数论模板 - 佩尔方程

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佩尔方程

输入一个数k,求方程ans^2 = k*n*n+1 中ans的最小整数解(n是大于等于一的整数,无上限)

C语言模板：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;int can[1005] = {0};int a[10005][605]= {0};int x[6005], y[6005], t[6005];int h1,h2;int bb, ee, xx, yy, c, n;void Pell(int ji,int many,int ma,int kk){    if (ji < kk)        Pell(ji + 1, a[ma][(ji-1)%a[ma][600]+1], ma, kk);    else    {        h1 = 1;        h2 = 1;        x[1] = many;        y[1] = 1;        return;    }    for (int i = 1; i <= h1; i++)        t[i] = x[i];    for (int i = 1; i <= h2; i++)        x[i] = y[i];    for (int i = 1; i <= h1; i++)        y[i] = t[i];    c = h1;    h1 = h2;    h2 = c;    for (int i = 1; i <= h2; i++)        if (i <= h1)            x[i] += many * y[i];        else            x[i] = many * y[i];    if (h2 > h1)        h1 = h2;    for (int i = 1; i < h1; i++)        if (x[i] >= 10)        {            x[i+1] += x[i] / 10;            x[i] %= 10;        }    while(x[h1] >= 10)    {        x[h1+1] = x[h1] / 10;        x[h1] %= 10;        h1++;    }    x[0] = h1;}void solve(){    int i, j;    for (j = 1; j <= 31; j++)        can[j*j] = true;    for(i = 1; i <= 1000; i++)    {        if(!can[i])        {            a[i][600]=1;            bb = 1;            ee = (int)sqrt((double)i);            a[i][0] = ee;            ee =- ee;            xx = bb;            yy = ee;            xx =- yy;            yy = i - yy * yy;            n=0;            while((xx - yy) * (xx - yy) < i || xx >= 0)            {                xx -= yy;                n++;            }            a[i][1] = n;            c = xx, xx = yy, yy = c;            while(xx != bb || yy != ee)            {                a[i][600]++;                c = xx;                xx =- yy;                yy = i - yy * yy;                yy = yy / c;                n = 0;                while ((xx - yy) * (xx - yy) < i || xx >= 0)                {                    xx -= yy;                    n++;                }                a[i][a[i][600]] = n;                c = xx, xx = yy, yy = c;            }        }    }}int main(){    int k;    solve();    while(scanf("%d",&k)!=EOF)///ans^2 = k * n ^ 2 + 1,求ans的最小值(k <= 1000)    {        if(!can[k])        {            if(a[k][600] % 2)                Pell(1, a[k][0], k, a[k][600]*2);            else                Pell(1, a[k][0], k, a[k][600]);            for (int j = x[0]; j >= 1; j--)                printf("%d",x[j]);            printf("\n");        }        else            printf("no solution\n");    }    return 0;}

Java模板（Java处理数据更大）：

import java.io.*;import java.util.*;import java.math.*;public class Main{    public static BigInteger[] pell(BigInteger n)    {///求解pell方程，返回两个大数，即一个两个元素的大数数组，分别为x和y        BigInteger[]p = new BigInteger[10];        BigInteger[]q = new BigInteger[10];        BigInteger[]h = new BigInteger[10];        BigInteger[]g = new BigInteger[10];        BigInteger[]a = new BigInteger[10];        p[0] = BigInteger.ZERO;q[0] = BigInteger.ONE;        p[1] = BigInteger.ONE;q[1] = BigInteger.ZERO;        a[0] = sqrt(n);        a[2] = a[0];        g[1] = BigInteger.ZERO;h[1] = BigInteger.ONE;        while(true){            g[2] = a[2].multiply(h[1]).subtract(g[1]);            h[2] = n.subtract(g[2].multiply(g[2])).divide(h[1]);            a[3] = g[2].add(a[0]).divide(h[2]);            p[2] = a[2].multiply(p[1]).add(p[0]);            q[2] = a[2].multiply(q[1]).add(q[0]);            if(p[2].multiply(p[2]).subtract(n.multiply(q[2].multiply(q[2]))).equals(BigInteger.ONE)==true               &&p[2].equals(BigInteger.ZERO)==false&&q[2].equals(BigInteger.ZERO)==false)            return new BigInteger[] {p[2],q[2]};            g[0] = g[1];h[0] = h[1];            g[1] = g[2];h[1] = h[2];            a[2] = a[3];            p[0] = p[1];p[1] = p[2];            q[0] = q[1];q[1] = q[2];        }    }    static BigInteger sqrt(BigInteger n)    {///对于大数开根号        BigInteger low = BigInteger.ZERO,high = n,ans = null;        while(low.compareTo(high)<=0){            BigInteger mid = low.add(high).shiftRight(1);            if(mid.multiply(mid).compareTo(n)<=0){                ans = mid;                low = mid.add(BigInteger.ONE);            }            else high = mid.subtract(BigInteger.ONE);        }        return ans;    }    public static void main(String[]args) throws IOException    {        /*        System.setIn(new BufferedInputStream(new FileInputStream("in.txt")));///文件重定向输入        System.setOut(new PrintStream(new FileOutputStream("out.txt")));   ///文件重定向输出        */        Scanner cin = new Scanner(new BufferedInputStream(System.in));        while(cin.hasNext()){            int n = cin.nextInt(),ok = 1;            for(int i = 1;i<=31;++i)                if(i*i==n){ok = 0;break;}            if(ok==0){System.out.println(-1);continue;}            BigInteger r[] = pell(BigInteger.valueOf(n));            System.out.println(r[0]);        }    }}

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