poj2115

构造出模线性方程c * x = b - a mod (2 ^ k)

很容易解。

利用LRJ书上的方法。

#include <iostream>
 
using namespace std;
 
#define LL long long int
 
LL ext_gcd(LL a, LL b, LL& x, LL& y)
{
	LL t, ret;
	if (!b){
		x = 1, y = 0;
		return a;
	}
	ret = ext_gcd(b, a%b, x, y);
	t = x, x = y, y = t - a / b*y;
	return ret;
}
//ax = b (mod n)
void gcd(LL a, LL b, LL &d, LL &x, LL &y)
{
	if (!b)
	{
		d = a, x = 1, y = 0;
	}
	else
	{
		gcd(b, a %b, d, y, x);
		y -= x * (a / b);
	}
}
LL modular_linear_equation(LL a, LL b, LL n)
{
	long long x, y, e, d;
	gcd(a, n, d, x, y);
	if (b % d)  return -1;
	e = b / d * x % n + n;
	return e % (n / d);
}
int main()
{
	////c * x = b - a mod (2 ^ k)
	int a, b, c, k;
	while (cin >> a >> b >> c >> k && (a || b || c || k))
	{
		LL num = modular_linear_equation(c, b - a, 1LL << k);
		if (num == -1)
		{
			cout << "FOREVER" << endl;
			continue;
		}
		cout << num << endl;
	}
}