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 andB and hypotenuse length C, satisfies the formula

A² +B² =C²

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 lengthC is integral?

Input  

Each line contains a single integer A, such that 2 ≤ A < 1048576 = 2²⁰. 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 andB has a hypotenuse with integral length.

Sample Input

3122104857410485750

Sample Output

1201175

Judge Tips  

A Hint and a Warning: Our hint is that you need not consider any value forB that is greater than(A²-1)/2, because for any such right triangle, hypotenuse C satisfiesB <C < B + 1, and thus cannot have integral length.

Our warning is that for values of A ≈ 2²⁰, there could be solutions withB ≈ 2³⁹, and thus values ofC² >B² ≈ 2⁷⁸.

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;    }}