HDU4143 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

 

/*
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<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\n",ans);
		else
			printf("-1\n");
	}
	return 0;
}


/*
代码二：---优化下：因为要满足x最小，则y-x与y+x要尽量接近，所以从sqrt(n)向前枚举的第一个满足条件的一定是最优解
#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",(n/i-i)/2);
        else
            printf("-1\n");    
    }
    return 0;
}*/