程序查找在给定范围内具有奇数除数的数的计数

给定两个整数A和B。任务是计算间隔[A，B]中有多少个因数为奇数的数。

例子：

Input : A = 1, B = 10
Output : 3

Input : A = 5, B = 15
Output : 1

天真的方法：
一种简单的方法是遍历范围[A，B]之间的所有数字，并检查它们的除数是否为奇数。
以下是上述想法的实现：

C++

// C++ program to find count of numbers having
// odd number of divisors in given range
  
#include 
using namespace std;
  
// Function to count numbers having odd
// number of divisors in range [A, B]
int OddDivCount(int a, int b)
{
    // variable to odd divisor count
    int res = 0;
    // iterate from a to b and count their
    // number of divisors
    for (int i = a; i <= b; ++i) {
  
        // variable to divisor count
        int divCount = 0;
        for (int j = 1; j <= i; ++j) {
            if (i % j == 0) {
                ++divCount;
            }
        }
  
        // if count of divisor is odd
        // then increase res by 1
        if (divCount % 2) {
            ++res;
        }
    }
    return res;
}
  
// Driver code
int main()
{
    int a = 1, b = 10;
    cout << OddDivCount(a, b) << endl;
  
    return 0;
}

Java

// Java program to find count of numbers having
// odd number of divisors in given range
  
import java.io.*;
  
class GFG {
    // Function to count numbers having odd
    // number of divisors in range [A, B]
    static int OddDivCount(int a, int b)
    {
        // variable to odd divisor count
        int res = 0;
        // iterate from a to b and count their
        // number of divisors
        for (int i = a; i <= b; ++i) {
  
            // variable to divisor count
            int divCount = 0;
            for (int j = 1; j <= i; ++j) {
                if (i % j == 0) {
                    ++divCount;
                }
            }
  
            // if count of divisor is odd
            // then increase res by 1
            if ((divCount % 2) != 0) {
                ++res;
            }
        }
        return res;
    }
  
    // Driver code
  
    public static void main(String[] args)
    {
  
        int a = 1, b = 10;
        System.out.println(OddDivCount(a, b));
    }
    // This code is contributed by ajit.
}

Python 3

# Python3 program to find count 
# of numbers having odd number
# of divisors in given range
  
# Function to count numbers 
# having odd number of divisors 
# in range [A, B]
def OddDivCount(a, b):
  
    # variable to odd divisor count
    res = 0
      
    # iterate from a to b and count 
    # their number of divisors
    for i in range(a, b + 1) :
  
        # variable to divisor count
        divCount = 0
        for j in range(1, i + 1) :
            if (i % j == 0) :
                divCount += 1
  
        # if count of divisor is odd
        # then increase res by 1
        if (divCount % 2) :
            res += 1
    return res
  
# Driver code
if __name__ == "__main__":
    a = 1
    b = 10
    print(OddDivCount(a, b))
  
# This code is contributed 
# by ChitraNayal

C#

// C# program to find count of numbers having
// odd number of divisors in given range
using System;
  
class Geeks {
  
    // Function to count numbers having odd
    // number of divisors in range [A, B]
    static int OddDivCount(int a, int b)
    {
        // variable to odd divisor count
        int res = 0;
        // iterate from a to b and count their
        // number of divisors
        for (int i = a; i <= b; ++i) {
  
            // variable to divisor count
            int divCount = 0;
            for (int j = 1; j <= i; ++j) {
                if (i % j == 0) {
                    ++divCount;
                }
            }
  
            // if count of divisor is odd
            // then increase res by 1
            if ((divCount % 2) != 0) {
                ++res;
            }
        }
        return res;
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int a = 1, b = 10;
        Console.WriteLine(OddDivCount(a, b));
    }
}

输出：

3

时间复杂度： O(n ² )

更好的方法：
数字可以由其主要因子与适当乘方的乘积来表示。这些能力可以用来获取整数具有的因数。如果数字为num并且可以表示为(a ^p1 )*(b ^p2 )*(c ^p3 )
那么num个因子的计数为(p1 +1)*(p2 +1)*(p3 +1)

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count
// of divisors of a number
int divisor(int a)
{
    int div = 1, count = 0;
    for (int i = 2; i <= sqrt(a); i++) {
  
        // Count the powers of the current
        // prime i which divides a
        while (a % i == 0) {
            count++;
            a = a / i;
        }
  
        // Update the count of divisors
        div = div * (count + 1);
  
        // Reset the count
        count = 0;
    }
  
    // If the remaining a is prime then a^1
    // will be one of its prime factors
    if (a > 1) {
        div = div * (2);
    }
    return div;
}
  
// Function to count numbers having odd
// number of divisors in range [A, B]
int OddDivCount(int a, int b)
{
    // To store the count of elements
    // having odd number of divisors
    int res = 0;
  
    // Iterate from a to b and find the
    // count of their divisors
    for (int i = a; i <= b; ++i) {
  
        // To store the count of divisors of i
        int divCount = divisor(i);
  
        // If the divisor count of i is odd
        if (divCount % 2) {
            ++res;
        }
    }
    return res;
}
  
// Driver code
int main()
{
    int a = 1, b = 10;
    cout << OddDivCount(a, b);
  
    return 0;
}
// This code is contributed by jrolofmeister

Java

// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to return the count
// of divisors of a number
static int divisor(int a)
{
    int div = 1, count = 0;
    for (int i = 2; i <= Math.sqrt(a); i++) 
    {
          
        // Count the powers of the current
        // prime i which divides a
        while (a % i == 0) 
        {
            count++;
            a = a / i;
        }
  
        // Update the count of divisors
        div = div * (count + 1);
  
        // Reset the count
        count = 0;
    }
  
    // If the remaining a is prime then a^1
    // will be one of its prime factors
    if (a > 1) 
    {
        div = div * (2);
    }
    return div;
}
  
// Function to count numbers having odd
// number of divisors in range [A, B]
static int OddDivCount(int a, int b)
{
    // To store the count of elements
    // having odd number of divisors
    int res = 0;
  
    // Iterate from a to b and find the
    // count of their divisors
    for (int i = a; i <= b; ++i)
    {
  
        // To store the count of divisors of i
        int divCount = divisor(i);
  
        // If the divisor count of i is odd
        if (divCount % 2 == 1) 
        {
            ++res;
        }
    }
    return res;
}
  
// Driver code
public static void main(String[] args) 
{
    int a = 1, b = 10;
    System.out.println(OddDivCount(a, b));
}
} 
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
  
# Function to return the count
# of divisors of a number
def divisor(a):
  
    div = 1;
    count = 0;
    for i in range(2, int(pow(a, 1 / 2)) + 1):
  
        # Count the powers of the current
        # prime i which divides a
        while (a % i == 0):
            count += 1;
            a = a / i;
              
        # Update the count of divisors
        div = div * (count + 1);
  
        # Reset the count
        count = 0;
      
    # If the remaining a is prime then a^1
    # will be one of its prime factors
    if (a > 1): 
        div = div * (2);
      
    return div;
  
# Function to count numbers having odd
# number of divisors in range [A, B]
def OddDivCount(a, b):
  
    # To store the count of elements
    # having odd number of divisors
    res = 0;
  
    # Iterate from a to b and find the
    # count of their divisors
    for i in range(a, b + 1):
          
        # To store the count of divisors of i
        divCount = divisor(i);
  
        # If the divisor count of i is odd
        if (divCount % 2): 
            res += 1;
  
    return res;
  
# Driver code
if __name__ == '__main__':
    a, b = 1, 10;
    print(OddDivCount(a, b));
  
# This code is contributed by PrinciRaj1992

C＃

// C# implementation of the approach
using System;
using System.Collections.Generic;
      
class GFG
{
  
// Function to return the count
// of divisors of a number
static int divisor(int a)
{
    int div = 1, count = 0;
    for (int i = 2; 
             i <= Math.Sqrt(a); i++) 
    {
          
        // Count the powers of the current
        // prime i which divides a
        while (a % i == 0) 
        {
            count++;
            a = a / i;
        }
  
        // Update the count of divisors
        div = div * (count + 1);
  
        // Reset the count
        count = 0;
    }
  
    // If the remaining a is prime then a^1
    // will be one of its prime factors
    if (a > 1) 
    {
        div = div * (2);
    }
    return div;
}
  
// Function to count numbers having odd
// number of divisors in range [A, B]
static int OddDivCount(int a, int b)
{
    // To store the count of elements
    // having odd number of divisors
    int res = 0;
  
    // Iterate from a to b and find the
    // count of their divisors
    for (int i = a; i <= b; ++i)
    {
  
        // To store the count of divisors of i
        int divCount = divisor(i);
  
        // If the divisor count of i is odd
        if (divCount % 2 == 1) 
        {
            ++res;
        }
    }
    return res;
}
  
// Driver code
public static void Main(String[] args) 
{
    int a = 1, b = 10;
    Console.WriteLine(OddDivCount(a, b));
}
} 
  
// This code is contributed by Princi Singh

输出：

3

时间复杂度： O(n * logn)

请参考本文以了解O(1)方法。