青蛙的约会

题解：显然可以列出方程 ( x + k ⋅ m ) % L = ( y + k ⋅ n ) % L (x+k\cdot m)\%L=(y+k\cdot n)\%L (x+k⋅m)%L=(y+k⋅n)%L
然后可以转换为 ( m − n ) ⋅ a + L ⋅ b = y − x (m-n)\cdot a+L\cdot b=y-x (m−n)⋅a+L⋅b=y−x，最后求最小正整数解。 e x g c d exgcd exgcd了呀～

代码

#include<iostream>

using namespace std;
typedef long long LL;

void exgcd(LL a,LL b,LL &x,LL &y, LL &d) 
{
	if(b == 0) {
		x = 1, y = 0;
		d = a;
		return;
	}
	exgcd(b, a % b, y, x, d);
	y -= (a / b) * x;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
#endif
	LL a,b,x,y,d,p,q,m,n,c,L;
	cin >> p >> q >> m >> n >> L;
	a = m - n, b = L, c = q - p;
	exgcd(a,b,x,y,d);
	if(c % d) {
		cout << "Impossible" << endl;
		return 0;
	}
	x = x * (c / d);
	LL ans = (x % L + L) % L;
/* LL k = (y - x) / (a + b); x += k * b, y -= k * a; if(y < 0) { x -= b, y += a; }*/
	cout << ans << endl;

    return 0;
}