数论-扩展欧几里得算法

扩展欧几里得算法

已知(a, b)求解(p, q), 使得p * a + q * b = GCD(a, b)；

一定有解;

因为GCD(a, b) = GCD(b, a%b)

所以p * a + q * b = GCD(a, b)

= GCD(b, a%b)

= p * b + q * a % b

= p * b + q * (a - a/b * b)

= p * b + q * a - q * a/b * b

= q * a + (p - a/b * q) * b

就把a与b的线性组合转换为b，a%b的线性组合

不断缩小，当b=0时，p=1，q=1

递归求解

【code】ax+by=gcd(a,b) 求解x，y

#include <iostream>using namespace std;//Extended_GCDint Extended_GCD(int a, int b, int &p, int &q) {if(b == 0) {p = 1;q = 0;return a;}int gcd = Extended_GCD(b, a%b, p, q), tmp;tmp = p;p = q;q = tmp - a/b * q;return gcd;}int main() {int gcd, a, b, x, y;cin >> a >> b;gcd = Extended_GCD(a, b, x, y);cout << gcd << ' ' << x << ' ' << y << endl;return 0;}