/* 扩展欧几里得算法 给予二整数 a 与 b, 必存在有整数 x 与 y 使得ax + by = gcd(a,b) 设aX' + bY' = gcd(a, b) bX'' + (a mod b)Y'' = gcd(b, a mod b) bX'' + (a mod b)Y'' = gcd(a, b) bX'' + (a mod b)Y'' = aX' + bY' bX'' + [a - b * (a / b)]Y'' = aX' + bY' bX'' + aY'' - b * (a / b)Y'' = aX' + bY' aY'' + b[X'' - Y'' * (a / b)] = aX' + b Y' X' = Y'' Y' = X'' -Y'' * (a / b) */ public class Main { public static void main(String[] args) { System.out.println(); } static int x, y; static int extgcd(int a, int b) { int d = a; if (b == 0) { x = 1; y = 0; } else { d = extgcd(b, a % b); int t = x; x = y; y = t - y * (a / b); } return d; } }