佩尔方程

Posted 2021-01-23

1.连分数
#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
using namespace std;  
typedef long long ll;  
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;  
    while(~scanf("%lld",&n)){  
        if(pell_minimum_solution(n,x,y)){  
            printf("%lld^2-%lld*%lld^2=1	",x,n,y);  
            printf("%lld-%lld=1
",x*x,n*y*y);  
        }  
    }  
2.暴力

递推式
x[n]=x[n-1]*x[1]+d*y[n-1]*y[1]; 
y[n]=x[n-1]*y[1]+y[n-1]*x[1];