打印给定范围内的所有完美方块

原文:https://www . geesforgeks . org/print-all-perfect-squares-from-the-给定范围/

给定一个范围【L，R】，任务是打印给定范围内的所有完美方块。
例:

输入: L = 2，R = 24
输出: 4 9 16
输入: L = 1，R = 100
输出: 1 4 9 16 25 36 49 64 81 100

天真的做法:从 L 开始到 R 检查当前元素是否是完美的正方形。如果是，那么打印出来。
以下是上述办法的实施情况:

C++

// C++ implementation of the approach
#include
using namespace std;
// Function to print all the perfect
// squares from the given range
void perfectSquares(float l, float r)
{
    // For every element from the range
    for (int i = l; i <= r; i++) {
        // If current element is
        // a perfect square
        if (sqrt(i) == (int)sqrt(i))
            cout <    }
}
// Driver code
int main()
{
    int l = 2, r = 24;
    perfectSquares(l, r);
    return 0;
}


Java 语言(一种计算机语言，尤用于创建网站)

//Java implementation of the approach
import java.io.*;
class GFG
{
// Function to print all the perfect
// squares from the given range
static void perfectSquares(int l, int r)
{
    // For every element from the range
    for (int i = l; i <= r; i++)
    {
        // If current element is
        // a perfect square
        if (Math.sqrt(i) == (int)Math.sqrt(i))
            System.out.print(i + " ");
    }
}
// Driver code
public static void main (String[] args)
{
    int l = 2, r = 24;
    perfectSquares(l, r);
}
}
// This code is contributed by jit_t


Python 3

# Python3 implementation of the approach
# Function to print all the perfect
# squares from the given range
def perfectSquares(l, r):
    # For every element from the range
    for i in range(l, r + 1):
        # If current element is
        # a perfect square
        if (i**(.5) == int(i**(.5))):
            print(i, end=" ")
# Driver code
l = 2
r = 24
perfectSquares(l, r)
# This code is contributed by mohit kumar 29


C

// C# implementation of the approach
using System;
class GFG
{
// Function to print all the perfect
// squares from the given range
static void perfectSquares(int l, int r)
{
    // For every element from the range
    for (int i = l; i <= r; i++)
    {
        // If current element is
        // a perfect square
        if (Math.Sqrt(i) == (int)Math.Sqrt(i))
            Console.Write(i + " ");
    }
}
// Driver code
public static void Main(String[] args)
{
    int l = 2, r = 24;
    perfectSquares(l, r);
}
}
// This code is contributed by 29AjayKumar


java 描述语言

Output: 

4 9 16


它是含氧(氮)的溶液。此外，平方根数量的使用导致计算费用。
有效方法:这种方法基于这样一个事实，即在数字 L 之后的第一个完美正方形肯定是 ⌈sqrt(L)⌉ 的正方形。简单来说 L 的平方根会非常接近我们要找的平方根的数字。因此，该数字将为 pow(ceil(sqrt(L))，2) 。
第一个完美的正方形对这个方法很重要。现在原来的答案就隐藏在这个图案上面即0 1 4 9 16 25
0 和 1 的差是 1
1 和 4 的差是 3
4 和 9 的差是 5 以此类推……
意思就是两个完美正方形的差总是奇数。
现在，问题来了，要得到下一个数字必须加上什么，答案是 (sqrt(X) * 2) + 1 ，其中 X 是已知的完美正方形。
让当前完美方块为 4 ，那么下一个完美方块肯定会是 4 + (sqrt(4) * 2 + 1) = 9 。这里加数 5 ，下一个要加的数将是 7 然后是 9 以此类推……这样就组成了一系列奇数。
加法比执行乘法或求每个数的平方根计算成本低。
以下是上述方法的实施:

C++

// C++ implementation of the approach
#include
using namespace std;
// Function to print all the perfect
// squares from the given range
void perfectSquares(float l, float r)
{
    // Getting the very first number
    int number = ceil(sqrt(l));
    // First number's square
    int n2 = number * number;
    // Next number is at the difference of
    number = (number * 2) + 1;
    // While the perfect squares
    // are from the range
    while ((n2 >= l && n2 <= r)) {
        // Print the perfect square
        cout <        // Get the next perfect square
        n2 = n2 + number;
        // Next odd number to be added
        number += 2;
    }
}
// Driver code
int main()
{
    int l = 2, r = 24;
    perfectSquares(l, r);
    return 0;
}


Java 语言(一种计算机语言，尤用于创建网站)

// Java implementation of the approach
class GFG
{
// Function to print all the perfect
// squares from the given range
static void perfectSquares(float l, float r)
{
    // Getting the very first number
    int number = (int) Math.ceil(Math.sqrt(l));
    // First number's square
    int n2 = number * number;
    // Next number is at the difference of
    number = (number * 2) + 1;
    // While the perfect squares
    // are from the range
    while ((n2 >= l && n2 <= r))
    {
        // Print the perfect square
        System.out.print(n2 + " ");
        // Get the next perfect square
        n2 = n2 + number;
        // Next odd number to be added
        number += 2;
    }
}
// Driver code
public static void main(String[] args)
{
    int l = 2, r = 24;
    perfectSquares(l, r);
}
}
// This code is contributed by 29AjayKumar


Python 3

# Python3 implementation of the approach
from math import ceil, sqrt
# Function to print all the perfect
# squares from the given range
def perfectSquares(l, r) :
    # Getting the very first number
    number = ceil(sqrt(l));
    # First number's square
    n2 = number * number;
    # Next number is at the difference of
    number = (number * 2) + 1;
    # While the perfect squares
    # are from the range
    while ((n2 >= l and n2 <= r)) :
        # Print the perfect square
        print(n2, end= " ");
        # Get the next perfect square
        n2 = n2 + number;
        # Next odd number to be added
        number += 2;
# Driver code
if __name__ == "__main__" :
    l = 2; r = 24;
    perfectSquares(l, r);
# This code is contributed by AnkitRai01


C

// C# implementation of the approach
using System;
class GFG
{
// Function to print all the perfect
// squares from the given range
static void perfectSquares(float l, float r)
{
    // Getting the very first number
    int number = (int) Math.Ceiling(Math.Sqrt(l));
    // First number's square
    int n2 = number * number;
    // Next number is at the difference of
    number = (number * 2) + 1;
    // While the perfect squares
    // are from the range
    while ((n2 >= l && n2 <= r))
    {
        // Print the perfect square
        Console.Write(n2 + " ");
        // Get the next perfect square
        n2 = n2 + number;
        // Next odd number to be added
        number += 2;
    }
}
// Driver code
public static void Main(String[] args)
{
    int l = 2, r = 24;
    perfectSquares(l, r);
}
}
// This code is contributed by Rajput Ji


java 描述语言

Output: 

4 9 16