Codeforces Round #368 (Div. 2) C. Pythagorean Triples 数学

C. Pythagorean Triples

题目连接：

http://www.codeforces.com/contest/707/problem/C

Description

Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.

For example, triples (3, 4, 5), (5, 12, 13) and (6, 8, 10) are Pythagorean triples.

Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.

Katya had no problems with completing this task. Will you do the same?

Input

The only line of the input contains single integer n (1 ≤ n ≤ 109) — the length of some side of a right triangle.

Output

Print two integers m and k (1 ≤ m, k ≤ 1018), such that n, m and k form a Pythagorean triple, in the only line.

In case if there is no any Pythagorean triple containing integer n, print  - 1 in the only line. If there are many answers, print any of them.

Sample Input

3

Sample Output

4 5

Hint

题意

给你直角三角形的一边长度，让你找到两个整数，是其他两条边的边长，问你能否找到。

题解：

其实是公式题，随便百度了一下，就找到结论了= =

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    long long n;
    cin>>n;
    if(n<=2){
        cout<<"-1"<<endl;
        return 0;
    }
    if(n%2)
    {
        long long a=(n-1LL)/2LL;
        cout<<2LL*a*a+2LL*a<<" "<<2LL*a*a+2LL*a+1LL<<endl;
        return 0;
    }
    else
    {
        long long a=n/2;
        cout<<a*a-1<<" "<<a*a+1<<endl;
    }
}