拓展欧几里得算法

int extgcd(ll a,ll b,ll& x,ll& y)
{
    ll d=a;
    if(b!=0)
    {
        d=extgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }
    else x=1,y=0;
    return d;

#include<iostream>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mod 1000000007
ll x,y;
ll m,n,L;
int extgcd(ll a,ll b,ll& x,ll& y)
{
    ll d=a;
    if(b!=0)
    {
        d=extgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }
    else x=1,y=0;
    return d;
}
ll solve(ll a,ll b,ll c)
{
    ll gcd=extgcd(a,b,x,y);
    if(c%gcd!=0)return -1;
    x*=c/gcd;
    b/=gcd;
    if(b<0)
        b=-b;
    ll ans=x%b;
    if(ans<0)ans+=b;
    return ans;
}
int main()
{
     while(cin>>x>>y>>m>>n>>L)
    {
        ll ans=solve(m-n,L,y-x);
        if(ans==-1)cout<<"Impossible"<<endl;
        else cout<<ans<<endl;
    }

    return 0;
}