遇到一道扩展欧几里得题（个人觉得）

但是他们说是同余最短路问题 ，我不知道这个算法，以后在学吧。

题目：https://ac.nowcoder.com/acm/contest/4853/D

#include <iostream>
#include <algorithm>
using namespace std;
using ll = long long;
void exgcd(ll a, ll b, ll &x, ll &y){
    if(b == 0){
        x = 1, y = 0;
        return;
    }
    exgcd(b,a%b,x,y);
    ll t = x;
    x = y;
    y = t - (a/b) * y;
}
ll gcd(ll a, ll b){
    return b == 0?a:gcd(b,a%b);
}
int main(){
    ll a, b, c, k;
    cin >> a >> b >> c >> k;
    ll e = gcd(a,b);
    for(int i = 0; i <= k /c; ++ i){
        ll d = k - i*c, x, y;
        if(d%e)continue;
        exgcd(a,b,x,y);
        ll n = d/e;
        ll m = b/e;
        x *= n, y *= n;//还原
        x = (x%m + m) % m;//变为正数；
        y = (d - x * a)/b;
        if(x>=0 && y >= 0){//保证大于零
        cout << x << " " << y << " " << i << endl;
        break;
        }
    }
    return 0;
}