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

最新推荐文章于 2019-02-28 19:10:00 发布

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#codeforces #CFRound #368 Div. 2 #CF368Pythagorean Tri #勾股定理 #规律

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本篇介绍了一个算法问题，即给定直角三角形一边长n，如何找到对应的毕达哥拉斯三元组（n, m, k）。通过分析n为奇数和偶数的不同情况，给出了具体的计算公式。

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 thatn, m andk 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

Output

4 5

Input

Output

8 10

Input

Output

-1

Input

Output

144 145

Input

Output

2244 2245

解：给定直角三角形的一条边长，求得另外两条边的长度。根据题数据量太大，考虑找规律递推。
       得到输入分为偶数和奇数。
       当n为奇数时：(n*n-1)/2         (n*n+1)/2
       当n为偶数时：(n/2)*(n/2)-1      (n/2)*(n/2)+1
（附：此题中涉及一条知识点就是：毕达哥拉斯三元数组）
开始思路不太对，认为偶数时一直除以2直到为奇数时，利用公式在乘以倍数即可。后来WA之后在原来基础修改。

#include<stdio.h>
#include<string.h>

 
int main()
{
	long long n;
	while(~scanf("%I64d",&n)){
		
		long long x=1;
		long long z=0; 
		z=n;
		if(n<3) printf("-1\n");
		else if(n%2!=0){
				printf("%I64d %I64d\n",(n*n-1)/2,(n*n+1)/2);
		} else{
			while(n%2==0){
				n=n/2;
				x=x*2;
			}
			if(n==1){
				
				printf("%I64d %I64d\n",(z/2)*(z/2)-1,(z/2)*(z/2)+1);
			} else{
				printf("%I64d %I64d\n",x*(n*n-1)/2,x*(n*n+1)/2);
			}
		}
	}
	return 0;
}