is C mod d=0

Make A=a ' *d, B=b ' *d

Then gcd (a ', B ') =1

The original equation can be converted to a ' x+b ' y=c/d

By extending Euclidean solution a ' x ' +b ' y ' = 1 to derive x ' and Y '

and X=x ' *c/d y=y ' *c/d

But this is just one of the set of solutions MoD b conditions all the Jiewei

xi=x0+i* (B/D) (i=1,2,3,.....,d-1) x0 for any solution

⑧ Linear mode equations

Solve a series of X=ai (mod bi) (2<=i<=n)

Consider the simple case first, n=2.

i.e. X=A1 (mod B1);

X=A2 (mod b2);

namely X=a1+b1*y1

X=a2+b2*y2

A1+b1y1=a2+b2y2-B1Y1-B2Y2=A2-A1

It's easy to transform into a linear mode equation.

Likewise made D=GCD (B1,B2);

The case of the non-solution of the linear equation Group (A2-A1) MoD d<>0

The same linear mode equation extends the value of y1,y2 by expanding Euclidean

The x=a1+b1*y1=a2+b1*y2 also solved the solution of two equations.

The minimum nonnegative integer solution for X is x= ((x mod LCM (B1,B2)) +LCM (B1,B2)) mod LCM (B1,B2);

Generalized to multiple equation sets

The two equations are combined, i.e., the x value of the two equations is Q,

The combined new equation is x=q (mod LCM (B1,B2));

All equations 22 are combined and eventually the final answer is reached.

⑨ Chinese remainder theorem

Also for a series of X=ai (mod bi) (and all bi 22 coprime)

X has a unique solution under the Die B1*b2...bn

B=b1*b2*......*bn

For each bi to find a pi,pi=b/bi

and Pi=1 (mod bi)

That is to solve Pi*xi+bi*yi=1, gcd (PI,BI) = 1, expand the Euclidean solution.

And the final answer is x=p1*x1* A1+p2*x2*a2+......+pn*xn*an

⑩ multiplication Inverse element

Ax=1 (mod b) x exists when and only if GCD (A, b) =1

X is the multiplication inverse of a

In general, when calculating the division of a modulo, it is converted to multiply the inverse of the divisor.

11. About calculating division on the modulo

By Fermat theorem, when a,p coprime a^p=a (mod p)--a^ (p-2) =a^-1 (mod p)

(1/a) mod p ie a^-1 mod p=a^ (p-2) (mod p) so I found (1/a) mod p

12. The modulus of processing is not prime and the division is processed

To pi^ci the modulus into the product of a

Processed separately

Merging the solutions with the Chinese remainder theorem

13. Euler functions

φ (n) is the number of coprime to n in a number less than or equal to N.

N=p1^q1*p2^q2*p3^q3*...*pk^qk (P is an inter-heterogeneous mass factor)

φ (n) =n* (1-1/P1) * (1-1/P2) * (1-1/P3) *...* (1-1/PK)

φ (n) is an integrable function if (a, b) =1 thenφ (AB) =φ (a) φ (b)

Euler's theorem: (A, a) =1 a^φ (b) =1 (mod b)

14. Approximate and

To the approximate and

A=p1^k1*p2^k2*...*pn^kn

S = (1+P1+P1^2+P1^3+...P1^K1) * (1+P2+P2^2+P2^3+....P2^K2) * (1+p3+ p3^3+...+ p3^k3) * .... * (1+PN+PN^2+PN^3+...PN^KN)

Number Theory Summary 2012-09-05