A summary of number theory-EPIC masterpiece (This is a flag)

1, Fast power

A fast algorithm for calculating a^b, for example, 3^5, we write 5 binary 101,3^5=3^1*1+3^2*2+3^4*1

1 ll Fast (ll A,ll b) {ll ans=1;  for (; b;b>>=1, A=mul (a,a))if(b&1) Ans=mul (ans,a); return ans;} // A quick power to a line

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2. Fast Multiply

When the modulus is large, the direct multiply will explode long long, requiring fast multiplication.

that is, using floating-point computation multiples, the difference is equivalent to the result of calculating the remainder modulus 2^63, and then mold it (because the remainder does not exceed a long long)

1 Long Long ll; 2 ll Mul (ll X,ll y) {return ((x*y-(LL) (((longdouble) x*y+0.5)/ MoD) *mod)%mod+mod)%mod;} // One line Quick multiply

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3, congruence principle (GCD)

Theorem: gcd (A, B) =gcd (b,a MoD)

Proof: A can be expressed as a=kb+r, then r=a MoD b

Assuming D is a number of conventions for a, B, then there are d|a, d|b, and r = a-kb, so d|r, so D is the number of conventions (B,a mod b)

Assuming D is (b,a mod b) The number of conventions, then d|b, D|r, and A=kb+r, so D is also (A, b) of the number of conventions

therefore (b) and (b,a mod b) The number of conventions is the same, and its greatest common divisor is necessarily equal, to be proven.

1 int gcd (A, b) {return !b?a:gcd (b,a%b);} // One line GCD

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4. Throw graph equation

bezout theorem: Drop graph equation (two yuan once, below) Ax+by=m has the solution and only if the m| (A, B)

Extended gcd: An integer solution for AX+BY=GCD (A, A, b) of the equation of loss graphs

Proof: A*X1+B*Y1=GCD (A, B)

b*x2+ (a mod b) *y2=gcd (b,a mod b)

Because gcd (b) =gcd (b,a mod b)

a*x1+b*y1= b*x2+ (a mod b) *y2 = b*x2+ (a-a/b*b) *y2 = a*y2+b* (x2-a/b*y2)

So X1=y2,y1=x2-a/b*y2

End state: B=0,A=GCD (A, B), gcd (A, B) *x=gcd (A, b), x=1

The process of extending Euclid can be understood as the process of backtracking upward from the last state until a set of solutions to the original equation is obtained.

  1  void  exgcd (int  a,int  b,int  &x,int  &y)  2  { 3  if  (B==0 ) {x=1 ; Y=0 ; return  ;}  4  exgcd (b,a%b,x,y);  5  int  t=x;x=y;y=t-a/b*Y;  6 }  

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General solution of the equation of lost graphs: if (b) =d, Ax+by=m has a set of solutions (X0,Y0) General Solution: x=x0+ (b/d) K, y=y0-(A/DK) is to try to get the positive and negative mutual cancellation, ( true • Primary School Olympiad content)

Doubts: Most beginners may have such a question (anyway I have just learned), to solve the equation is ax+by=m, and the above algorithm is the solution of AX+BY=GCD (A, B)

In fact, AX+BY=GCD (A, b) can become AX*K+BY*K=GCD (A, b) *k, so that gcd (a, b) *k=m, find K, and then X*k is the solution of ax+by=m.

5 , multiplication inverse element

Leave a hole.

6. Chinese remainder theorem

Leave a hole.

References--Wang Mengdi Henan Provincial Team Training Lecture notes on the basis of number theory

A summary of number theory-EPIC masterpiece (This is a flag)