CodeForces-707c[数学构造]

Pythagorean Triples

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.

Example
Input
3
Output
4 5
Input
6
Output
8 10
Input
1
Output
-1
Input
17
Output
144 145
Input
67
Output
2244 2245

题意：给出直角三角形的一边，求出另外的两条边。

分析：给出的边可能是直角边，也可能是斜边。故答案有多种。题目只要求输出一种即可。那么，就假设该边为直角边。
当 n 为奇数时 （2k+1, 2*k^2+k^2, 2*k^2+k^2+1）;
当 n 为偶数时 （2k, k^2-1, k^2+1）;

#include<iostream>#include<cstdio>using namespace std;typedef long long ll;int main(){    ll n;    cin>>n;    if(n<=2)    {        puts("-1");        return 0;    }    if(n%2==1)    {        n=(n-1)/2;        printf("%I64d %I64d\n",2*n*n+2*n,2*n*n+2*n+1);    }    else    {        n/=2;        printf("%I64d %I64d\n",n*n-1,n*n+1);     }    return 0;}