Extended Euclidean algorithm

Prove:

The problem with mathematical language is that the integer x, y makes ax+by=1. It is not difficult to find, if GCD (A, B)!=1 must have no solution, on the contrary, if gcd (A, B) =1 that there must be an integer pair (x, y) satisfies the AX+BY=GCD (A, B), you can use the extended Euclidean algorithm to solve the answer

Suppose we have obtained b*xt+ (a%b) yt=gcd (b,a%b) integer solution XT, YT

The Euclidean algorithm shows that gcd (b,a%b) =GCD (A, B)

∴b*xt+ (a%b) *YT=GCD (A, b) ①

∵a%b=a-(A/b) *b② Note: because in C + +, the number of two int types is not directly divided by special processing, the result is a value of two number of quotient down rounding

∴ ② into ①: a*yt+b* (xt-(A/b) *yt) =GCD (A, B)

The code is as follows:

int extgcd (int A,int b,int &x,int &y) {int d= A; if (b!=0) {    D=extgcd (b,a%b,y,x);    Y-= (A/b) *x;} Else {x=1; y=0;} return D;}

Extended Euclidean algorithm