Solution to greatest common divisor-Euclidean algorithm and its expansion (Juchenchu equation)

The most famous of the greatest common divisor is the Euclidean rolling division method, which has two forms (recursion and non-recursion, in fact, any recursion can be written as non-recursive), the following to see the implementation of GCD code:

/*** a B greatest common divisor ***/
LongLonggcdLongLongALongLongb) {
if(b = =0)
returnA
Else
returnGCD (b, a% B);
}evidence Process (excerpt from Wikipedia:en. ) Wikipedia. org/wiki/could conversion method )

Let say

Want to prove

First

Available and known.
Indicates that D is the common cause of b,r, but
So

Available and known.
Indicates that e is the public reason for a, B, but
So

To be informed

The extended Euclidean algorithm is an integer solution such as a * x + b * y = (A, b), which can be modeled after the Euclidean algorithm. The procedure is as follows:

/*** extended Euclidean algorithm a*x + b*y = GCD (A, b) of a set of integer solutions, the result exists in X, y ***/
voidEXTEND_GCD (Long LongA,Long LongB,Longlong& x,Long Long&y) {
        if(b = =0) {
x =1;
y =0;
                return;
        }
EXTEND_GCD (b, a% B, x, y);
        Long Longtmp = x;
x = y;
y = tmp-a/b * y;
}

The above procedure only obtains a set of solutions, it is obvious that the solution is not unique: x increases B, y reduces a must be a set of solutions to the original equation:

A * (x + b) + b * (y-a) = a * x + b * y = (A, b).

However, in application, it is often not so simple that the indefinite equation A * x + b * y = n will be solved many times. This is the time to apply the above algorithm:

1. (A, B), set C = (A, b), if! C|n, there is no integer solution. Because the left and right sides are divided by the C, you can know that the left is an integer, and a non-integer on the other, so the contradiction.
2. Divide the left and right sides at the same time by C, set the new equation to a ' * x + b ' * y = n ', apply the above algorithm to the solution of a ' * x + b ' * y = 1 (by the first step know (a ', B ') = 1). Set the result to X ', Y '.
3. x = X ' * n ', y = y ' * n ' is the equation A * x + b * y = n. This is a good understanding, will be a ' * x + b ' * y = 12 edges simultaneously expand n ' Times on the line.
4. x = X ' * n ' + t * b, y = y ' * n '-T * A (T is an integer) is the original equation A * x + b * y = n all solutions.

linear congruence Equation implementation code:

/* Extended Euclidean theorem extended Euclidean theorem: for two not all 0 integers a, b, there must be a set of solution x, Y, making ax+by==gcd (A, b);    Expand Euclid to achieve the following side of the program, X and y with global variables to save int gcd (int a,int b) {int t,d;        if (b==0) {x=1; y=0; At this time b==0, that is to say gcd (a,0) ==a. The original becomes AX+BY==A=GCD (A, B)-and x==1,y==0 return A; Returns the value of a, B greatest common divisor} d=gcd (b,a%b);    Euclid to seek greatest common divisor application t=x;    X=y;  Y=t-(A/b) *y;     Unknown place 2 return D; Returns the value of a, B greatest common divisor}//Unknown 2 explanation ax+by==gcd (A, B) (1) Description rule, X, Y represents the value at the first recursion, X1,y1 represents the value at the second recursion. Then GCD (A, B) ==gcd (B,A%B), at the same time, are substituting 1, there is ax+by==b*x1+ (a%b) *y1. Transform the right side of the b*x1+ (a%b) *y1==b*x1+ (A-(A/b) *b) *y1==a*y1+b* (x1-(A/b) *y1) and eventually get Ax+by==a*y1+b* (x1-(A/b) *y1) that is:    The x of the previous depth equals the y1 of the next depth, and y of the previous depth is equal to the x1-(A/b) *y1 of the next depth.    * It is important to note that the division used in the derivation above is integer division. Therefore, a set of solutions of the indefinite ax+by==gcd (A, B) are obtained, X, Y. So what should be the solution to the general infinitive ax+by==c? It is very simple, x1=x* (A, B), y1=y* (A, b), a little further, the solution of such a group of solutions is generally not solve any problem, we now have to get all the solution, then what is the solution? Suppose D=gcd (A, B). So x=x0+b/d*t;  Y=y0-a/d*t, where T is an arbitrarily constant integer.  */#include <stdio.h> #include <stdlib.h>/* the minimum positive integer solution for ax≡1 (mod b) of the congruence equation for x. That is to ask ax=mb+R 1=nb+r Deformation ax+ (n-m) b=1, this equation expands Euclid's application ax+by=gcd (A, B), (n-m equivalent to Y) in fact ax≡1 (b) the necessary condition for the solution is gcd (A, a, and A/b), that is, gcd (A, b) = 1; use expand Euclid It is known that Ax+by=1 (x, y is an integer) */int extragcd (int a,int b,int *x,int *y) {int d,t;if (b==0)//recursive call termination condition, when the remainder is 0 stop {*x=1;*y=0, according to Euclid's Euclidean rule ; return A;}     ELSE{D=EXTRAGCD (b,a%b,x,y); t=*x; Based on the value of the next x1,y1, reverse the previous x, Y value *x=*y;*y=t-a/b* (*y); return D;}} int main () {int a,b,x,y;int ans;freopen ("mod.in", "R", stdin), Freopen ("Mod.out", "w", stdout), scanf ("%d%d", &a, &AMP;B); ans=extragcd (a,b,&x,&y); if (ans!=1) return 0;//according to if X is a solution to the equation, then all the solutions of the equation are x+k*b k for integer x=x%b;    guarantees the smallest positive integer solution x, and x belongs to {0,1,2,3...b-1} while (x<=0) x+=b;    printf ("%d\n", X);  return 0;}

Solution to greatest common divisor-Euclidean algorithm and its expansion (Juchenchu equation)