poj1061-青蛙的约会

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题意：两只蛤蟆分别在操场的x,y处，他们分别以m,n的速度向同一个方向跳，问最终他们是否可以相遇
     0.设出同余方程
     1.exgcd解决

//#include <bits/stdc++.h>
#include <iostream>
#define X 10005
#define inF 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO  ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
//typedef unsigned long long Ull; //2^64
const int maxn = (int)1e6 + 10;
const int MOD = (int)1e9 + 7;
const ll inf = 9223372036854775807;
const int N = 47;
ll primer[maxn];
ll a[maxn];
int ans[maxn], num[maxn];
void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y) { if (!b) { x = 1; y = 0; d = a; } else { ex_gcd(b, a%b, d, y, x); y -= x * (a / b); }; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return b / gcd(a, b)*a; }
ll inv_exgcd(ll a, ll m) { ll d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
ll inv1(ll b) { return b == 1 ? 1 : (MOD - MOD / b)*inv1(MOD%b) % MOD; }
void e_gcd(ll a, ll m,ll &d,ll &x,ll &y)
{
	if (!m) x = 1, y = 0, d = a;
	else
		e_gcd(m, a%m, d, y, x),y -= a / m * x;
}
// a=m=0时是无解的
int main()
{
	IO;
	ll x, y, n, m, l;
	cin >> x >> y >> m >> n >> l;
	ll a = m - n, b = y - x;
	if(a<0)a=-a,b=-b;
	if(!a)
	{
	    cout<<"Impossible"<<endl;
	    return 0;
	}
	m = l;
	ll d,xx,yy;
	e_gcd(a,m,d,xx,yy);
	if (b%d!=0)
		cout<<"Impossible"<<endl;
	else
	{
        cout<<((xx*b/d)%(m/d) + m/d) % (m/d)<<endl;
 	}
	return 0;
}