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脑补知识:佩尔方差
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
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
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原文地址:http://www.cnblogs.com/theskulls/p/4786969.html