查找数字是否为理想平方的优化方法 [英] Optimized way to find if a number is a perfect square

问题描述

我在分配作业中遇到一个问题，以确定一个数字是否完美的正方形:

I had a question in my assignment to find whether a number was perfect square or not:

完美平方是代数结构的元素，等于 另一个元素的平方.

Perfect square is an element of algebraic structure which is equal to the square of another element.

例如:4、9、16等.

For example: 4, 9, 16 etc.

我的朋友做的是，如果n是数字，他们循环n - 1次计算n * n:

What my friends did is, if n is the number, they looped n - 1 times calculating n * n:

// just a general gist
int is_square = 0;
for (int i = 2; i < n; i++)
{
  if ((i * i) == n)
  {
    std::cout << "Yes , it is";
    is_square = 1;
    break;
  }
}
if (is_square == 0)
{
  std::cout << "No, it is not";
}

我想出了一个解决方案，如下所示:

I came up with a solution as shown below:

 if (ceil(sqrt(n)) == floor(sqrt(n)))
 {
   std::cout << "Yes , it is";
 }
 else
 {
   std::cout << "no , it is not";
 }

它正常工作.

与其他解决方案相比，它可以称为 更优化的解决方案 吗?

Can it be called as more optimized solution than others?

推荐答案

久经考验的事实仍然存在:

The tried and true remains:

double sqrt(double x); // from lib

bool is_sqr(long n) {
    long root = sqrt(n);
    return root * root == n;
}

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