Codeforces Round #131 (Div. 2) / 214A System of Equations(枚举&优化)

本文链接：https://blog.youkuaiyun.com/synapse7/article/details/10300375

本博客介绍了一个用于解决特定数学系统方程的高效算法。通过输入两个整数n和m，读者可以了解如何计算满足给定系统的整数对的数量。文中详细解释了算法的实现，并提供了优化后的代码，使得解决方案在时间和空间复杂度上更为高效。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?

You are given a system of equations:

$\text{[math]}$

You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.

Input

A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.

Output

On a single line print the answer to the problem.

   Sample test(s)
 
     input
   
9 3

     output
   
1

     input
   
14 28

     output
   
1

     input
   
4 20

     output
   
0

Note

In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.

数据范围很小，直接暴力搞定。

完整代码：

/*30ms,0KB*/

#include<cstdio>
#include<cmath>

int main()
{
	int n, m, a, b, count = 0;
	scanf("%d%d", &n, &m);
	int sqrtm = (int)sqrt((double)m);
	for (b = 0; b <= sqrtm; ++b)
	{
		a = m - b * b;
		if (a * a + b == n)
			++count;
	}
	printf("%d", count);
	return 0;
}

优化：

经测试得，只有当m=n=1时才有两解(1,0)和(0,1)，其他情况至多有一解。

/*30ms,0KB*/

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

int main()
{
	int n, m, a, b;
	scanf("%d%d", &n, &m);
	if (n == m)
	{
		if (n == 1)
			putchar('2');
		else
		{
			double ans = (sqrt((double)4 * n + 1) - 1) / 2;
			putchar(fabs(ans - (int)ans < 1e-9) ? '1' : '0');
		}
	}
	else
	{
		bool has = false;
		if (n < m)
		{
			int minb = min(n, (int)sqrt((double)m));
			for (b = 0; b <= minb; ++b)
			{
				a = m - b * b;
				if (a * a + b == n)
				{
					has = true;
					break;
				}
			}
		}
		else
		{
			int mina = min(m, (int)sqrt((double)n));
			for (a = 0; a <= mina; ++a)
			{
				b = n - a * a;
				if (b * b + a == m)
				{
					has = true;
					break;
				}
			}
		}
		putchar(has ? '1' : '0');
	}
	return 0;
}