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📜  两个给定数字之间的完美平方数

📅  最后修改于: 2021-04-27 05:30:02             🧑  作者: Mango

给定两个给定的数字a和b,其中1 <= a <= b,找到a和b之间的完美平方数(包括a和b)。
例子

Input :  a = 3, b = 8
Output : 1
The only perfect in given range is 4.

Input : a = 9, b = 25
Output : 3
The three squares in given range are 9, 
16 and 25

方法1 :一种幼稚的方法是检查a和b之间的所有数字(包括a和b),并在遇到完美正方形时将计数加一。

下面是上述想法的实现:

C++
// A Simple Method to count squares between a and b
#include 
using namespace std;
 
int countSquares(int a, int b)
{
    int cnt = 0; // Initialize result
 
    // Traverse through all numbers
    for (int i = a; i <= b; i++)
 
        // Check if current number 'i' is perfect
        // square
        for (int j = 1; j * j <= i; j++)
            if (j * j == i)
                cnt++;
 
    return cnt;
}
 
// Driver code
int main()
{
    int a = 9, b = 25;
    cout << "Count of squares is "
         << countSquares(a, b);
    return 0;
}


Java
// Java program to count squares between a and b
class CountSquares {
 
    static int countSquares(int a, int b)
    {
        int cnt = 0; // Initialize result
 
        // Traverse through all numbers
        for (int i = a; i <= b; i++)
 
            // Check if current number 'i' is perfect
            // square
            for (int j = 1; j * j <= i; j++)
                if (j * j == i)
                    cnt++;
        return cnt;
    }
}
 
// Driver Code
public class PerfectSquares {
    public static void main(String[] args)
    {
        int a = 9, b = 25;
        CountSquares obj = new CountSquares();
        System.out.print("Count of squares is " + obj.countSquares(a, b));
    }
}


Python
# Python program to count squares between a and b
 
def CountSquares(a, b):
 
    cnt = 0 # initialize result
 
    # Traverse through all numbers
    for i in range (a, b + 1):
        j = 1;
        while j * j <= i:
            if j * j == i:
                 cnt = cnt + 1
            j = j + 1
        i = i + 1
    return cnt
 
# Driver Code
a = 9
b = 25
print "Count of squares is:", CountSquares(a, b)


C#
// C# program to count squares
// between a and b
using System;
 
class GFG {
 
    // Function to count squares
    static int countSquares(int a, int b)
    {
        // Initialize result
        int cnt = 0;
 
        // Traverse through all numbers
        for (int i = a; i <= b; i++)
 
            // Check if current number
            // 'i' is perfect square
            for (int j = 1; j * j <= i; j++)
                if (j * j == i)
                    cnt++;
        return cnt;
    }
 
    // Driver Code
    public static void Main()
    {
        int a = 9, b = 25;
        Console.Write("Count of squares is " + countSquares(a, b));
    }
}
 
// This code is contributed by Sam007


PHP


Javascript


C++
// An Efficient Method to count squares between a and b
#include 
using namespace std;
 
// An efficient solution to count square between a
// and b
int countSquares(int a, int b)
{
    return (floor(sqrt(b)) - ceil(sqrt(a)) + 1);
}
 
// Driver code
int main()
{
    int a = 9, b = 25;
    cout << "Count of squares is "
         << countSquares(a, b);
    return 0;
}


Java
// An Efficient method to count squares between
// a and b
class CountSquares {
    double countSquares(int a, int b)
    {
        return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1);
    }
}
 
// Driver Code
public class PerfectSquares {
    public static void main(String[] args)
    {
        int a = 9, b = 25;
        CountSquares obj = new CountSquares();
        System.out.print("Count of squares is " + (int)obj.countSquares(a, b));
    }
}


Python
# An Efficient Method to count squares between a
# and b
import math
def CountSquares(a, b):
    return (math.floor(math.sqrt(b)) - math.ceil(math.sqrt(a)) + 1)
 
# Driver Code
a = 9
b = 25
print "Count of squares is:", int(CountSquares(a, b))


C#
// C# program for efficient method
// to count squares between a & b
using System;
 
class GFG {
 
    // Function to count squares
    static double countSquares(int a, int b)
    {
        return (Math.Floor(Math.Sqrt(b)) - Math.Ceiling(Math.Sqrt(a)) + 1);
    }
 
    // Driver Code
    public static void Main()
    {
        int a = 9, b = 25;
        Console.Write("Count of squares is " + (int)countSquares(a, b));
    }
}
 
// This code is contributed by Sam007.


PHP


Javascript


输出 :

Count of squares is 3

该解决方案的时间复杂度上限为O((ba)* sqrt(b))。
方法2(有效)我们可以简单地取’a’的平方根和’b’的平方根,然后使用来计算它们之间的理想平方

floor(sqrt(b)) - ceil(sqrt(a)) + 1

We take floor of sqrt(b) because we need to consider 
numbers before b.

We take ceil of sqrt(a) because we need to consider 
numbers after a.


For example, let b = 24, a = 8.  floor(sqrt(b)) = 4, 
ceil(sqrt(a)) = 3.  And number of squares is 4 - 3 + 1
= 2. The two numbers are 9 and 16.

下面是上述想法的实现:

C++

// An Efficient Method to count squares between a and b
#include 
using namespace std;
 
// An efficient solution to count square between a
// and b
int countSquares(int a, int b)
{
    return (floor(sqrt(b)) - ceil(sqrt(a)) + 1);
}
 
// Driver code
int main()
{
    int a = 9, b = 25;
    cout << "Count of squares is "
         << countSquares(a, b);
    return 0;
}

Java

// An Efficient method to count squares between
// a and b
class CountSquares {
    double countSquares(int a, int b)
    {
        return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1);
    }
}
 
// Driver Code
public class PerfectSquares {
    public static void main(String[] args)
    {
        int a = 9, b = 25;
        CountSquares obj = new CountSquares();
        System.out.print("Count of squares is " + (int)obj.countSquares(a, b));
    }
}

Python

# An Efficient Method to count squares between a
# and b
import math
def CountSquares(a, b):
    return (math.floor(math.sqrt(b)) - math.ceil(math.sqrt(a)) + 1)
 
# Driver Code
a = 9
b = 25
print "Count of squares is:", int(CountSquares(a, b))

C#

// C# program for efficient method
// to count squares between a & b
using System;
 
class GFG {
 
    // Function to count squares
    static double countSquares(int a, int b)
    {
        return (Math.Floor(Math.Sqrt(b)) - Math.Ceiling(Math.Sqrt(a)) + 1);
    }
 
    // Driver Code
    public static void Main()
    {
        int a = 9, b = 25;
        Console.Write("Count of squares is " + (int)countSquares(a, b));
    }
}
 
// This code is contributed by Sam007.

的PHP


Java脚本


输出 :

Count of squares is 3

该解决方案的时间复杂度为O(Log b)。对于n的平方根的典型实现需要花费等于O(Log n)的时间。