hdu--5675 ztr loves math

最新推荐文章于 2025-05-26 17:47:58 发布

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最新推荐文章于 2025-05-26 17:47:58 发布

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本文探讨了形如n=x²-y²的方程是否有正整数解的问题，通过构造等式推导出当n为奇数或4的倍数时，方程存在正整数解，并给出了特例说明。

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ztr loves math

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 41 Accepted Submission(s): 24

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

Sample Output

  
   False
True
True
True


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

Source

BestCoder Round #82 (div.2)

分析：可以构造两组等式

(k + 1)^2−k^2= 2k + 1
(k + 1)^2−(k−1)^2 = 4k 得出结论，当z为奇数或者4的倍数时，方程一定有正整数解

特例：1 和 4 不行。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
int main()
{
#ifdef OFFLINE
	freopen("t.txt", "r", stdin);
#endif
	ll i, j, k, n, m, t;
	scanf("%lld", &t);
	while (t--){
		scanf("%lld", &n);
		if (n == 1 || n == 4)
			puts("False");
		else if (n % 2 || n % 4 == 0)
			puts("True");
		else
			puts("False");
	}
	return 0;
}