题目

链接

https://www.luogu.com.cn/problem/CF707C

字面描述

题面翻译

给出一个数字，要你求出另外的两个数使得这三个数构成勾股数

题目描述

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?

输入格式

The only line of the input contains single integer $ n $ ( $ 1<=n<=10^{9} $ ) — the length of some side of a right triangle.

输出格式

Print two integers $ m $ and $ k $ ( $ 1<=m,k<=10^{18} $ ), 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.

样例 #1

样例输入 #1

3

样例输出 #1

4 5

样例 #2

样例输入 #2

6

样例输出 #2

8 10

样例 #3

样例输入 #3

1

样例输出 #3

-1

样例 #4

样例输入 #4

17

样例输出 #4

144 145

样例 #5

样例输入 #5

67

样例输出 #5

2244 2245

提示

Illustration for the first sample.

思路

纯数学题

体面很好理解，就是给你一个数$n$求另外2个数能与它构成勾股数。

勾股数： $a^2+b^2=c^2$
$n$可能为$a,b,c$

给2个数求1个很好办，几个$if$的事情，但1个求2个没有任何关系的量这个问题对我来说很未知

数学上遇到一个未知的问题先根据数的界或性质分类（引用至胡老师语录（初中数学老师））

∵本人有一点点因式分解的基础
∴突然想到了平方差公式

我可以把$n$当成$a$
∴$a^2=c^2-b^2$
∴$a^2=(c-b)(c+b)$

已经推到这里，但是还不能得出任何结论，根据数性，我将$a$分成2类：

1. 奇数
2. 偶数

若$a为奇数$
$\because b、c均为整数$
$\therefore (c-b)=1,(c+b)=a^2$
$\therefore b=(a^2-1)/2,c=(a^2+1)/2$
$a=1时无解$
ok,奇数搞定

若$a为偶数$
根据等式性质,假设$a$有$x$个2可提，设$cnt=a/(2^x)$

奇数的前面已经推出来了，只需要代入奇数的推一遍，最后在乘上$2^x$
$\therefore b=(cn{t}^2-1)/2\cdot 2^x,c=(cn{t}^2+1)/2\cdot 2^x$

但是如果$cnt=1$就不妙了，$\because a=1无解$
应最朴素的方式观察一下:3,4,5 这一组例子 非常的好观察到用$n$表示4就可以了；
$\therefore b=3\cdot 2$^x-2$,c=5\cdot 2$^x-2

时间复杂度分析
$n$为奇数时, 时间复杂度:$o(1)$
$n$为偶数时, 时间复杂度:$o(log2(n))$

代码实现

#include<bits/stdc++.h>
#define ll long long
using namespace std;

ll x;
int main(){
	scanf("%lld",&x);
	if(x%2==1){
		if(x==1){
			printf("-1\n");
			return 0;
		}
		x*=x;
		printf("%lld %lld\n",x/2,x/2+1);
	}
	else {
		if(x==2){
			printf("-1\n");
			return 0;
		}
		ll cnt=0;
		while(x%2==0){
			++cnt;
			x/=2;
		}
		if(x!=1){
			x*=x;
			ll r=(ll)pow(2,cnt);
			printf("%lld %lld\n",x/2*r,(x/2+1)*r);
		}
		else{
			cnt-=2;
			ll r=(ll)pow(2,cnt);
			printf("%lld %lld\n",3*r,5*r);
		}
	}
	return 0;
}