Codeforces 707C.Pythagorean Triples

239

题目


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.


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



题解


对于完全平方数,有
2n n2-1 n2+1
2n+1 2n2+2n 2n2+2n+1
两种形式
分别计算即可

当然,奇数直接 3 4 5 等比例扩大也行

代码


/*
By:OhYee
Github:OhYee
HomePage:http://www.oyohyee.com
Email:oyohyee@oyohyee.com

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxn = 100005;
const int INF = 0x7FFFFFFF;


bool  Do() {
    __int64 n;
    if(scanf("%I64d",&n) == EOF)
        return false;
    if(n == 1 || n == 2)
        printf("-1\n");
    else
        if(n % 2) {
            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 true;
}

int main() {
    while(Do());
    return 0;
}
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