题目135：求方程x^2 - y^2 - z^2=n的解的个数

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Given the positive integers, x, y, and z, are consecutive terms of an arithmetic progression, the least value of the positive integer, n, for which the equation, x² − y² − z² = n, has exactly two solutions is n = 27:

34² − 27² − 20² = 12² − 9² − 6² = 27

It turns out that n = 1155 is the least value which has exactly ten solutions.

How many values of n less than one million have exactly ten distinct solutions?

题目：

给定等差数列中的三个数字 x，y 和 z，最小 x² − y² − z² = n 有两个解的最小的 n 是 27：

34² − 27² − 20² = 12² − 9² − 6² = 27

事实上 n = 1155 是使得上述方程具有 10 个解的最小值。

一百万以下有多少个数使得上述方程有 10 个不同的解？

jerryxjr1220

1. """
   
2. (z+2d)^2-(z+d)^2-z^2=n
   
3. 3d^2+2dz-z^2=n
   
4. (3d-z)*(d+z)=n
   
5. z<3d => d > z/3
   
6. """
   
7. from numba import jit
   
8. import numpy as np
   
9. @jit
   
10. def solve(target, limit):
    
11.         master = np.zeros(limit+1,dtype='int16')
    
12.         for z in range(1, limit):
    
13.                 for d in range(z//3+1, limit):
    
14.                         if (3*d-z)*(d+z)>limit:
    
15.                                 break
    
16.                         master[(3*d-z)*(d+z)] += 1
    
17.         count = 0
    
18.         for value in master:
    
19.                 if value == target:
    
20.                         count += 1
    
21.         return count
    
22. print(solve(10, 1000000))

复制代码

4989
[Finished in 5.2s]

guosl

/*
答案：4989
耗时：1.41674 (4核)
          6.37623 (单线程)
解题思路：令方程为：(x+d)^2 - x^2 - (x-d)^2 = n
=> (x+d)(3d-x)=n
所以对n进行因数分解为：n=m1 x m2
d=(m1+m2)/4,x=m1 - d
*/
#include <iostream>
#include <set>
#include <cmath>
#include <omp.h>
using namespace std;

int main(void)
{
  double t = omp_get_wtime();
  int nCount = 0;
#pragma omp parallel for reduction(+:nCount) schedule(dynamic,64)
  for (int n = 1155; n <= 1000000; ++n)
  {
    set<int> s;//记录所以整数解
    int d1 = (int)sqrt((double)n);
    for (int m1 = 1; m1 <= d1; ++m1) //枚举n的因数分解
    {
      if (n % m1 == 0)
      {
        int m2 = n / m1;
        if ((m1 + m2) % 4 != 0) //检查d是否是一个整数
          continue;
        int d = (m1 + m2) / 4;
        if (m1 - d > 0)
          s.insert(m1 - d);//得到一个整数解
        if (m2 - d > 0)
          s.insert(m2 - d);//得到另一个整数解
      }
    }
    if (s.size() == 10)
      ++nCount;
  }
  t = omp_get_wtime() - t;
  cout << nCount << endl << t << endl;
  return 0;
}