佩尔方程最小解模板

java代码：

import java.math.BigInteger;
import java.util.Scanner;

public class Main
{
    public static void solve(int n)
    {
        BigInteger N, p1, p2, q1, q2, a0, a1, a2, g1, g2, h1, h2,p,q;
        g1 = q2 = p1 = BigInteger.ZERO;
        h1 = q1 = p2 = BigInteger.ONE;
        a0 = a1 = BigInteger.valueOf((int)Math.sqrt(1.0*n));
        BigInteger ans=a0.multiply(a0);
        if(ans.equals(BigInteger.valueOf(n)))
        {
            System.out.println("No solution!");
            return;
        }
        N = BigInteger.valueOf(n);
        while (true)
        {
            g2 = a1.multiply(h1).subtract(g1);
            h2 = N.subtract(g2.pow(2)).divide(h1);
            a2 = g2.add(a0).divide(h2);
            p = a1.multiply(p2).add(p1);
            q = a1.multiply(q2).add(q1);
            if (p.pow(2).subtract(N.multiply(q.pow(2))).compareTo(BigInteger.ONE) == 0) break;
            g1 = g2;h1 = h2;a1 = a2;
            p1 = p2;p2 = p;
            q1 = q2;q2 = q;
        }
        System.out.println(p+" "+q);
    }

    public static void main(String[] args)
    {
        Scanner cin = new Scanner(System.in);
        int t=cin.nextInt();
        while(t>0)
        {
            int n=cin.nextInt();
            if(n==0)
            {
                System.out.println("1 1");
                continue;
            }
            solve(n);
            t--;
        }
    }
}  

c++代码：

#include<bits/stdc++.h>
#define ll long long 
#define dbg(x) cout<<#x<<" == "<<x<<endl
int rd(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){    if(ch=='-'){f=-1;} ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';    ch=getchar();}return x*f;}
ll rdll(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){    if(ch=='-'){f=-1;} ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';    ch=getchar();}return x*f;}
using namespace std;
const int maxn=1e5+5;
const int mod=128;
const int Mod=13331;
ll a[20000];
bool pell_minimum_solution(ll n,ll &x0,ll &y0){
    ll m=(ll)sqrt((double)n);
    double sq=sqrt(n);
    int i=0;
    if(m*m==n)return false;//当n是完全平方数则佩尔方程无解
    a[i++]=m;
    ll b=m,c=1;
    double tmp;
    do{
        c=(n-b*b)/c;
        tmp=(sq+b)/c;
        a[i++]=(ll)(floor(tmp));
        b=a[i-1]*c-b;
        //printf("%lld %lld %lld
",a[i-1],b,c);
    }while(a[i-1]!=2*a[0]);
    ll p=1,q=0;
    for(int j=i-2;j>=0;j--){
        ll t=p;
        p=q+p*a[j];
        q=t;
        //printf("a[%d]=%lld %lld %lld
",j,a[j],p,q);
    }
    if((i-1)%2==0){x0=p;y0=q;}
    else{x0=2*p*p+1;y0=2*p*q;}
    return true;
}
 
int main(){
    ll n,x,y;
    int t=rd();
    while(t--){
        n=rdll();
        if(pell_minimum_solution(n,x,y)){
            printf("%d %d
",x,y); 
//            printf("%lld^2-%lld*%lld^2=1	",x,n,y);
//            printf("%lld-%lld=1
",x*x,n*y*y);
        }
    }
}

 详解博客：https://blog.csdn.net/wh2124335/article/details/8871535

补充：

import java.io.*;
import java.math.*;
import java.util.*;
 
public class pei {
    static long [] a = new long [1000];
    static BigInteger x;
    static BigInteger y;
    
    public static void main(String [] args)
    {
        Scanner cin = new Scanner(System.in);
        int t=cin.nextInt();
        while(t>0){
            long n=cin.nextInt();
            if(pell_solution(n)){
                            
                System.out.println(x+" "+y);
            }else{
                System.out.print(""no solution",");
               }
        }
    }
    public static boolean pell_solution(long D){
        double sq=Math.sqrt((double)D);
        long m=(long) Math.floor(sq);
        int i=0;
        if(m*m==D)return false;
        a[i++]=m;
        long b=m,c=1;
        double tmp;
        do{
            c=(D-b*b)/c;
            tmp=(sq+b)/c;
            a[i++]=(long)(Math.floor(tmp));
            b=a[i-1]*c-b;
        }while(a[i-1]!=2*a[0]);
        BigInteger p=new BigInteger("1");
        BigInteger q=new BigInteger("0");
        BigInteger t;
        for(int j=i-2;j>=0;j--){
            t=p;
            p=q.add(p.multiply(BigInteger.valueOf(a[j])));
            q=t;
        }
        if((i-1)%2==0){
            x=p;y=q;
        }else{
            x=BigInteger.valueOf(2).multiply(p).multiply(p).add(BigInteger.ONE);
            y=BigInteger.valueOf(2).multiply(p).multiply(q);
        }
        return true;
    }
}