A Simple Problem

 

Problem Description

For a given positive integer n, please find the smallest positive integer x that we can find an integer y such that y^2 = n +x^2.

 

 

Input

The first line is an integer T, which is the the number of cases.
Then T line followed each containing an integer n (1<=n <= 10^9).

 

 

Output

For each integer n, please print output the x in a single line, if x does not exit , print -1 instead.

 

 

Sample Input

2 2 3

 

 

Sample Output

-1 1

 

 

Author

HIT

Problem analyse:

This is very interesting math problem.In the equation " y^2 = n +x^2",  "x"and "y"are all variable, so it's hard to handle it.

So first we can use the equation to reduce the varibles:

y^2 = n +x^2 → n=(y-x)*(y+x)  and then we assume that(1) i=y-x;  and (2)y+x=n/i;  

with the equations (1) and (2),we can get a new equation x=(n/i-i)/2;

Now the problem is gonna easy!

  

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        int i;
        i=(int)sqrt(double(n));
        for(i;i>0;i--)
        {
            if(n%i==0&&(n/i-i)%2==0&&n/i!=i)
            {
                cout<<(n/i-i)/2<<endl;
                break;
            }
        }
        if(i==0)
            cout<<"-1"<<endl;
    }
    return 0;
}