HDU-4143 A Simple Problem

A Simple Problem
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1959    Accepted Submission(s): 518
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
 

/*
    ：
  n，    x     y^2 = n +x^2 ，      (y-x)*(y+x)=n，  y-x  ，  y-x    y+x；
*/
//   ：
#include<stdio.h>
int main()
{
	int T,i,a,n,tmp,ans;
	scanf("%d",&T);
	while(T--)
	{
		ans=0x7fffffff;
		scanf("%d",&n);
		for(i=1;i*i<n;++i)
		{
			if(n%i==0)
			{
				a=n/i;
				if(((a+i)&1)==0&&((a-i)&1)==0)
				{
					tmp=(a-i)>>1;
					if(tmp<ans)
						ans=tmp;
				}
			}
		}
		if(ans!=0x7fffffff)
			printf("%d
",ans);
		else
			printf("-1
");
	}
	return 0;
}


/*
   ：---   ：     x  ， y-x y+x     ，   sqrt(n)                   
#include <stdio.h>
#include <math.h>

int main() 
{
    int T,i,flag,n;
    scanf("%d",&T);
    while(T--)
    {
        flag=0;
        scanf("%d",&n);
        for(i=sqrt(n);i>=1;--i)
        {
            if(n%i==0&&(n/i-i)%2==0&&n/i!=i)
            {
                flag=1;
                break;
            }
        }
        if(flag)
            printf("%d
",(n/i-i)/2);
        else
            printf("-1
");    
    }
    return 0;
}*/