hdu 2281 - 佩尔方程

题目链接:https://vjudge.net/problem/HDU-2281

 

解题思路:

原式可以化为x^2 = (n+1)(2*n+1)/6,(4n+3)^2 - 48*x^2 = 1

然后就是佩尔方程求解了。

x^2 - n*y^2 = 1

x[i+1] = x[1]*x[i] + n* y[1]*y[i]

y[i+1] = x[1]*y[i] + y[1]*x[i]

#include 
using namespace std;
typedef long long ll;
typedef __int128 hll;
const int mx = 1e2 + 10;
vector  g;
ll ans[mx];
int main()
{
	hll x = 7,y = 1;
	while((x-3)/4<=1e18){
		if((x-3)%4==0) ans[g.size()] = y,g.push_back((x-3)/4);
		hll x1 = x,y1 = y;
		x = 7*x1 + 48*y1;
		y = 7*y1 + x1;
	}
	while(scanf("%lld",&x)&&x)
	{
		int k = upper_bound(g.begin(),g.end(),x) - g.begin() - 1;
		printf("%lld %lld\n",g[k],ans[k]);
	}
	return 0;
}