Bezu Theorem: If ab is an integer, then there must be an integer such that xy ax + by = gcd (a, b)
That is, if ax + by = c solvable, then c must be a multiple of gcd (c necessarily divisible by gcd, c% gcd == 0)
If ax + by = 1 has a solution, then the gcd (a, b) = 1.
Euclid was removed divided by seeking the greatest common divisor ab
gcd(a,b)=gcd(b,a%b)
int gcd(int a,int b){
if(b==0) return a;
return gcd(b,a%b);
}ax + by = gcd (a, b) solvable
Recursive boundary when b = 0, a = gcd function returns a i.e. a * 1 (1,0) when the set of solutions for boundary reaches recursive + b * 0 = gcd
Recursive i.e. gcd not change with the number of layers varies recursive gcd (a, b) = gcd (b, a% b) in a recursive process
ax1+by1=gcd=bx2+(a%b)y2
Simultaneous (a / b) * b + a% b = a to give
x1 = y2, y1 = x2- (a / b) y2 at this time to establish a relationship between this layer and the next layer of recursion, x1 y1 is this layer solution, x2, y2 are the next layer solution
// ax+by=gcd
int exgcd(int a,int b,int &x,int &y){
if(b==0){
x=1,y=0;
return a;
}
int g=exgcd(b,a%b,x,y);
int x2=x,y2=y;
x=y2;
y=x2-(a/b)*y2;
return g;
}Can be expanded Euclidean algorithm ax + by a set of solutions (x0, y0) by = gcd may be negative. So how general solution is obtained by special solution
Suppose x0 + s1, y0-s2 another set of equations for the solution of the original, the simultaneous ax0 + by0 = gcd have a * s1 = b * s2
s1 / s2 = b / a by b / gcd with a / gcd prime order s1 and s2 takes a minimum value, s1 = b / gcd, s2 = a / gcd
Solution is obtained through equation x = x0 + b / gcd * k y = y0-a / gcd * k k is an integer
x is the smallest non-negative integer solution (x% (b / gcd) + b / gcd) / (b / gcd)
If gcd = 1 ab prime i.e., the formula can be simplified
