Project Euler 66: Diophantine equation

题目链接

思路：

连分数求佩尔方程最小特解

参考博客

模板：

LL a[20000];
bool min_pell(LL d, LL &x, LL &y) {LL m = floor(sqrt(d+0.5)); if(m*m == d) return false;int cnt = 0;a[cnt++] = m;LL b = m, c = 1;double sq = sqrt(d);do {c = (d - b*b)/c;a[cnt++] = (LL)floor((sq+b)/c);b = a[cnt-1]*c - b;}while(a[cnt-1] != 2*a[0]);LL p = 1, q = 0;for (int j = cnt-2; j >= 0; --j) {LL t = p;p = a[j]*p + q;q = t;} if(cnt%2) x = p, y = q;else x = 2*p*p+1, y = 2*p*q;return true;
}

由于某些解超出long long范围，所以用到java大数

代码：

import java.math.*;
import java.util.*;public class Main {public static long a[] = new long [200000]; public static BigInteger x, y;public static boolean min_pell(long d) {long m = (long)Math.floor(Math.sqrt(d+0.5)); if(m*m == d) return false;int cnt = 0;a[cnt++] = m;long b = m, c = 1;double sq = Math.sqrt(d);do {c = (d - b*b)/c;a[cnt++] = (long)Math.floor((sq+b)/c);b = a[cnt-1]*c - b;}while(a[cnt-1] != 2*a[0]);BigInteger p = BigInteger.ONE, q = BigInteger.ZERO;for (int j = cnt-2; j >= 0; --j) {BigInteger t = p;p = p.multiply(BigInteger.valueOf(a[j])).add(q);q = t;} if(cnt%2 != 0) {x = p;y = q;}else {x = p.multiply(p).multiply(BigInteger.valueOf(2)).add(BigInteger.ONE);y = p.multiply(q).multiply(BigInteger.valueOf(2));}return true;}public static void main(String[] args) {// TODO Auto-generated method stubBigInteger mx = BigInteger.valueOf(0), ans = BigInteger.valueOf(5);for (long d = 1; d <= 1000; ++d) {if(!min_pell(d)) continue;//System.out.println(x);if(x.compareTo(mx) > 0) {mx = x;ans = BigInteger.valueOf(d);}}System.out.println(ans);}
}

转载于:https://www.cnblogs.com/widsom/p/10415547.html

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