思路：

连分数求佩尔方程最小特解

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);            }}