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

C. Pythagorean Triples
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
input
3
output
4 5
input
6
output
8 10
input
1
output
-1
input
17
output
144 145
input
67
output
2244 2245
Note
给你直角三角形的其中一条边问另外两边。
a^2+b^2=c^2 显然有a=s*v;b=(s^2-v^2)/2;c=(s^2+v^2)/2;
所以分奇偶讨论：当n为奇数时显然a=1*n；b=(n^2-1)/2;c=(n^2+1)/2;
n为偶数时，要分成n=2^k或n=2^k*m 对于前者显然有一组已知解 4 3 5
所以只需要构造3*2^(k-2)与5*2^(k-2),对于后者可转化为奇数的情况。
最后要特判n=1与n=2的情况。

#include<stdio.h>#include<iostream>#include<algorithm>#include<string>#include<string.h>#include<math.h>using namespace std;#define F(x,a,b) for (int x=a;x<=b;x++)#define ll long longint main(){    ll n;ll a,b;    scanf("%I64d",&n);    if (n==1||n==2){printf("-1"); return 0;}    if (n&1){a=(n*n+1)/2;b=(n*n-1)/2;printf("%I64d %I64d",a,b);}else    {        ll t=1,p=n;        while (!(p&1)){t=t<<1;p=p>>1;}        if (p!=1) {a=(p*p+1)/2*t;b=(p*p-1)/2*t;printf("%I64d %I64d",a,b);return 0;}        cout<<3*(t>>2)<<" "<<5*(t>>2);    }}