拓展欧几里得 edgcd 模板+简易推论

LL exgcd(LL a,LL b, LL &x, LL &y) {
	if(b == 0) {
		x=1,y=0;
		return a;
	}
	LL d = exgcd(b, a%b, x, y);
	//x=x1,y=y1
	LL z = x;//z=x1
	x = y;//x=y1
	y = z - y * (a / b);
	return d;
}

求

a

x

+

b

y

=

=

g

c

d

(

x

,

y

)

ax + by == gcd(x,y)

ax+by==gcd(x,y) 的最小整数解

假设

a

x

+

b

y

=

=

g

c

d

(

x

,

y

)

ax+by == gcd(x,y)

ax+by==gcd(x,y)
可得

a

x

+

b

y

=

=

g

c

d

(

y

,

x

m

o

d

y

)

ax+by == gcd(y,xmody)

ax+by==gcd(y,xmody)
构造

b

x

1

+

(

a

m

o

d

b

)

y

1

=

=

g

c

d

(

y

,

x

m

o

d

y

)

bx1+(amodb) y1 == gcd(y,xmody)

bx1+(amodb)y1==gcd(y,xmody)
可得

b

x

1

+

(

a

−

(

a

/

b

×

b

)

)

y

1

=

=

g

c

d

(

y

,

x

m

o

d

y

)

bx1+(a-(a/b \times b)) y1 == gcd(y,xmody)

bx1+(a−(a/b×b))y1==gcd(y,xmody)
化简左项

b

x

1

+

(

a

y

1

−

(

a

/

b

×

b

)

y

1

)

bx1+(ay1-(a/b \times b)y1)

bx1+(ay1−(a/b×b)y1)

=

=

b

x

1

+

a

y

1

−

(

a

/

b

×

b

×

y

1

)

== bx1+ay1-(a/b \times b \times y1)

==bx1+ay1−(a/b×b×y1)

=

=

a

y

1

+

b

(

x

1

−

a

/

b

×

b

×

y

1

)

== ay1+b(x1-a/b \times b \times y1)

==ay1+b(x1−a/b×b×y1)
得

x

=

y

1

,

y

=

x

1

−

a

/

b

×

b

×

y

1

x = y1, y = x1-a/b \times b \times y1

x=y1,y=x1−a/b×b×y1