Project Euler 66: Diophantine equation

题目链接

思路：

连分数求佩尔方程最小特解

参考博客

模板：

LL a[20000];
bool min_pell(LL d, LL &x, LL &y) {
    LL m = floor(sqrt(d+0.5)); 
    if(m*m == d) return false;
    int cnt = 0;
    a[cnt++] = m;
    LL b = m, c = 1;
    double sq = sqrt(d);
    do {
        c = (d - b*b)/c;
        a[cnt++] = (LL)floor((sq+b)/c);
        b = a[cnt-1]*c - b;
    }while(a[cnt-1] != 2*a[0]);
    LL p = 1, q = 0;
    for (int j = cnt-2; j >= 0; --j) {
        LL t = p;
        p = a[j]*p + q;
        q = t;
    } 
    if(cnt%2) x = p, y = q;
    else x = 2*p*p+1, y = 2*p*q;
    return true;
}

由于某些解超出long long范围，所以用到java大数

代码：

import java.math.*;
import java.util.*;

public class Main {
    public static long a[] = new long [200000]; 
    public static BigInteger x, y;
    public static boolean min_pell(long d) {
        long m = (long)Math.floor(Math.sqrt(d+0.5)); 
        if(m*m == d) return false;
        int cnt = 0;
        a[cnt++] = m;
        long b = m, c = 1;
        double sq = Math.sqrt(d);
        do {
            c = (d - b*b)/c;
            a[cnt++] = (long)Math.floor((sq+b)/c);
            b = a[cnt-1]*c - b;
        }while(a[cnt-1] != 2*a[0]);
        BigInteger p = BigInteger.ONE, q = BigInteger.ZERO;
        for (int j = cnt-2; j >= 0; --j) {
            BigInteger t = p;
            p = p.multiply(BigInteger.valueOf(a[j])).add(q);
            q = t;
        } 
        if(cnt%2 != 0) {
            x = p;
            y = q;
        }
        else {
            x = p.multiply(p).multiply(BigInteger.valueOf(2)).add(BigInteger.ONE);
            y = p.multiply(q).multiply(BigInteger.valueOf(2));
        }
        return true;
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        BigInteger mx = BigInteger.valueOf(0), ans = BigInteger.valueOf(5);
        for (long d = 1; d <= 1000; ++d) {
            if(!min_pell(d)) continue;
            //System.out.println(x);
            if(x.compareTo(mx) > 0) {
                mx = x;
                ans = BigInteger.valueOf(d);
            }
        }
        System.out.println(ans);
        
    }
}