Pythagorean Triples
题目链接：
http://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?
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##Input
<big>
The only line of the input contains single integer n (1 ≤ n ≤ 109) — the length of some side of a right triangle.
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##Output
<big>
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.
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##Examples
<big>
input
3
output
4 5
input
6
output
8 10
input
1
output
-1
input
17
output
144 145
input
67
output
2244 2245
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##Source
<big>
Codeforces Round #368 (Div. 2)
</big>
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##题意：
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构造一个直角三角形以N作为一条边长.(直角边和斜边皆可)
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<br/>
##题解：
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N作为直角边时，N^2为两个平方数的差：
易知任意两平方数之差必定为奇数或者是4的倍数. [参考](http://www.cnblogs.com/Sunshine-tcf/p/5698800.html)
N作为斜边时，N^2为两个平方数的和：
由于N的规模较小，可以直接枚举找到 a^2+b^2 = N^2. (枚举规模sqrt(n))
<br/>
upd：这个问题其实有[公式](http://www.zybang.com/question/712e6230d81ecf844ef145786bf6918a.html)：
奇数：（2n+1, 2n^2+2n, 2n^2+2n+1）
偶数：（2n, n^2-1, n^2+1）
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<br/>
##代码：
``` cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <list>
#define LL long long
#define eps 1e-8
#define maxn 101000
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;
int main(int argc, char const *argv[])
{
//IN;
LL n;
while(scanf("%I64d", &n) != EOF)
{
n = n * n;
if(n!=1 && n!=4) {
if(n & 1) {
LL cur = (n - 1) >> 1;
printf("%I64d %I64d
", cur,cur+1);
continue;
}
if(n % 4 == 0) {
LL cur = (n - 4) >> 2;
printf("%I64d %I64d
", cur,cur+2);
continue;
}
}
bool flag = 0;
for(LL i=1; i*i<=n; i++) {
LL cur = n - i * i;
if(cur < 1) continue;
LL sqt = (LL)sqrt(cur);
if(sqt * sqt == cur) {
printf("%I64d %I64d
", i, sqt);
flag = 1;
break;
}
}
if(!flag) printf("-1
");
}
return 0;
}