hdu 5675 ztr loves math

ztr loves math

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 483    Accepted Submission(s): 216

Problem Description

ztr loves research Math.One day,He thought about the "Lower Edition" of triangle equation set.Such as  n=x2−y2 .

He wanted to know that ,for a given number n,is there a positive integer solutions?

 

Input

There are T test cases. 
The first line of input contains an positive integer  T(T<=106)  indicating the number of test cases.

For each test case:each line contains a positive integer , n<=1018 .

 

Output

If there be a positive integer solutions,print  True ,else print  False

 

Sample Input

  
  

   
   4
6
25
81
105
  
  

 

Sample Output

  
  

   
   False
True
True
True


   
   

    
    

     
     Hint
    
    
For the fourth case,$105 = 13^{2}-8^{2}$
   
   
   
    
  
  

 

Source

BestCoder Round #82 (div.2)

z=x^2-y^2,
令z=k^2+2*k*y   (y+k)^2=x^2;
若z为奇数，则令另k=1，z=1+2*y,(y>=1),所以z取大于1的所有奇数。
若z为偶数，显然k也必须偶数。z=4+4*y,则z取大于4且必须是4的倍数的数。

#include<stdio.h>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<map>
#include<cmath>
#define mv(a,b) memset(a,b,sizeof(a));
using namespace std;
typedef long long LL;
const int maxn=10000+10;
const double eps=1e-8;
const double PI =acos(-1.0);
int pos[maxn];
int a,b;
int fa[maxn];
// char a[maxn],b[maxn];
int main()
{
 //    ios_base::sync_with_stdio(false);
 //    freopen("input.txt","r",stdin);
	// freopen("output.txt","w",stdout);
	int T;
    scanf("%d",&T);
    while(T--)
    {
        LL m;
        scanf("%I64d",&m);
        if(m==1||m==4)
        {
        	printf("False\n");
        	continue;
        }
        if(m%4==0||m%2!=0)
            printf("True\n");
        else
            printf("False\n");
    }
}