欧几里得算法

1.最大公约数

设：$gcd(a, \ b) = k$

$$a = kc,\ b = kd（c∈N^*, d∈N^*）$$
$$∴c、d互质（c、d的公约数只有1）且c > d$$
$$∴a - b = k(c - d)$$
$$∴gcd(b, \ a - b) = gcd(kd, \ k(c - d))$$

假设 d、(c - d)不互质
即 $d|(c - d)$
∴ 设$c - d = k'd$
∴ $c = d(k' + 1)$
与题设“c、d互质”不符，
∴ d与(c - d)互质

$$∴gcd(kd,\ k(c - d)) = k$$
$$∴gcd(a,\ b) = gcd(b,\ a - b)$$
$$∴gcd(a,\ b) = gcd(b,\ a\ \%\ b)$$

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2.ax + by = gcd(a, b)

$$ax + by = gcd(a,\ b)$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ = gcd(b,\ a \ \%\ b)$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ = bx' + (a\ \%\ b)y'$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ = bx' + (a - [a / b] * b)y'$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ = bx' + ay' - [a / b] * by'$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ = ay' + b(x' - [a / b] * y')$$

$$∴x = y',\ y = x' - [a / b] * y'$$

终止条件：
$b = 0$时：$a * 1 + b * 0 = a$
即$x = 1,\ y = 0$
递归求解

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3. ax + by = c

用扩展欧几里得先求出$ax' + by' = gcd(a,\ b)$的一组解, x’, y’及gcd(a, b)的值

若$c\ mod\ gcd(a,\ b) ≠ 0$
方程无解（整数范围内）

令：

$$c = gcd(a,\ b) * k$$
$$∴k = c / gcd(a,\ b)$$
$$∴ax + by = c$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = k * gcd(a,\ b)$$
$$∴ax + by = akx' + bky'$$
$$∴ax = akx', by = bky'$$
$$∵a != 0 且 b != 0$$
$$∴x = kx', y = ky'$$
$$∵k = c / gcd(a,\ b)$$
$$∴x = x' * c / gcd(a,\ b)$$
$$\ \ \ y = y' * c / gcd(a,\ b)$$

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4. x的非负最小值

用扩展欧几里得先求出$ax' + by' = gcd(a,\ b)$的一组解,
$x'$, $y'$及$gcd(a,\ b)$的值

$$lcm(a,\ b) = a * b / gcd(a,\ b)$$

$$ax + lcm(a,\ b) + by - lcm(a,\ b) = c$$
$$ax + a * b / gcd(a,\ b) + by - a * b / gcd(a,\ b) = c$$
$$a(x + b / gcd(a,\ b)) + b(y - a / gcd(a,\ b)) = c$$
$∴$$x$ $+$ 或 $-$ 任意倍数的$b / gcd(a,\ b)$均有对应的y的整数解

$$设 t = b / gcd(a,\ b)$$
$x = ((x' * c / gcd(a,\ b))\ \%\ t + t)\ \%\ t$ 为方程的最小非负解.