Pythagorean Triples

最新推荐文章于 2025-05-05 21:49:29 发布

原创最新推荐文章于 2025-05-05 21:49:29 发布 · 597 阅读

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本文介绍了一种快速找到包含指定整数边长的勾股数的方法。对于给定的整数n，通过判断n的奇偶性，分别给出了两种计算勾股数中另两边的公式，并提供了AC代码实现。

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
Note

Illustration for the first sample.

这题......想了好久，也没想出来

答案就是：给出最小一条边a，其他两边分别是：1° a是偶数 b=n^2/4-1 c=b+2

2° a是奇数 b=(n^2-1)/2 c=b+1

.....

恩，就是这样，伟大的数学~

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<set>
#include<queue>
#include<cmath>
#include<cstdlib>
using namespace std;
#define INF 999999
typedef long long ll;
const int maxn=1500;
long long int n,m,k;

int main()
{
    scanf("%lld",&n);
    if(n%2==0){
        m=((n/2)*(n/2))-1;
        k=m+2;
    }
    else{
        m=((n*n)-1)/2;
        k=m+1;
    }
    if(m==0||k==0)
        printf("-1\n");
    else
        printf("%lld %lld\n",m,k);
    return 0;
}