给定一个数字，检查它是否为无正方形。如果没有一个素因数将其除以一次以上，则该数字被认为是无平方的，即，除以n的素因数的最大乘方为1。前几个平方自由数是1，2，3，5，6，7，7，10，11，13，14，15，17，19，21，22，23，26，29，30，31，33，34， 35、37、38、39 …
例子：
Input : n = 10
Output : Yes
10 can be factorized as 2*5. Since
no prime factor appears more than
once, it is a square free number.
Input : n = 20
Output : No
20 can be factorized as 2 * 2 * 5.
Since prime factor appears more than
once, it is not a square free number.这个想法很简单，我们一步一步找到所有主要因素。对于每个素数，我们检查其平方是否也除以n。如果是，则返回false。最后，如果我们找不到一个可以被一次整除的素数，则返回false。
C++
// C++ Program to print
// all prime factors
# include
using namespace std;
// Returns true if n is a square free
// number, else returns false.
bool isSquareFree(int n)
{
if (n % 2 == 0)
n = n/2;
// If 2 again divides n, then n is
// not a square free number.
if (n % 2 == 0)
return false;
// n must be odd at this point. So we can
// skip one element (Note i = i +2)
for (int i = 3; i <= sqrt(n); i = i+2)
{
// Check if i is a prime factor
if (n % i == 0)
{
n = n/i;
// If i again divides, then
// n is not square free
if (n % i == 0)
return false;
}
}
return true;
}
/* Driver program to test above function */
int main()
{
int n = 10;
if (isSquareFree(n))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java Program to print
// all prime factors
class GFG {
// Returns true if n is a square free
// number, else returns false.
static boolean isSquareFree(int n)
{
if (n % 2 == 0)
n = n / 2;
// If 2 again divides n, then n is
// not a square free number.
if (n % 2 == 0)
return false;
// n must be odd at this point. So we can
// skip one element (Note i = i +2)
for (int i = 3; i <= Math.sqrt(n); i = i + 2)
{
// Check if i is a prime factor
if (n % i == 0)
{
n = n / i;
// If i again divides, then
// n is not square free
if (n % i == 0)
return false;
}
}
return true;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 10;
if (isSquareFree(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by prerna saini.Python3
# Python3 Program to print
# all prime factors
from math import sqrt
# Returns true if n is
# a square free number,
# else returns false.
def isSquareFree(n):
if n % 2 == 0:
n = n / 2
# If 2 again divides n,
# then n is not a square
# free number.
if n % 2 == 0:
return False
# n must be odd at this
# point. So we can skip
# one element
# (Note i = i + 2)
for i in range(3, int(sqrt(n) + 1)):
# Check if i is a prime
# factor
if n % i == 0:
n = n / i
# If i again divides, then
# n is not square free
if n % i == 0:
return False
return True
# Driver program
n = 10
if isSquareFree(n):
print ("Yes")
else:
print ("No")
# This code is contributed by Shreyanshi Arun.C#
// C# Program to print
// all prime factors
using System;
class GFG {
// Returns true if n is a square free
// number, else returns false.
static bool isSquareFree(int n)
{
if (n % 2 == 0)
n = n / 2;
// If 2 again divides n, then n is
// not a square free number.
if (n % 2 == 0)
return false;
// n must be odd at this point. So we can
// skip one element (Note i = i +2)
for (int i = 3; i <= Math.Sqrt(n); i = i + 2)
{
// Check if i is a prime factor
if (n % i == 0)
{
n = n / i;
// If i again divides, then
// n is not square free
if (n % i == 0)
return false;
}
}
return true;
}
// Driver program
public static void Main()
{
int n = 10;
if (isSquareFree(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m.PHP
Javascript
输出：
Yes时间复杂度： O(sqrt(N))
在最坏的情况下，如果该数字是一个完美的平方，那么将存在sqrt(n)/ 2次迭代。