码迷,mamicode.com
首页 > 其他好文 > 详细

欧拉工程第66题:Diophantine equation

时间:2015-09-06 20:09:11      阅读:255      评论:0      收藏:0      [点我收藏+]

标签:

题目链接

脑补知识:佩尔方差

 

技术分享

上面说的貌似很明白,最小的i,对应最小的解

然而我理解成,一个循环的解了,然后就是搞不对,后来,仔细看+手工推导发现了问题。i从0开始变量,知道第一个满足等式的解就是最小解。

 

问题转化为求根号n的连分数问题,分子就是x,分母就是y

 

要求的分子,分母,问题又转化为:根号n的连分数表示,对,求出其连分数表示就OK了

技术分享

 

先求出a的序列是什么?

 

第64题,就是求a的序列的。

 

a求出来了,要求出分子分母的表达式。

 

第65题,就是已经知道了a的序列,求分子,当然也可以求分母的

 

分子,分母求出来了,在验证:X*X-D*Y*Y=1时候就是最小解

 

问题真是一环套一环的。

 

Python程序:

import time as time 

start = time.time()

def getD(N):
    x_max ,y_max= 0,0 
    D = 0
    x,y = 0,0
    for S in range(2,N+1):
        x,y = resolve(S)
        if x>x_max:
            x_max ,y_max= x,y 
            D = S 
    return D,x_max,y_max

def resolve(S):
    m = 0 
    d = 1 
    a0 = int(S**0.5)
    if a0*a0 == S :return -1,-1;
    a= a0
    li = [a]
    x,y = 1,1
    while x*x-S*y*y!=1:
        m = d*a - m
        d = (S - m*m)/d 
        a = int((a0 + m)/d)
        li.append(a)
        x = getX(li)
        y = getY(li)
#     print li
    return x,y;

def getX(li):
    x0 = 1
    x1 = li[0]
    li = li[1:]
    for l in li:
        x = l * x1 + x0
        x0 = x1
        x1 = x      
    return x 

def getY(li):
    y0 = 0
    y1 = 1
    li = li[1:]
    for l in li:
        y = l * y1 + y0 
        y0 = y1 
        y1 = y 
    return y 

if __name__ == __main__:

    start = time.time()
    N = 1000
    D ,x_max,y_max= getD(N)
    print "running time={0}seconds,D={1},x_max={2},y_max={3}".format(time.time()-start,D,x_max,y_max)

 

求的是最小解X的最大值时候的D,答案是661

 

然而:

x_max=16421658242965910275055840472270471049

y_max=638728478116949861246791167518480580

这个值好大的


附一:python程序:

from math import sqrt
from time import time

def prefect_sqrt(n):
    return int(sqrt(n))**2 == n 

def floor_root(n):
    return int(sqrt(n))

def chakravals(n):
    x_max = 0 
    for d in range(2,n+1):
        if not prefect_sqrt(d):
            p1 = floor_root(d)
            q1 = 1 
            m1 = p1**2 - d 
#             print -p1,(-p1) % abs(m1),(-p1) % abs(m1)
            if (-p1) % abs(m1) ==0:
                x1 = abs(m1)
            else:
                x1 = (-p1) % abs(m1)
                
            while m1!=1:
                p0 = p1
                q0 = q1
                m0 = m1
                x0 = x1
                
                p1 = (p0 * x0 +d *q0)/abs(m0)
                q1 = (p0 + x0)/abs(m0)
                
                m1 = (x0**2 -d)/m0 
                
                if (-x0)%abs(m1) ==0:
                    x1 = abs(x0)
                else:
                    x1 = (-x0)%abs(m1)
            if p1>x_max:
                x_max = p1 
                d_max = d 
                print "d= %04d  x = %d"%(d_max,x_max)
    print 
    
if __name__==__main__:
    start = time()
    chakravals(1000)
    end = time()
    print "time elapse=%f"%(end - start)

Java程序:

这个跑的好慢的

package project61;
import java.math.*;  


public class P66   
{  
    public static final int precision = 500;  
    public static void main( String args[] )  
    {  
        BigInteger Max = new BigInteger("0");   
        int ans = 0;  
          
outer:  for (int D_i = 2; D_i <= 1000; D_i++)  
        {  
            BigDecimal D = new BigDecimal(D_i);  
            BigDecimal SD = calculation.Sqrt(D);  
              
            BigDecimal SD_i = SD.setScale(0, BigDecimal.ROUND_FLOOR);  
            if (SD_i.multiply(SD_i).equals(D))  
                continue;  
              
            int a[] = calculation.toContinuedFraction(SD, 100);  
              
            for (int i = 1; i < 100; i++)  
            {  
                Fraction temp = new Fraction(a[i],1);  
                  
                for (int j = i - 1; j >= 0; j--)  
                    temp = Fraction.Compute(a[j], temp);  
                  
                BigInteger y_2 = temp.denominator.multiply(temp.denominator);  
                BigInteger x_2 = temp.numerator.multiply(temp.numerator);  
                BigInteger result = x_2.subtract(y_2.multiply(D.toBigIntegerExact())).subtract(BigInteger.ONE);  
                  
                if (result.equals(BigInteger.ZERO))  
                {  
                    if (temp.numerator.compareTo(Max) > 0)  
                    {  
                        Max = temp.numerator;  
                        ans = D_i;  
                    }  
                      
                    continue outer;  
                }  
              
            }  
            System.out.print("Warning!\n");  
              
        }  
        System.out.print(ans+"\n");  
    }  
}  
  
class Fraction  
{  
    public BigInteger numerator;  
    public BigInteger denominator;  
      
    private Fraction()  
    {  
          
    }  
    public Fraction (int numerator, int denominator)  
    {  
        this.numerator = BigInteger.valueOf(numerator);  
        this.denominator = BigInteger.valueOf(denominator);  
    }  
    public static Fraction Compute(int p1, Fraction p2)  
    {  
        Fraction ans = new Fraction();  
        ans.numerator = p2.denominator.add(p2.numerator.multiply(BigInteger.valueOf(p1)));  
        ans.denominator = p2.numerator.add(BigInteger.ZERO);  
        return ans;  
    }  
}  
  
  
class calculation  
{  
    private static final BigDecimal N0 = new BigDecimal(0);  
    private static final BigDecimal N1 = new BigDecimal(1);   
    private static final BigDecimal N2 = new BigDecimal(2);  
      
    public static BigDecimal Sqrt(BigDecimal In)  
    {  
        BigDecimal N = new BigDecimal(1);  
        while(true)  
        {  
            BigDecimal NN = N.multiply(N);  
            NN = NN.add(In);  
            NN = NN.divide(N2);  
            NN = NN.divide(N, P66.precision, BigDecimal.ROUND_FLOOR);  
              
            if (NN.equals(N))  
                break;  
              
            N = NN;  
        }  
          
        return N;  
    }  
    public static int[] toContinuedFraction(BigDecimal In, int l)  
    {  
        int ans[] = new int[l];  
          
          
        BigDecimal temp = In.add(N0);  
        for (int i = 0; i < l; i++)  
        {  
            ans[i] = Integer.valueOf(temp.setScale(0, BigDecimal.ROUND_FLOOR).toString()).intValue();  
            temp = temp.subtract(temp.setScale(0, BigDecimal.ROUND_FLOOR));  
            temp = N1.divide(temp, P66.precision, BigDecimal.ROUND_FLOOR);  
        }  
        return ans;  
    }  
}

 

最后两个程序在网上复制过来的

欧拉工程第66题:Diophantine equation

标签:

原文地址:http://www.cnblogs.com/theskulls/p/4786969.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!