A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F, sits in the opposite corner. By travelling on the surfaces of the room the shortest "straight
line" distance from S to F is 10 and the path is shown on the diagram.

However, there are up to three "shortest" path candidates for any given cuboid and the shortest route doesn't always have integer length.

It can be shown that there are exactly 2060 distinct cuboids, ignoring rotations, with integer dimensions, up to a maximum size of M by M by M, for which the shortest route has integer
length when M = 100. This is the least value of M for which the number of solutions first exceeds two thousand; the number of solutions when M = 99 is 1975.

Find the least value of M such that the number of solutions first exceeds one million.

高中做过的题目。把立方体各面展开，这个路径实际上是一个直角三角形的斜边。

要使得的这个路径最小。如果矩阵个边长分别为a<=b <=c

最短路径为sqrt((a+b)^2+c^2)

把a,b视为总体，记做ab

则ab范围是[2,2M]

在寻找到开方后结果为整数的ab和c后

假设ab<c：a,b是能够平均分ab的

假设ab>=c：b的取值到大于ab/2而且满足b<=c,ab-b<=c 得到b的取值个数为(c-(ab+1)/2)+1

#include <iostream>

#include <string>

#include <cmath>

using namespace std;

int main()

{

	int c = 1;

	int count = 0;

	while (count < 1000000)

	{

		c++;

		for (int ab = 2; ab <= 2 * c; ab++)

		{

			int path=ab*ab + c*c;

			int tmp = int(sqrt(path));

			if (tmp*tmp == path)

			{

				count += (ab >= c) ?

1+(c-(ab+1)/2) : ab / 2;

			}

		}

	}

	cout << c << endl;

	system("pause");

	return 0;

}