佩尔方程最小解模板

 1 import java.math.BigInteger;
 2 import java.util.Scanner;
 3 
 4 public class Main
 5 {
 6     public static void solve(int n)
 7     {
 8         BigInteger N, p1, p2, q1, q2, a0, a1, a2, g1, g2, h1, h2,p,q;
 9         g1 = q2 = p1 = BigInteger.ZERO;
10         h1 = q1 = p2 = BigInteger.ONE;
11         a0 = a1 = BigInteger.valueOf((int)Math.sqrt(1.0*n));
12         BigInteger ans=a0.multiply(a0);
13         if(ans.equals(BigInteger.valueOf(n)))
14         {
15             System.out.println("No solution!");
16             return;
17         }
18         N = BigInteger.valueOf(n);
19         while (true)
20         {
21             g2 = a1.multiply(h1).subtract(g1);
22             h2 = N.subtract(g2.pow(2)).divide(h1);
23             a2 = g2.add(a0).divide(h2);
24             p = a1.multiply(p2).add(p1);
25             q = a1.multiply(q2).add(q1);
26             if (p.pow(2).subtract(N.multiply(q.pow(2))).compareTo(BigInteger.ONE) == 0) break;
27             g1 = g2;h1 = h2;a1 = a2;
28             p1 = p2;p2 = p;
29             q1 = q2;q2 = q;
30         }
31         System.out.println(p+" "+q);
32     }
33 
34     public static void main(String[] args)
35     {
36         Scanner cin = new Scanner(System.in);
37         int t=cin.nextInt();
38         while(t>0)
39         {
40             int n=cin.nextInt();
41             if(n==0)
42             {
43                 System.out.println("1 1");
44                 continue;
45             }
46             solve(n);
47             t--;
48         }
49     }
50 }  

 1 #include<bits/stdc++.h>
 2 #define ll long long 
 3 #define dbg(x) cout<<#x<<" == "<<x<<endl
 4 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;}
 5 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;}
 6 using namespace std;
 7 const int maxn=1e5+5;
 8 const int mod=128;
 9 const int Mod=13331;
10 ll a[20000];
11 bool pell_minimum_solution(ll n,ll &x0,ll &y0){
12     ll m=(ll)sqrt((double)n);
13     double sq=sqrt(n);
14     int i=0;
15     if(m*m==n)return false;//当n是完全平方数则佩尔方程无解
16     a[i++]=m;
17     ll b=m,c=1;
18     double tmp;
19     do{
20         c=(n-b*b)/c;
21         tmp=(sq+b)/c;
22         a[i++]=(ll)(floor(tmp));
23         b=a[i-1]*c-b;
24         //printf("%lld %lld %lld\n",a[i-1],b,c);
25     }while(a[i-1]!=2*a[0]);
26     ll p=1,q=0;
27     for(int j=i-2;j>=0;j--){
28         ll t=p;
29         p=q+p*a[j];
30         q=t;
31         //printf("a[%d]=%lld %lld %lld\n",j,a[j],p,q);
32     }
33     if((i-1)%2==0){x0=p;y0=q;}
34     else{x0=2*p*p+1;y0=2*p*q;}
35     return true;
36 }
37  
38 int main(){
39     ll n,x,y;
40     int t=rd();
41     while(t--){
42         n=rdll();
43         if(pell_minimum_solution(n,x,y)){
44             printf("%d %d\n",x,y); 
45 //            printf("%lld^2-%lld*%lld^2=1\t",x,n,y);
46 //            printf("%lld-%lld=1\n",x*x,n*y*y);
47         }
48     }
49 }