检查给定的号码是否被电源隔离

给定一个整数N，素数分解为n1 ^p1 * n2 ^p2 ……任务是检查整数N是否与功率隔离。

An integer is said to be power-isolated if n1 * p1 * n2 * p2 ….. = N. 

例子：

Input: N = 12
Output: Power-isolated Integer.

Input: N = 18
Output: Not a power-isolated integer.

方法：对于要进行幂隔离的整数，其素数和幂的乘积等于整数本身。因此，要计算相同的值，您还必须找到给定整数的所有素数因子以及它们各自的幂。稍后，计算其乘积并检查乘积是否等于整数。
算法：

• Find the prime factor with their factor and store them in key-value pair.
• Later calculate product of all factors and their powers.
• if product is equal to integer, print true else false.

下面是上述算法的实现：

C++

// C++ program to find whether a number
// is power-isolated or not
#include 
using namespace std;
 
void checkIfPowerIsolated(int num)
{
    int input = num;
    int count = 0;
    int factor[num + 1]={0};
 
    // for 2 as prime factor
    if(num % 2 == 0)
    {
        while(num % 2 == 0)
        {
            ++count;
            num/=2;
        }
        factor[2] = count;
    }
 
    // for odd prime factor
    for (int i = 3; i*i <= num; i += 2)
    {
        count = 0;
        while(num % i == 0)
        {
            ++count;
            num /= i;
        }
        if(count > 0)
            factor[i] = count;
    }
         
    if(num > 1)
        factor[num] = 1;
         
    // calculate product of powers and prime factors
    int product = 1;
    for(int i = 0; i < num + 1; i++)
    {
        if(factor[i] > 0)
            product = product * factor[i] * i;
    }
         
    // check result for power-isolation
    if (product == input)
        cout << "Power-isolated Integer\n";
    else
        cout << "Not a Power-isolated Integer\n";
}
 
// Driver code
int main()
{
    checkIfPowerIsolated(12);
    checkIfPowerIsolated(18);
    checkIfPowerIsolated(35);
    return 0;
}
 
// This code is contributed by mits

Java

// Java program to find whether a number
// is power-isolated or not
class GFG
{
     
static void checkIfPowerIsolated(int num)
{
    int input = num;
    int count = 0;
    int[] factor= new int[num+1];
 
    // for 2 as prime factor
    if(num % 2 == 0)
    {
        while(num % 2 == 0)
        {
            ++count;
            num/=2;
        }
        factor[2] = count;
    }
 
    // for odd prime factor
    for (int i = 3; i*i <= num; i += 2)
    {
        count = 0;
        while(num % i == 0)
        {
            ++count;
            num /= i;
        }
        if(count > 0)
            factor[i] = count;
    }
         
    if(num > 1)
        factor[num] = 1;
         
    // calculate product of powers and prime factors
    int product = 1;
    for(int i = 0; i < num + 1; i++)
    {
        if(factor[i] > 0)
            product = product * factor[i] * i;
    }
         
    // check result for power-isolation
    if (product == input)
        System.out.print("Power-isolated Integer\n");
    else
        System.out.print("Not a Power-isolated Integer\n");
}
 
// Driver code
public static void main(String[] args)
{
    checkIfPowerIsolated(12);
    checkIfPowerIsolated(18);
    checkIfPowerIsolated(35);
}
}
 
// This code is contributed by Code_Mech.

Python3

# Python3 program to find whether a number
# is power-isolated or not
 
def checkIfPowerIsolated(num):
 
    input1 = num;
    count = 0;
    factor = [0] * (num + 1);
 
    # for 2 as prime factor
    if(num % 2 == 0):
        while(num % 2 == 0):
            count += 1;
            num //= 2;
        factor[2] = count;
 
    # for odd prime factor
    i = 3;
    while(i * i <= num):
        count = 0;
        while(num % i == 0):
            count += 1;
            num //= i;
        if(count > 0):
            factor[i] = count;
        i += 2;
         
    if(num > 1):
        factor[num] = 1;
     
    # calculate product of powers and prime factors
    product = 1;
    for i in range(0, len(factor)):
        if(factor[i] > 0):
            product = product * factor[i] * i;
         
    # check result for power-isolation
    if (product == input1):
        print("Power-isolated Integer");
    else:
        print("Not a Power-isolated Integer");
 
# Driver code
checkIfPowerIsolated(12);
checkIfPowerIsolated(18);
checkIfPowerIsolated(35);
 
# This code is contributed by mits

C#

// C# program to find whether a number
// is power-isolated or not
using System;
 
class GFG
{
static void checkIfPowerIsolated(int num)
{
    int input = num;
    int count = 0;
    int[] factor= new int[num+1];
 
    // for 2 as prime factor
    if(num % 2 == 0)
    {
        while(num % 2 == 0)
        {
            ++count;
            num/=2;
        }
        factor[2] = count;
    }
 
    // for odd prime factor
    for (int i = 3; i*i <= num; i += 2)
    {
        count = 0;
        while(num % i == 0)
        {
            ++count;
            num /= i;
        }
        if(count > 0)
            factor[i] = count;
    }
         
    if(num > 1)
        factor[num] = 1;
         
    // calculate product of powers and prime factors
    int product = 1;
    for(int i = 0; i < num + 1; i++)
    {
        if(factor[i] > 0)
            product = product * factor[i] * i;
    }
         
    // check result for power-isolation
    if (product == input)
        Console.Write("Power-isolated Integer\n");
    else
        Console.Write("Not a Power-isolated Integer\n");
}
 
// Driver code
static void Main()
{
    checkIfPowerIsolated(12);
    checkIfPowerIsolated(18);
    checkIfPowerIsolated(35);
}
}
 
// This code is contributed by mits

PHP

1)
           $factor[$num] = 1;
        // calculate product of powers and prime factors 
        $product  = 1;
        foreach ($factor as $primefactor => $power) {
            $product = $product * $primefactor * $power;
        }
         
        // check result for power-isolation
        if ($product ==  $input)
            print_r("Power-isolated Integer\n");
        else
            print_r("Not a Power-isolated Integer\n");
}
 
// driver code
checkIfPowerIsolated(12);
checkIfPowerIsolated(18);
checkIfPowerIsolated(35);
 
?>

Javascript

 
// Javascript program to find whether a number
// is power-isolated or not
 
function checkIfPowerIsolated(num)
{
    let input = num;
    let count = 0;
    let factor = new Array(0);
 
    // for 2 as prime factor
    if(num % 2 == 0)
    {
        while(num % 2 == 0)
        {
            ++count;
            num/=2;
        }
        factor[2] = count;
    }
 
    // for odd prime factor
    for (let i = 3; i*i <= num; i += 2)
    {
        count = 0;
        while(num % i == 0)
        {
            ++count;
            num /= i;
        }
        if(count > 0)
            factor[i] = count;
    }
         
    if(num > 1)
        factor[num] = 1;
         
    // calculate product of powers and prime factors
    let product = 1;
    for(let i = 0; i < num + 1; i++)
    {
        if(factor[i] > 0)
            product = product * factor[i] * i;
    }
         
    // check result for power-isolation
    if (product == input)
        document.write("Power-isolated Integer" + "<br>");
    else
        document.write("Not a Power-isolated Integer" + "<br>");
}
 
// Driver code
 
    checkIfPowerIsolated(12);
    checkIfPowerIsolated(18);
    checkIfPowerIsolated(35);
 
// This code is contributed by Mayank Tyagi
 

输出：

Power-isolated Integer
Not a Power-isolated Integer
Power-isolated Integer

时间复杂度： O(num)

辅助空间： O(num)