检查N的阶乘是否可被前N个自然数的平方和除

给定一个正整数N，则任务是找到事实(N)是否是由总和整除(N)，其中的事实(N)是阶乘的N和总和(N)= 1 ² + 2 ² + 3 ² + … + N ² 。
例子：

Input: N = 5 
Output: No 
fact(N) = 120, sum(N) = 55 
And, 120 is not divisible by 55
Input: N = 7 
Output: Yes 
 

方法：

1. 在这里重要的是要首先实现所有数字平方和的封闭公式。前N个自然数的平方和。
2. 现在，由于n是N阶乘和求和的公共因子，我们可以将其删除。
3. 现在，对于值(N +1)*(2N +1)中的每个素数P，说值中有X个P因子，然后找到因式(N – 1)中P的因子个数，说它们是Y。如果Y
4. 要计算P因数(N)中的因数，我们可以简单地使用Lengendre公式。
5. 在第4点中，将质数2、3的计数加1以在求和公式中占6。
6. 单独检查“值”中的所有素数P，如果都满足条件3，则答案为“是”。
7. 第2点将帮助我们将时间复杂度降低N倍。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#define ll long long int
using namespace std;
 
// Function to count number of times
// prime P divide factorial N
bool checkfact(int N, int countprime, int prime)
{
    int countfact = 0;
    if (prime == 2 || prime == 3)
        countfact++;
    int divide = prime;
 
    // Lengendre Formula
    while (N / divide != 0) {
        countfact += N / divide;
        divide = divide * divide;
    }
 
    if (countfact >= countprime)
        return true;
    else
        return false;
}
 
// Function to find count number of times
// all prime P divide summation
bool check(int N)
{
 
    // Formula for summation of square after removing n
    // and constant 6
    int sumsquares = (N + 1) * (2 * N + 1);
    int countprime = 0;
 
    // Loop to traverse over all prime P which divide
    // summation
    for (int i = 2; i <= sqrt(sumsquares); i++) {
        int flag = 0;
 
        while (sumsquares % i == 0) {
            flag = 1;
            countprime++;
            sumsquares /= i;
        }
 
        if (flag) {
            if (!checkfact(N - 1, countprime, i))
                return false;
            countprime = 0;
        }
    }
 
    // If Number itself is a Prime Number
    if (sumsquares != 1)
        if (!checkfact(N - 1, 1, sumsquares))
            return false;
 
    return true;
}
 
// Driver Code
int main()
{
    int N = 5;
    if (check(N))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GfG
{
 
// Function to count number of times
// prime P divide factorial N
static boolean checkfact(int N, int countprime,
                                    int prime)
{
    int countfact = 0;
    if (prime == 2 || prime == 3)
        countfact++;
    int divide = prime;
 
    // Lengendre Formula
    while (N / divide != 0)
    {
        countfact += N / divide;
        divide = divide * divide;
    }
 
    if (countfact >= countprime)
        return true;
    else
        return false;
}
 
// Function to find count number of times
// all prime P divide summation
static boolean check(int N)
{
 
    // Formula for summation of square after removing n
    // and constant 6
    int sumsquares = (N + 1) * (2 * N + 1);
    int countprime = 0;
 
    // Loop to traverse over all prime P which divide
    // summation
    for (int i = 2; i <= Math.sqrt(sumsquares); i++)
    {
        int flag = 0;
 
        while (sumsquares % i == 0)
        {
            flag = 1;
            countprime++;
            sumsquares /= i;
        }
 
        if (flag == 1)
        {
            if (!checkfact(N - 1, countprime, i))
                return false;
            countprime = 0;
        }
    }
 
    // If Number itself is a Prime Number
    if (sumsquares != 1)
        if (!checkfact(N - 1, 1, sumsquares))
            return false;
 
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5;
    if (check(N))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Prerna Saini

Python3

# Python 3 implementation of the approach
from math import sqrt
 
# Function to count number of times
# prime P divide factorial N
def checkfact(N, countprime, prime):
    countfact = 0
    if (prime == 2 or prime == 3):
        countfact += 1
    divide = prime
 
    # Lengendre Formula
    while (int(N / divide ) != 0):
        countfact += int(N / divide)
        divide = divide * divide
 
    if (countfact >= countprime):
        return True
    else:
        return False
 
# Function to find count number of times
# all prime P divide summation
def check(N):
     
    # Formula for summation of square after
    # removing n and constant 6
    sumsquares = (N + 1) * (2 * N + 1)
    countprime = 0
 
    # Loop to traverse over all prime P
    # which divide summation
    for i in range(2, int(sqrt(sumsquares)) + 1, 1):
        flag = 0
 
        while (sumsquares % i == 0):
            flag = 1
            countprime += 1
            sumsquares /= i
 
        if (flag):
            if (checkfact(N - 1,
                countprime, i) == False):
                return False
            countprime = 0
 
    # If Number itself is a Prime Number
    if (sumsquares != 1):
        if (checkfact(N - 1, 1,
            sumsquares) == False):
            return False
 
    return True
 
# Driver Code
if __name__ == '__main__':
    N = 5
    if(check(N)):
        print("Yes")
    else:
        print("No")
         
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to count number of times
// prime P divide factorial N
static bool checkfact(int N, int countprime,
                              int prime)
{
    int countfact = 0;
    if (prime == 2 || prime == 3)
        countfact++;
    int divide = prime;
 
    // Lengendre Formula
    while (N / divide != 0)
    {
        countfact += N / divide;
        divide = divide * divide;
    }
 
    if (countfact >= countprime)
        return true;
    else
        return false;
}
 
// Function to find count number of times
// all prime P divide summation
static bool check(int N)
{
 
    // Formula for summation of square
    // after removing n and constant 6
    int sumsquares = (N + 1) * (2 * N + 1);
    int countprime = 0;
 
    // Loop to traverse over all prime P
    // which divide summation
    for (int i = 2; i <= Math.Sqrt(sumsquares); i++)
    {
        int flag = 0;
 
        while (sumsquares % i == 0)
        {
            flag = 1;
            countprime++;
            sumsquares /= i;
        }
 
        if (flag == 1)
        {
            if (!checkfact(N - 1, countprime, i))
                return false;
            countprime = 0;
        }
    }
 
    // If Number itself is a Prime Number
    if (sumsquares != 1)
        if (!checkfact(N - 1, 1, sumsquares))
            return false;
 
    return true;
}
 
// Driver Code
public static void Main()
{
    int N = 5;
    if (check(N))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed
// by Akanksha Rai

PHP

= $countprime)
        return true;
    else
        return false;
}
 
// Function to find count number of times
// all prime P divide summation
function check($N)
{
 
    // Formula for summation of square
    // after removing n and constant 6
    $sumsquares = ($N + 1) * (2 * $N + 1);
    $countprime = 0;
 
    // Loop to traverse over all prime P
    // which divide summation
    for ($i = 2; $i <= sqrt($sumsquares); $i++)
    {
        $flag = 0;
 
        while ($sumsquares % $i == 0)
        {
            $flag = 1;
            $countprime++;
            $sumsquares = (int)($sumsquares / $i);
        }
 
        if ($flag == 1)
        {
            if (checkfact($N - 1, $countprime, $i))
                return false;
            $countprime = 0;
        }
    }
 
    // If Number itself is a Prime Number
    if ($sumsquares != 1)
        if (checkfact($N - 1, 1, $sumsquares))
            return false;
 
    return true;
}
 
// Driver Code
$N = 5;
if (check($N))
    echo("Yes");
else
    echo("No");
 
// This code is contributed by Code_Mech
?>

Javascript

// javascript implementation of the approach
 
// Function to count number of times
// prime P divide factorial N
function checkfact(N , countprime, prime)
{
    var countfact = 0;
    if (prime == 2 || prime == 3)
        countfact++;
    var divide = prime;
 
    // Lengendre Formula
    while (N / divide != 0)
    {
        countfact += N / divide;
        divide = divide * divide;
    }
 
    if (countfact >= countprime)
        return true;
    else
        return false;
}
 
// Function to find count number of times
// all prime P divide summation
function check(N)
{
 
    // Formula for summation of square after removing n
    // and constant 6
    var sumsquares = (N + 1) * (2 * N + 1);
    var countprime = 0;
 
    // Loop to traverse over all prime P which divide
    // summation
    for (i = 2; i <= Math.sqrt(sumsquares); i++)
    {
        var flag = 0;
 
        while (sumsquares % i == 0)
        {
            flag = 1;
            countprime++;
            sumsquares /= i;
        }
 
        if (flag == 1)
        {
            if (!checkfact(N - 1, countprime, i))
                return false;
            countprime = 0;
        }
    }
 
    // If Number itself is a Prime Number
    if (sumsquares != 1)
        if (!checkfact(N - 1, 1, sumsquares))
            return false;
 
    return true;
}
 
// Driver Code
var N = 5;
if (check(N))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by Princi Singh

输出：

No