poj 1061 青蛙的约会（扩展欧几里得）

http://poj.org/problem?id=1061

思路：设它们跳了t次相遇，那么有 (x+t*m)-(y+t*n) = z*l(z是一个整数，表示它们路程差是l的z倍)，变形得

(n-m)*t + z*l = (x-y);

令 a = n-m; b = l; c = x-y;

那么原式变为 a*t + z*b = c;

扩展欧几里得模板，求解形如a*x + b*y = gcd(a,b)方程。

LL extend_gcd(LL a, LL b, LL &x, LL &y)
{
	if(b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}

	LL d = extend_gcd(b,a%b,x,y);
	LL t = x;
	x = y;
	y = t-a/b*y;
	return d;
}

方程a*x + b*y = c有解的前提是 gcd(a,b) | c,在这个基础上方程有d=gcd(a,b)个不同的解。当中基础解x0 = x'*(c/d)%b(当中x'为a*x'+b*y' = gcd(a,b)的解)；通解为xi = x0 + i * (b/d)。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#define LL long long
#define _LL __int64
#define eps 1e-8

using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 10;



LL extend_gcd(LL a, LL b, LL &x, LL &y)
{
	if(b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}

	LL d = extend_gcd(b,a%b,x,y);
	LL t = x;
	x = y;
	y = t-a/b*y;
	return d;
}

int main()
{
	LL x,y,m,n,l;
	LL a,b,c,d;
	while(~scanf("%lld %lld %lld %lld %lld",&x,&y,&m,&n,&l))
	{
		a = n-m;
		b = l;
		c = x-y;

		LL t,z;
		d = extend_gcd(a,b,t,z);
		if(c%d != 0)
		{
			printf("Impossible\n");
			continue;
		}
		t = t*(c/d);
		t = (t%b+b)%b;
		printf("%lld\n",t);
	}
	return 0;
}