【Codeforces Round #368 (Div. 2)】Codeforces 707C Pythagorean Triples

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.

構造解。貌似方法很多。

我的方法很麻煩，是用a=pq,b=(p^2-q^2)/2,c=(p^2+q^2)/2推導的。

一種比較簡單的方法是對于一個奇數，直接取n^2/2和n^2/2+1，對于一個偶數，将它不斷除以二得到奇數再擴大回原來的倍數。

#include<cstdio>
#include<cstring>
#include<cmath>
int main()
{
    long long i,j,k,m,n,p,q,x,y,z;
    scanf("%I64d",&x);
    for (p=;p*p<=x;p++)
      if (x%p==)
      {
        q=x/p;
        i=;
        j=;
        if ((q*q+p*p)%2==&&p!=q)
        {
            printf("%I64d %I64d\n",(q*q-p*p)/,(q*q+p*p)/);
            return ;
        }
      }
    for (y=;y*y<=*x;y+=)
      if (x%y==)
      {
        p=x/y-y/2;
        q=x/y+y/;
        if (!(p&&q)) continue;
        printf("%I64d %I64d\n",p*q,(p*p+q*q)/);
        return ;
      }
    for (p=;p*p<=*x;p++)
    {
        q=sqrt(*x-p*p)+.;
        if (q*q==*x-p&&p!=q&&p&&q)
        {
            printf("%I64d %I64d\n",p*q,(q*q-p*p)/);
            return ;
        }
    }
    printf("-1\n");
}