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Pell_equation.cpp
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Pell_equation.cpp
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//
// ペル方程式
//
// verified
// AOJ 2116 Subdividing a Land
// http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2116
//
// Project Euler 140
// https://projecteuler.net/problem=140
//
/*
ペル方程式 x^2 - Dy-2 = ±1 の最小解 (x0, y0), (x0-, y0-) を返す
解なしのときは(-1, -1)を返す
一般解は、
1 : (1, 0) から始めて (x0, y0) を次々と mul していく
-1 : (x0-, y0-) から始めて (x0, y0) を次々と mul していく
*/
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
// Pell's Equation
pair<long long, long long> Pell(long long D, int c = 1) {
if (D == 1) return make_pair(-1, -1);
long long a = D, b = 1;
while (a > b) {
a /= 2;
b *= 2;
}
a = b + 1;
while (a > b) {
a = b;
b = (b + D/b)/2;
}
if (a*a == D) return make_pair(-1, -1);
long long a0 = a;
bool parity = false;
b = 1;
long long x2 = 1, x = a, y2 = 0, y = 1, q;
while (true) {
b = (D - a*a) / b;
q = (a0 + a) / b;
a = q * b - a;
parity = !parity;
if (b == 1) break;
long long tx = x, tx2 = x2, ty = y, ty2 = y2;
x2 = tx, x = tx * q + tx2;
y2 = ty, y = ty * q + ty2;
}
long long x0 = x, y0 = y;
if (!parity) {
if (c == 1) return make_pair(x, y);
else return make_pair(-1, -1);
}
else if (c == -1) return make_pair(x, y);
long long tx = x, ty = y;
x = x0 * tx + D * y0 * ty, y = tx * y0 + x0 * ty;
return make_pair(x, y);
}
pair<long long, long long> PellMul(pair<long long, long long> p, pair<long long, long long> q, long long D) {
long long f = p.first * q.first + D * p.second * q.second;
long long s = p.first * q.second + p.second * q.first;
return make_pair(f, s);
}
//------------------------------//
// Examples
//------------------------------//
void AOJ2116() {
long long N;
int id = 0;
while (cin >> N, N) {
N *= 2;
long long sq = (long long)(sqrt(N) + 0.5);
if (sq * sq == N) cout << "Case " << ++id << ": " << sq << " " << 1 << endl;
else {
auto res = Pell(N);
cout << "Case " << ++id << ": " << res.first << " " << res.second << endl;
}
}
}
int main() {
AOJ2116();
}