HUNNU11351：Pythagoras's Revenge

http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11351&courseid=0

Problem description

The famous Pythagorean theorem states that a right triangle, having side lengthsA and B and hypotenuse length C, satisfies the formula

A ² + B ² = C ²

It is also well known that there exist some right triangles in which all three side lengths are integral, such as the classic:

Further examples, both having A=12, are the following:

The question of the day is, given a fixed integer value for A, how many distinct integersB > A exist such that the hypotenuse length C is integral?

Input

Each line contains a single integer A, such that

2 ≤ A < 1048576 = 2
²⁰
. The end of the input is designated by a line containing the value 0.

Output

For each value of A, output the number of integers B > A such that a right triangle having side lengthsA and B has a hypotenuse with integral length.

Sample Input

3
12
2
1048574
1048575
0

Sample Output

1
2
0
1
175

Judge Tips

A Hint and a Warning: Our hint is that you need not consider any value forB that is greater than ( A ²-1)/2 , because for any such right triangle, hypotenuse C satisfies B < C < B + 1 , and thus cannot have integral length.

Our warning is that for values of A ≈ 2 ²⁰ , there could be solutions with B ≈ 2 ³⁹ , and thus values of C ² > B ² ≈ 2 ⁷⁸.

You can guarantee yourself 64-bit integer calculations by using the type long long in C++ or long in Java. But neither of those types will allow you to accurately calculate the value ofC² for such an extreme case. (Which is, after all, what makes thisPythagoras's revenge!)

题意：给出一条最短的直角边，要求另外两边都是整数的直角三角形的个数

思路：根据勾股定理a^2+b^2=c^2

可得:a^2=(c-b)(c+b)

令x = c-b;

y=c+b;

又y-x=2*b;

所以b=(y-x)/2；

并且需要b>a，所以只需要对x进行枚举即可

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    while (true)
    {
        long long a;
        cin >> a;
        if (a == 0) break;

        long long count;
        count = 0;
        for (long long x=1; x <= a/2; x++)
        {
            if (a*a % x == 0)
            {
                long long y = a*a / x;
                if ((y-x)%2 == 0)
                {
                    long long b = (y-x)/2;
                    if (b > a)
                    {
                        count++;
                    }
                }
            }
        }
        cout << count << endl;
    }
}