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): 474    Accepted Submission(s): 142

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

 

Source

2011百校联动“菜鸟杯”程序设计公开赛

 

Recommend

lcy

 

解法：

x^2+n=y^2

n=y^2-x^2=(y-x)*(y+x)

所以，可以枚举y-x的值来确定y+x

然后通过不断更新x的值来获得最小的ans

 

我的代码：

import java.math.*;
import java.io.*;
import java.util.*;

public class Main {
	public static void main(String arg[])
	{
		Scanner cin=new Scanner(System.in);
		int t,T,n,i,ans;
		int a,b,x,y;
		T=cin.nextInt();
		for(t=1;t<=T;t++)
		{
			n=cin.nextInt();
			ans=1999999999;
			for(i=1;i*i<=n;i++)
			{
				if(n%i==0)
				{
					a=i;
					b=n/i;
					if(b-a>0)
					{
						if((a+b)%2==0)
						{
							if((b-a)%2==0)
							{
								x=(b-a)/2;
								y=(a+b)/2;
								if(x<ans)
									ans=x;
							}
						}
					}
				}
			}
			if(ans>=1999999999)
				System.out.println("-1");
			else
				System.out.println(ans);
		}
	}
}