J - 青蛙的约会（扩展欧几里得）

https://vjudge.net/contest/218366#problem/J

第一步追及公式要写对：y+nk-(x+mk)=pL => (n-m)k+lp=x-y

可以看出扩展欧几里得原型，这里注意扩展欧几里得求出的是任意解，非最优，要推出最小解k。

(n-m)x+ly=gcd => (n-m)(x*(x-y)/gcd) + l*y*(x-y)/gcd = x-y

则k = x*(x-y)/gcd（某一解非最小），由于k每次可转移t = l/gcd

最小解为(k%t+t)%t。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cstdlib>
 #include<cmath>
 #include<vector>
 #include<stack>
 #include<queue>
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 #define IO ios::sync_with_stdio(false);cin.tie(0);
 #define INF 0x3f3f3f3f
 #define MAXN 500010
 const int MOD=1e9+;
 typedef long long ll;
 using namespace std;
 ll ext_gcd(ll a, ll b, ll &x, ll &y)
 {
     if(b == ){
         x = ;
         y = ;
         return a;
     }
     ll r = ext_gcd(b, a%b, x, y);
     ll t = x;
     x = y;
     y = t-(a/b)*y;
     return r;
 }
 int main()
 {
     ll n, m, x, y, l, k, p;
     cin >> x >> y >> m >> n >> l;
     ll r = ext_gcd(n-m, l, k, p);//k和p为任意解，非最优
     if((x-y)%r!=){
         cout << "Impossible" << endl;
     }
     else{
         ll t = l/r;//k每次转移量
         ll ans = k*(x-y)/r;//从扩展欧几里得标准公式转到我们所列公式求得的k值
         cout << (ans%t+t)%t << endl;
     }
     return ;
 }