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

  1 #include 
  2 #include 
  3 using namespace std;
  4 using ll = long long;
  5 void exgcd(ll a, ll b, ll &x, ll &y){
  6     if(b == 0){
  7         x = 1, y = 0;
  8         return;
  9     }
 10     exgcd(b,a%b,x,y);
 11     ll t = x;
 12     x = y;
 13     y = t - (a/b) * y;
 14 }
 15 ll gcd(ll a, ll b){
 16     return b == 0?a:gcd(b,a%b);
 17 }
 18 int main(){
 19     ll a, b, c, k;
 20     cin >> a >> b >> c >> k;
 21     ll e = gcd(a,b);
 22     for(int i = 0; i <= k /c; ++ i){
 23         ll d = k - i*c, x, y;
 24         if(d%e)continue;
 25         exgcd(a,b,x,y);
 26         ll n = d/e;
 27         ll m = b/e;
 28         x *= n, y *= n;//还原
 29         x = (x%m + m) % m;//变为正数；
 30         y = (d - x * a)/b;
 31         if(x>=0 && y >= 0){//保证大于零
 32         cout << x << " " << y << " " << i << endl;
 33         break;
 34         }
 35     }
 36     return 0;
 37 }