Extended Euclidean algorithm

For a, B is not all 0, there is an integer x and y make gcd (A, A, a) =x*a+y*b; Integer... That is, you can make it negative.

Code:

1 intEXGCD (intAintBint&x,int&y)2 {3   if(b==0)4   {5x=1;6y=0;7     returnA;8   }9   intR=EXGCD (b,a%b,x,y);Ten   intt=x; Onex=y; Ay=t-a/b*y; -   returnR; -}

extended Euclidean algorithm

　　

First, let's take a look at how he was formed. Push up from the last layer: first b=0 get greatest common divisor a, return a; Now look at a layer, from the R that sentence to continue to run down, you can know, since then, every time from the R sentence to get the greatest common divisor, and then go down to execute ...

For a, b for the two layer, B is a, B greatest common divisor, then there is B = x1*a + y1*b; (i.e. x1=0; Y1=1;)

Similarly, for the three layer of a, B, two layers of B is also this layer a, B greatest common divisor have b ' = x2*a + y2*b; b ' =a%b;   B ' =x1*a ' +y1*b '; B ' =x1*b+y1* (A%B);

---> x1*b+y1* (a%b) = x2*a+ y2*b;

----> (* * *) This step is critical. is to dissolve a%b x1*b+y1* (a-a/b*b) = X2*a + y2*b (except in computer programming)

-----> Merge Similar Terms x1*b + A * y1-(a/b*b) *y1= x2*a+y2*b;

----> b* (x1-a/b*y1) +a*y1=x2*a+y2*b;

----->  x2=y1; y2=x1-a/b *y1; (La-la-la finally came out,,)

That is, the next layer of x is equal to the previous layer of Y, the next layer of Y is equal to the previous layer of the x-a/b* on the Y, here A/b is this layer of ... Although here is the Special column: Inverted three layer interpretation, but because there is no specific x, Y value, so the formula should be universal. This explains the reason of our code, at the same time, from the two layer to know x1=0; Y1=1;

The recursive way of writing first to the bottom, and then by the bottom of the X, y step up to get the top of x, Y, is we want to ask for X, y ...

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Extended Euclidean algorithm