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### 裴蜀定理(贝祖定理)
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#### 定理
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对任何整数 $a、b$ 和 $m$，关于未知数 $x$和 $y$ 的线性丢番图方程（称为裴蜀等式）：
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$$\LARGE ax+by=m$$
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* 有整数解时当且仅当 $m$是$a$及$b$的最大公约数$d=gcd(a,b)$的倍数。
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* 裴蜀等式有解时必然有无穷多个整数解，每组解$x,y$都称为**裴蜀数**
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* 可用扩展欧几里得算法求解**裴蜀数**
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#### 例子
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例如，$12$和$42$的最大公约数是$6$，则方程$12x + 42y = 6$有解。
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事实上有:
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* $(−3)×12+1×42=6$
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* $4×12+(−1)×42=6$
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* $11×12+(-3)*42=6$
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* ...
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用计算机帮着算一下：
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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int main() {
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for (int x = -100; x <= 100; x++)
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for (int y = -100; y <= 100; y++)
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if (2 * x + y * 7 == 1) cout << x << " " << y << endl;
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return 0;
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}
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```
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#### 证明
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看这个证明可能要好懂点
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例:
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$$\LARGE 2 * x + 4 * y = c$$
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那么$c$无论如何都是偶数，换句话来说，$c$一定是$2$的倍数$2 = gcd(2,4)$
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$$\LARGE 3 * x + 9 * y = c$$
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那么同理$c$一定是$3$的倍数，因为上述等于 : $3 * x + 3 * 3 * y = c$
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$3 * (x + 3 * y) = c$，外面乘了个$3$，那么$c$再怎么都是$3$的倍数，$3 = gcd(3,9)$
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那么这个定理有什么用呢。当$x，y$为整数解的时候，$c$一定是$gcd(a,b)$的倍数，那么我们就可以求得当为整数解的时候，$c$的最小值肯定就是$gcd(a,b)$ //最大公约数的$1$倍。
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[用几何方式证明裴蜀定理](https://www.bilibili.com/video/av200767266/)
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[点到直线距离的公式证明](https://www.toutiao.com/video/6815606337615954439/)
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[练习题 P4549 【模板】裴蜀定理](https://www.luogu.com.cn/problem/P4549)
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#### 实现代码
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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int main() {
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int n;
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scanf("%d", &n);
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int d = 0, a;
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for (int i = 1; i <= n; i++) {
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scanf("%d", &a);
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d = __gcd(d, abs(a));
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}
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printf("%d", d);
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return 0;
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}
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```
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---
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#### 引理
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<font color='blue' size=4><b>给定$a,b$，均是正整数且互质，那么由$ax+by，x>=0,y>=0$不能凑出的最大数是$ab-a-b$;</b></font>
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[证明](https://blog.csdn.net/qq_38973721/article/details/79752734)
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[练习题 蓝桥杯.买不到的数目](http://t.zoukankan.com/qwertiLH-p-9674097.html)
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**解法I：采用引理解题**
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```c++
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#include<bits/stdc++.h>
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using namespace std;
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int main(){
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int a, b;
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scanf("%d %d",&a,&b);
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printf("%d\n",a*b-a-b);
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return 0;
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}
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```
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**解法II:采用爆搜解题**
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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int p, q, res;
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/*
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测试数据：
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4 7
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根据公式，最大无法凑出的是:4*7-4-7=17
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*/
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bool dfs(int x) {
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if (x == 0) return true;
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if (x >= p && dfs(x - p)) return true;
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if (x >= q && dfs(x - q)) return true;
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return false;
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}
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int main() {
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scanf("%d %d", &p, &q);
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for (int i = 1; i <= 1000; i++)
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if (!dfs(i)) res = i;
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printf("%d\n", res);
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return 0;
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}
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```
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---
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