CodeForces - 707C (找规律)

这篇博客介绍了如何解决CodeForces上的707C问题,通过分析题目给出的样例,建议先编写程序输出前几项数据以揭示隐藏的规律。博主指出,初始的1和2不存在构成直角三角形的情况,并进一步提出了对偶数和奇数情况的分类讨论方法。

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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.

一个简单的规律题,可以先写个程序把前面的数据都输出来看看规律,然后1和2是没有直角三角型的,然后分类讨论偶数和奇数:

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
//typedef pair<int , int > p;
//const int maxn = 300 + 10;
//char s[maxn][maxn];
//int m, n;
//int sx, sy, gx, gy;
//int d[maxn][maxn];
//int dx[] = {0, 1 , -1,0};
//int dy[] = {1, 0, 0, -1};
int main()
{
    long long n;
    cin >> n;
    if(n == 1 || n == 2)
        cout << -1 << endl;
    else if(n & 1)
        {

            cout << (n * n - 1) / 2 << " " << (n * n + 1) / 2 << endl;
        }
    else
    {
        cout << (n * n / 2 - 2) / 2 << " " << (n * n / 2 + 2) / 2 << endl;
    }
}

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