给定整数N ，任务是计算给定数字的无平方除数的数量。
A number is said to be square-free, if no prime factor divides it more than once, i.e., the largest power of a prime factor that divides N is one.
例子：
Input: N = 72
Output: 3
Explanation: 2, 3, 6 are the three possible square free numbers that divide 72.
Input: N = 62290800
Output: 31
天真的方法：
对于每个整数N ，找到其因数，然后检查它是否为无平方数。如果它是一个无平方的数字，则增加计数或否则转到下一个数字。最后，打印计数，得出我们所需的N的无平方除数。
时间复杂度： O(N 3/2 )
高效方法：
请按照以下步骤解决问题：
- 根据无平方数的定义，可以理解，通过找出给定数N的所有素数，可以找出可以除N的所有可能的无平方数。
- 设N的素数为X。因此，使用这些X质数可以形成2个X – 1个无平方数。
- 由于这些X素数因子中的每一个都是N的因数，因此，这些X素数因子的任何乘积组合也是N的因子，因此将有2个X – 1平方的N除数。
Illustration:
- N = 72
- Prime factors of N are 2, 3.
- Hence, the three possible square free numbers generated from these two primes are 2, 3 and 6.
- Hence, the total square-free divisors of 72 are 3( = 22 – 1).
下面是上述方法的实现：
C++
// C++ Program to find the square
// free divisors of a given number
#include
using namespace std;
// The function to check
// if a number is prime or not
bool IsPrime(int i)
{
// If the number is even
// then its not prime
if (i % 2 == 0 && i != 2)
return false;
else {
for (int j = 3;
j <= sqrt(i); j += 2) {
if (i % j == 0)
return false;
}
return true;
}
}
// Driver Code
int main()
{
// Stores the count of
// distinct prime factors
int c = 0;
int N = 72;
for (int i = 2;
i <= sqrt(N); i++) {
if (IsPrime(i)) {
if (N % i == 0) {
c++;
if (IsPrime(N / i)
&& i != (N / i)) {
c++;
}
}
}
}
// Print the number of
// square-free divisors
cout << pow(2, c) - 1
<< endl;
return 0;
} Java
// Java program to find the square
// free divisors of a given number
import java.util.*;
class GFG{
// The function to check
// if a number is prime or not
static boolean IsPrime(int i)
{
// If the number is even
// then its not prime
if (i % 2 == 0 && i != 2)
return false;
else
{
for(int j = 3;
j <= Math.sqrt(i);
j += 2)
{
if (i % j == 0)
return false;
}
return true;
}
}
// Driver code
public static void main(String[] args)
{
// Stores the count of
// distinct prime factors
int c = 0;
int N = 72;
for(int i = 2;
i <= Math.sqrt(N); i++)
{
if (IsPrime(i))
{
if (N % i == 0)
{
c++;
if (IsPrime(N / i) &&
i != (N / i))
c++;
}
}
}
// Print the number of
// square-free divisors
System.out.print(Math.pow(2, c) - 1);
}
}
// This code is contributed by sanjoy_62Python3
# Python3 program to find the square
# free divisors of a given number
import math
# The function to check
# if a number is prime or not
def IsPrime(i):
# If the number is even
# then its not prime
if (i % 2 == 0 and i != 2):
return 0;
else:
for j in range(3, int(math.sqrt(i) + 1), 2):
if (i % j == 0):
return 0;
return 1;
# Driver code
# Stores the count of
# distinct prime factors
c = 0;
N = 72;
for i in range(2, int(math.sqrt(N)) + 1):
if (IsPrime(i)):
if (N % i == 0):
c = c + 1
if (IsPrime(N / i) and
i != (N / i)):
c = c + 1
# Print the number of
# square-free divisors
print (pow(2, c) - 1)
# This code is contributed by sanjoy_62C#
// C# program to find the square
// free divisors of a given number
using System;
class GFG{
// The function to check
// if a number is prime or not
static Boolean IsPrime(int i)
{
// If the number is even
// then its not prime
if (i % 2 == 0 && i != 2)
return false;
else
{
for(int j = 3;
j <= Math.Sqrt(i);
j += 2)
{
if (i % j == 0)
return false;
}
return true;
}
}
// Driver code
public static void Main(String[] args)
{
// Stores the count of
// distinct prime factors
int c = 0;
int N = 72;
for(int i = 2;
i <= Math.Sqrt(N); i++)
{
if (IsPrime(i))
{
if (N % i == 0)
{
c++;
if (IsPrime(N / i) &&
i != (N / i))
c++;
}
}
}
// Print the number of
// square-free divisors
Console.Write(Math.Pow(2, c) - 1);
}
}
// This code is contributed by shivanisinghss2110输出：
3
时间复杂度： O(N)
辅助空间： O(1)