Extended Euclid:
Before looking at the extended Euclidean we should understand gcd, which is the greatest common divisor algorithm
Let us look at the following code
int gcd(int a,int b){
if(b==0) return a;
else return gcd(b,a%b);
}
Of course it is written more simply written ternary operator, reducing the code length
int gcd(int a,int b){ return b==0?a:gcd(b,a%b); }
This method becomes Euclidean
The following reasoning
Two numbers are given assuming a = 16, b = 12;
Seeking the greatest common divisor of two numbers
Start the simulation calculation process {
b!=0
Returns a = 12, b = 4;
b!=0
Returns a = 4, b = 0
b=0
Gives a = 4
Indeed the greatest common divisor
}
That mathematical proof of it
Provided a / b == k .... r thus be obtained r = ak * b; where a is the dividend, b is the divisor, k is the supplier, r is the remainder
Setting c = gcd (a, b)
I.e., c is the greatest common divisor of a and b
We may assume that a = m * c, b = n * c;
Available equation r = ak * b = mc-k * nc = c (m-kn);
Then lists gcd (b, r);
Therefore, r = c (m-kn);
C can be obtained by a factor of r
Available m-kn with n mutual contradiction by the quality
Available gcd (b, r) = c;
Therefore, gcd (a, b) = gcd (b, r)
QED
Wonderful wonderful ah ~ ~ ~ ~ ~ ~
Then get to the point that
Extended Euclid
Algorithm itself can be used to solve variable equation
It can also be used to inverse yuan
Also can be used to solve linear congruence equation
First introduced first use
1. Indefinite extension Euclid solving equations
I.e. dualistic simple indeterminate equation ax + by = m;
But we can change the formula it (anyway formula ignorant)
According to Baidu available
ax + by = gcd (a, b) a solution, but zkc tell us chiefs solution is not unique
Provided ax + by = gcd (a, b) * k
得 ax+by=d, gcd(a,b)|d;
gcd (a, b) | d denotes d divisible gcd (a, b);
When x = 1, y = 0 when
a=gcd(a,b);
我们可以靠递归退回x,y到本身
即可得出答案
网上大佬的论证
当 b=0 时，gcd(a,b)=a，此时 x=1 , y=0
当 b!=0 时，
设 ax1+by1=gcd(a,b)=gcd(b,a%b)=bx2+(a%b)y2
又因 a%b=a-a/b*b
则 ax1+by1=bx2+(a-a/b*b)y2
ax1+by1=bx2+ay2-a/b*by2
ax1+by1=ay2+bx2-b*a/b*y2
ax1+by1=ay2+b(x2-a/b*y2)
解得 x1=y2 , y1=x2-a/b*y2
因为当 b=0 时存在 x , y 为最后一组解
而每一组的解可根据后一组得到
所以第一组的解 x , y 必然存在
代码
void exgcd(int a,int b){
if(b==0){
x=1;
y=0;
return;
}
exgcd(b,a%b);
k=x;
x=y;
y=k-a/b*y;
return;
}
end;
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