勾股数（知道最小边，求另外两个边）

#include <iostream>using namespace std;typedef long long ll;ll n;ll k;int main(){    cin >> n;    if(n<=2) cout  <<  "-1" <<endl;    else    {        if(n%2==0)        {            k=n/2;            cout << k*k-1  <<" " << k*k+1 <<endl;        }        else        {            k=(n-1)/2;            cout << 2*k*k+2*k << " " <<  2*k*k+2*k+1<<endl;        }    }////    最小边为奇数时　（2n+1,2n2+2n,2n2+2n+1）（n为正整数）//      最小边为偶数时 （2n,n2-1,n2+1）（n≥2且n为正整数）}