拓展欧几里得

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;

#define ll long long

const int INF = 1e9;
const ll MOD = 19940417;

//裴蜀定理：若 ax + by = n 有解，则 gcd(a， b) | n
//拓展欧几里得：求 ax + by = gcd(a, b) 的解，通过辗转相除法得到一组特解：x0 y0
// 得到 ax + by = c 的特解：
// x = x0 * c / gcd(a, b)
// y = y0 * c / gcd(a, b)
// 通解:
// X = x + b / gcd(a, b) * t
// Y = y - a / gcd(a, b) * t   (t = 0, 1, 2, 3 ...)
//最小非零解： x = (x % b + b) % b
// ax + by = gcd(a, b)
// ax + by + k * ab - k * ak = gcd(a, b)
// a(x + kb) + b(y - bk) = gcd(a, b)  ---->  x = (x % b + b) % b

ll exgcd(ll a, ll b, ll &x, ll &y)
{

    if(b == 0){//推理1，终止条件
        x = 1;
        y = 0;
        return a;
    }
    ll r = exgcd(b, a%b, x, y);
    //先得到更底层的x2,y2,再根据计算好的x2,y2计算x1，y1。
    //推理2，递推关系
    ll t = y;
    y = x - (a/b) * y;
    x = t;
    return r;
}

void solve()
{

    //ax同余1(mod prime)  求最小正整数解
    // ax + by = n ---> ax' + by' = gcd(a, b)
    // k = gcd(a, b)  k*(ax' + by') = n
    //以下 a, b, x, y对应上面 a, b, x', y'
    ll a, b, x, y;
    //以下求出mod不为质数的逆元
    // a / b % mod c
    // a / b * b^-1 ≡ x * b^-1(mod c)
    // b * b^-1 ≡ 1 (mod c)
    // b * b^-1 + y * c  = 1
    // a * x + b * y = 1
    exgcd(2, MOD, x, y);
    ll inv2 = x;
    exgcd(6, MOD, x, y);
    ll inv6 = x;

    cout << "inv2 = " << (inv2 + MOD) % MOD << endl;

    cout << "inv6 = " << (inv6 + MOD) % MOD << endl;
}

int main()
{

    solve();

    return 0;
}