Cattle-off practice match 51 C Pythagorean theorem

 

Links: https://ac.nowcoder.com/acm/contest/1083/C

Title Description

Given right-angled triangle in which the length of one side of n, is to construct your task remaining two sides, so that it can form three sides of a right triangle.

Enter a description:

An integer n.

Output Description:

The two other edges b, c. The answer is not unique, as long as any set of output that is reasonable, if you can not make construction output -1.

Example 1

Entry

copy

3

Export

copy

4 5

Example 2

Entry

copy

4

Export

copy

3 5

Remarks:

0 <= n-<= 1E9 

. 1 <= B, C <= 1e18 

n-, B, C are integers 


Number Pythagorean:

The first type

When a is greater than one odd + 1 2n, b = 2n² + 2n, c = 2n² + 2n + 1.

A is actually the number of squares split into two consecutive natural numbers , for example:

When n = 1 (a, b, c) = (3,4,5)

when n = 2 (a, b, c) = (5,12,13)

For n = 3 (a, b, c) = (7,24,25)

……

This is the most classic of a routine, and because of the inevitable two consecutive natural numbers relatively prime , so use this routine to get the Pythagorean are all relatively prime.

The second type

2, when a is larger than an even number 2n 4, b = n²-1, c = n² + 1

That is the half of a square, respectively, minus one and plus one, for example:

For n = 3 (a, b, c) = (6,8,10)

when n = 4 (a, b, c) = (8,15,17)

n = 5 when (a, b, c) = (10,24,26)

when n = 6 (a, b, c) = (12,35,37)

 

#include <iostream>

using namespace std;

int main()
{
    int n;
    cin >> n;
    if(n>1&&n&1)
    {
        long long t=(n-1)/2;
        long long  b=2*t*t+2*t;
        long long c=2*t*t+2*t+1;
        cout << b << " " << c << endl;
    }
    else if(n>=4&&n%2==0)
    {
        long long t=n/2;
        long long b=t*t-1;
        long long c=t*t+1;
        cout << b << " " << c << endl;
    }
    else
        cout << -1 << endl;
    return 0;
}