CodeForces - 707C Pythagorean Triples (数学)

最新推荐文章于 2021-08-23 08:41:57 发布

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#数学 #直角三角形 #思维扩散 #素数

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本文介绍了一种算法，用于根据给定的一条边长找到对应的勾股数三元组。讨论了勾股数的性质，并提供了当给定边长为斜边或直角边时寻找其余两边的具体方法。

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

Note

$\text{[math]}$

Illustration for the first sample.

给定N 求直接三角形另外两条边.

根据直角三角形的两条性质

当n*n为奇数时有 a+b=n*n ; a-b=1;

当n*n为偶数时,有 a+b=(n*n)/2; a-b=2;

联立解 a,b;

并且当 n=1 || n=2 时无解; A B C

分析: 可以发现从3 开始是 3,5,7,11,13,15 素数是可以组成的, 3-4-5 已知A B=(A*A-1)/2; C=B+1;

对于偶数, 一定可以变成素数的倍数, 那么队友 6-8-10- 6是3的倍数, 那么就可以转成成 3*2-4*2-5*2; k=6/3;

对于是2的倍数的, 要特殊处理, 因为2 无法构成直角三角形, 4 可以, 是2的倍数就一定是4 的倍数 4-3-5 成倍数关系就可以

例如 1024 是2 的倍数1024/2=512 1024/4=256 3*256=768 5*256=1280 1024-768-1280;

代码实现:

#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x,b,n) lower_bound(b+1,b+1+n,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define SHUT ios_base::sync_with_stdio(false); cout.setf(ios::fixed);cout.precision(20); cout.tie(nullptr); cin.tie(nullptr);
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define  FI(n) IO::read(n)
#define  Be IO::begin()

using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MAXN=1e5+5;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};

namespace IO {
	const int MT = 5e7;
	char buf[MT]; int c,sz;
	void begin(){
		c = 0;
		sz = fread(buf, 1, MT, stdin);//一次性输入
	}
	template<class T>
	inline bool read(T &t){
		while( c < sz && buf[c] != '-' && ( buf[c]<'0' || buf[c] >'9')) c++;
		if( c>=sz) return false;
		bool flag = 0; if( buf[c]== '-') flag = 1,c++;
		for( t=0; c<=sz && '0' <=buf[c] && buf[c] <= '9'; c++ ) t= t*10 + buf[c]-'0';
		if(flag) t=-t;
		return true;
	}
}
ll inv[maxn*2];
inline void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);};}
inline ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
inline ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
inline ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
inline ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
inline ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
inline ll inv2(ll b){return qpow(b,MOD-2);}
int main()
{
    ll n;
    cin>>n;
    ll p=n*n;
    if(n==1||n==2)
    {
        cout<<-1<<endl;
        return 0;
    }
    if(p%2==0)
    {
        cout<<(p/2+2)/2-2<<" "<<(p/2+2)/2<<endl;
    }
    else
    {
        cout<<(p-1)/2<<" "<<(p-1)/2+1<<endl;
    }
    return 0;
}

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