在给定范围内选择随机数时获得完美平方的可能性

给定两个表示范围的整数L和R ，任务是找到当在L到R的范围内选择一个随机数时获得完美平方数的概率。
例子：

Input: L = 6, R = 20 
Output: 0.133333 
Explanation: 
Perfect squares in range [6, 20] = {9, 16} => 2 perfect squares 
Total numbers in range [6, 20] = 15 
Probability = 2 / 15 = 0.133333
Input: L = 16, R = 25 
Output: 0.2 
 

方法：此问题的主要观察结果是，可用给定公式计算从0到数字范围内的理想平方数：

// Count of perfect squares in the range 0 to N is given as 
Count of perfect squares = Floor(sqrt(N)) 
 

同样，可以在上述公式的帮助下，按以下公式计算给定范围内的理想平方数：

Count of perfect Squares[L, R] = floor(sqrt(R)) – ceil(sqrt(L)) + 1 
Total numbers in the range = R – L + 1
$\text{Probability of getting perfect square} =\frac{floor(sqrt(R)) - ceil(sqrt(L)) + 1}{R - L + 1}$
 

下面是上述方法的实现：

C++

// C++ implementation to find the
// probability of getting a
// perfect square number
 
#include 
using namespace std;
 
// Function to return the probability
// of getting a perfect square
// number in a range
float findProb(int l, int r)
{
    // Count of perfect squares
    float countOfPS = floor(sqrt(r)) - ceil(sqrt(l)) + 1;
 
    // Total numbers in range l to r
    float total = r - l + 1;
 
    // Calculating probability
    float prob = (float)countOfPS / (float)total;
    return prob;
}
 
// Driver Code
int main()
{
    int L = 16, R = 25;
    cout << findProb(L, R);
 
    return 0;
}

Java

// Java implementation to find the
// probability of getting a
// perfect square number
 
class GFG{
 
// Function to return the probability
// of getting a perfect square
// number in a range
static float findProb(int l, int r)
{
 
    // Count of perfect squares
    float countOfPS = (float) (Math.floor(Math.sqrt(r)) -
                               Math.ceil(Math.sqrt(l)) + 1);
 
    // Total numbers in range l to r
    float total = r - l + 1;
 
    // Calculating probability
    float prob = (float)countOfPS / (float)total;
    return prob;
}
 
// Driver Code
public static void main(String[] args)
{
    int L = 16, R = 25;
    System.out.print(findProb(L, R));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 implementation to find 
# the probability of getting a
# perfect square number
import math
 
# Function to return the probability
# of getting a perfect square
# number in a range
def findProb(l, r):
     
    # Count of perfect squares
    countOfPS = (math.floor(math.sqrt(r)) -
                  math.ceil(math.sqrt(l)) + 1)
     
    # Total numbers in range l to r
    total = r - l + 1
 
    # Calculating probability
    prob = countOfPS / total
     
    return prob
     
# Driver code
if __name__=='__main__':
     
    L = 16
    R = 25
     
    print(findProb(L, R))
     
# This code is contributed by rutvik_56

C#

// C# implementation to find the probability
// of getting a perfect square number
using System;
 
class GFG{
 
// Function to return the probability
// of getting a perfect square
// number in a range
static float findProb(int l, int r)
{
     
    // Count of perfect squares
    float countOfPS = (float)(Math.Floor(Math.Sqrt(r)) -
                            Math.Ceiling(Math.Sqrt(l)) + 1);
 
    // Total numbers in range l to r
    float total = r - l + 1;
 
    // Calculating probability
    float prob = (float)countOfPS / (float)total;
    return prob;
}
 
// Driver Code
public static void Main(String[] args)
{
    int L = 16, R = 25;
     
    Console.Write(findProb(L, R));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript implementation to find the
// probability of getting a
// perfect square number
 
// Function to return the probability
// of getting a perfect square
// number in a range
function findProb(l, r)
{
 
    // Count of perfect squares
    var countOfPS = (Math.floor(Math.sqrt(r)) -
                     Math.ceil(Math.sqrt(l)) + 1);
 
    // Total numbers in range l to r
    var total = r - l + 1;
 
    // Calculating probability
    var prob = countOfPS / total;
    return prob;
}
 
// Driver code
var L = 16, R = 25;
 
// Function Call
document.write(findProb(L, R));
 
// This code is contributed by Khushboogoyal499
    

输出：

0.2

时间复杂度： O(1)