扩展欧几里得算法
int ex_gcd(int a, int b,int *x,int *y){
if(!b){
*x = 1;
*y = 0;
return a;
}
int ret = ex_gcd(b, a % b, y, x);
y = y - a / b * x;
}
证明://gcd(a,b)为a,b 最大公约数
ax + yb = gcd(a,b) ①
bx’ + a%by’ = gcd(b,a%b) ②
由a%b = a / b - b 带入②得:
bx’ + (a / b - b)y’ = gcd(b,a%b) ③
又由 gcd(b,a%b) = gcd(a,b) 整理③得:
ay’ + b(x’-(a / b)y’) = gcd(a,b) ④
于是乎有 x = y’; y = x’ - (a / b) x;
终止条件: b = 0时 x = 1; y = 0;gcd(x,y) = a;