给定一个偶数N ，任务是检查它是否为一个完美数而不找到其除数。
In number theory, an Even Perfect Number is a positive integer which is even or that is equal to the sum of its positive divisors, exccluding the number itself.
An even perfect number can be represented as P * (P + 1) / 2 where P is Mersenne Prime.
A Mersenne Prime is a prime number of form 2q – 1 where q is also a prime number.
For example: if N = 6,
If we choose q to be 2 (prime number) then mersenne prime (P) is 22 – 1 = 3.
Therefore, the Even perfect number formed by the formula is 3 * (3 + 1) / 2 = 6.
例子：
Input: N = 6
Output: Yes
Explanation:
The integer 6 can be written as 6 = 1 + 2 + 3. Hence, its perfect number.
Input: N =156
Output: No
Explanation:
The integer 156 cannot be written as a sum of its divisors. Hence, its not a perfect number.
方法：
- 找到给定数字的平方根，得到接近2 q – 1的数字。
- 从数字的平方根中找到q-1，然后检查2 q-1 *(2 q -1)是否给出输入的数字。如果不是，那么它不是一个完美的数字，否则请继续。
- 检查q是否为质数。如果不是素数，则2 q -1不能为素数，随后检查2 q -1是否为素数。
- 如果上述所有条件都成立，那么它是一个偶数个完美的数，否则不是。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
bool isPrime(long n);
// Function to check for perfect number
void check(long num)
{
// Find a number close to 2^q-1
long root = (long)sqrt(num);
// Calculate q-1
long poww = (long)(log(root) / log(2));
// Condition of perfect number
if (num == (long)(pow(2, poww) *
(pow(2, poww + 1) - 1)))
{
// Check whether q is prime or not
if (isPrime(poww + 1))
{
// Check whether 2^q - 1 is a
// prime number or not
if (isPrime((long)pow(2,
poww + 1) - 1))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
else
cout << "No" << endl;
}
else
cout << "No" << endl;
}
// Function to check for prime number
bool isPrime(long n)
{
if (n <= 1)
return false;
// Check whether it is equal to 2 or 3
else if (n == 2 || n == 3)
return true;
else
{
// Check if it can be divided by 2
// and 3 then it is not prime number
if (n % 2 == 0 || n % 3 == 0)
return false;
// Check whether the given number be
// divide by other prime numbers
for(long i = 5; i <= sqrt(n); i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
}
// Driver Code
int main()
{
long num = 6;
check(num);
return 0;
}
// This code is contributed by rutvik_56 Java
// Java program for the above approach
class GFG {
// Function to check for perfect number
private static void check(long num)
{
// Find a number close to 2^q-1
long root = (long)Math.sqrt(num);
// Calculate q-1
long pow
= (long)(Math.log(root)
/ Math.log(2));
// Condition of perfect number
if (num
== (long)(Math.pow(2, pow)
* (Math.pow(2, pow + 1) - 1))) {
// Check whether q is prime or not
if (isPrime(pow + 1)) {
// Check whether 2^q - 1 is a
// prime number or not
if (isPrime(
(long)Math.pow(
2, pow + 1)
- 1))
System.out.println("Yes");
else
System.out.println("No");
}
else
System.out.println("No");
}
else
System.out.println("No");
}
// Function to check for prime number
public static boolean isPrime(long n)
{
if (n <= 1)
return false;
// Check whether it is equal to 2 or 3
else if (n == 2 || n == 3)
return true;
else {
// Check if it can be divided by 2
// and 3 then it is not prime number
if (n % 2 == 0 || n % 3 == 0)
return false;
// Check whether the given number be
// divide by other prime numbers
for (long i = 5;
i <= Math.sqrt(n);
i += 6) {
if (n % i == 0
|| n % (i + 2) == 0)
return false;
}
return true;
}
}
// Driver code
public static void main(String args[])
{
long num = 6;
check(num);
}
}Python3
# Python3 program for the above approach
import math
# Function to check for perfect number
def check(num):
# Find a number close to 2^q-1
root = (int)(math.sqrt(num))
# Calculate q-1
poww = (int)(math.log(root) /
math.log(2))
# Condition of perfect number
if (num == (int)(pow(2, poww) *
(pow(2, poww + 1) - 1))):
# Check whether q is prime or not
if (isPrime(poww + 1)):
# Check whether 2^q - 1 is a
# prime number or not
if (isPrime((int)(pow(2,
poww + 1)) - 1)):
print("Yes")
else:
print("No")
else:
print("No")
else:
print("No")
# Function to check for prime number
def isPrime(n):
if (n <= 1):
return bool(False)
# Check whether it is equal to 2 or 3
elif (n == 2 or n == 3):
return bool(True)
else:
# Check if it can be divided by 2
# and 3 then it is not prime number
if (n % 2 == 0 or n % 3 == 0):
return bool(False)
# Check whether the given number be
# divide by other prime numbers
for i in range(5, sqrt(n + 1) + 1, 6):
if (n % i == 0 or n % (i + 2) == 0):
return bool(False)
return bool(True)
# Driver Code
num = 6
check(num)
# This code is contributed by divyeshrabadiya07C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check for perfect number
private static void check(long num)
{
// Find a number close to 2^q-1
long root = (long)Math.Sqrt(num);
// Calculate q-1
long pow = (long)(Math.Log(root) /
Math.Log(2));
// Condition of perfect number
if (num == (long)(Math.Pow(2, pow) *
(Math.Pow(2, pow + 1) - 1)))
{
// Check whether q is prime or not
if (isPrime(pow + 1))
{
// Check whether 2^q - 1 is a
// prime number or not
if (isPrime((long)Math.Pow(2, pow + 1) - 1))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
else
Console.WriteLine("No");
}
else
Console.WriteLine("No");
}
// Function to check for prime number
public static bool isPrime(long n)
{
if (n <= 1)
return false;
// Check whether it is equal to 2 or 3
else if (n == 2 || n == 3)
return true;
else
{
// Check if it can be divided by 2
// and 3 then it is not prime number
if (n % 2 == 0 || n % 3 == 0)
return false;
// Check whether the given number be
// divide by other prime numbers
for(long i = 5;
i <= Math.Sqrt(n);
i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
}
// Driver code
public static void Main(String []args)
{
long num = 6;
check(num);
}
}
// This code is contributed by amal kumar choubey输出：
Yes
时间复杂度： O(N 1/4 )
辅助空间： O(1)