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

 

/*解题思路：给定n，求最小的x使其满足 y^2 = n +x^2 ，分解因式可得(y-x)*(y+x)=n，枚举y-x即可，注意y-x一定小于y+x；*///代码一：#include
      
       int main(){	int T,i,a,n,tmp,ans;	scanf("%d",&T);	while(T--)	{		ans=0x7fffffff;		scanf("%d",&n);		for(i=1;i*i
       
        >1;					if(tmp
        
         #include 
         
          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",(n/i-i)/2);        else            printf("-1\n");        }    return 0;}*/