The solution of the number theory study note linear equation a*x + b*y = gcd (A, B)

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Cough cough!!! Write this note today, in case the dementia will not solve the equation later!!!

Begin!

~1~, first of all, saw a gcd (a, B), what the hell is this thing? What kind of thing is not important, the important thing she represents, in fact, GCD (A, A, b) represents a non-negative integer A and a (not 0) of the greatest common divisor, (number theory in general: Calculation of A and B of the maximum common factor of the lower effective method is my daughter's four-year teacher teaching method, the teacher asked the students to find Factor, and then find the largest number that appears in two tables at the same time. yes! A Good idea for elementary school students! ） 。 Now, we can't do that!

All right! Start with gcd (A, B)!

The following are the general formulas:

A = b * q1 + R1;

b = R1 *q2 + R2;

R1 = R2 * Q3 + R3;

R2 = R3 * Q4 + R4;

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When is this writing going to be the head ~_~ ...

Obviously there must be a place for the end, because there will definitely be a R for 0 (R2 >= R4 >= R6 ...), R1 >= R3 >= R5 ...), OK, keep writing down:

Rn-1 = Rn * qn+1 + rn+1;

Rn = rn+1 * qn+2 + rn+2; Suppose rn+2 = = 0

。。。。。。。。。。。。。 Ok! ^_^ ........

rn+2 = 0, then by the last formula Rn+1 divisible by Rn, then, rn+1 divisible Rn-1,,,, it is clear that this recursion can be pushed to the first formula, So,rn+1 division a,rn+1 divisible by B, that is, Rn+1 is a and B convention number, is not the largest It? First, write rn+1 as G, set D as any of the conventions of A and B, by the first

The formula (a = b * q1 + R1) can be known, D is divided into R1, and then into the second formula, D divide R2, push, push, push, push, and finally know that D evenly divisible by rn+1, that is, d is evenly divisible by G,

Because d is an arbitrary number of conventions, D is divisible by G, then D is greatest common divisor, then G is both the number of conventions and multiples of greatest common divisor, only: g is greatest common divisor.

in front ",,,, recursion. "Hidden a big secret, that is gcd (A, b) = gcd (b,a%b), this secret Ah, very wonderful!" Why do we have this conclusion? From the first formula ( < Span style= "color: #000000; font-size:16px; " > a = b * Q1 + R1 ) Obviously, gcd (A, b) = g; Again obviously gcd (b,r1) = g; Obviously, GCD (A, b) = gcd (b,a%b),

。。。。。。。。。。。 Ok. The problem was solved. The code is attached ....... .....

1 int gcd (int A,int  b)2{3      return b==0a:gcd (b,a%b); 4 }

GCD

~2~, Well, in fact, the front is just chatter! But I can bet five cents in front of things no tricky!

The following is a positive solution to this problem: a*x + b*y = gcd (A, b);

The following are the general formulas:

A = b * q1 + R1; R1 = a-q1 * b;

b = R1 *q2 + R2; R2 = b-q2 * R1;

R1 = R2 * Q3 + R3; R3 = r1-q3 * R2;

R2 = R3 * Q4 + R4;

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It is worth noting that the right side of the equation. It can be found that each R can be represented as multiples of A and B (R1 (1,-Q1)), and a struct A, B (OK, two) can be defined.

A.x, A.y respectively represents a, b coefficients, Initializes a = a = (1,0), b = b = (0,1), then R1 = a-q1*b; You can see that the R1 is also used below, and you can't use a.

Just make A = R1 much better! A = a-b*q1; See R2 again, R2 = B-Q2*R1 = B-q2*a; The same makes B = R2; b = b-a*q2; below R3 = A-q3*b;

A = a-q3*b; then: B = b-a*q ... Wait, when does it end? Before "Look at R2 again", consider such a question, if R1 = = g? Obviously this time will be terminated, pseudo code as follows:

while (1) {

Q = A/b;

R = a%b;

A = A-b*q;

(A, b) = (b,a%b);

(A, b) = (b,a)//This makes it easier to write code

if (a%b = = 0) break;

}

: 0 requires special award.

The code is as follows:

1 voidSolintAintBint&g)2 {3a.x = B.Y =1 ;4A.Y = b.x =0 ;5     if(A = =0) {6g =b;7a.x =1;8A.Y =0;9         return ;Ten     } One     if(b = =0) { Ag =A; -a.x =0; -A.Y =1; the         return ; -     } - Node C; -     intQ, R; +      while(1) { -Q = A/b; +r = a%b; Aa.x = a.x-b.x*Q; atA.Y = a.y-b.y*Q; -A =b; -b =R; -         if(A%b = =0) Break; -C =A; -A =B; inB =C; -     } tog =b; +}

EX_GCD

~3~. Not to be continued.

The solution of the number theory study note linear equation a*x + b*y = gcd (A, B)