给定整数N ，任务是检查前N个自然数的乘积是否可被前N个自然数的总和整除。
例子：
Input: N = 3
Output: Yes
Product = 1 * 2 * 3 = 6
Sum = 1 + 2 + 3 = 6
Input: N = 6
Output: No
天真的方法：找到前N个自然数的和与乘积，并检查乘积是否可除。
高效方法：我们知道，前N个自然数的和与乘积为sum =(N *(N + 1))/ 2和乘积= N！分别。现在要检查乘积是否可被总和整除，我们需要检查以下方程式的余数是否为0。
N! / (N *(N + 1) / 2)
2 * (N – 1)! / N + 1
i.e. every factor of (N + 1) should be in (2 * (N – 1)!). So, if (N + 1) is a prime then we are sure that the product is not divisible by the sum.
So ultimately just check if (N + 1) is prime or not.
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true if n is prime
bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function that return true if the product
// of the first n natural numbers is divisible
// by the sum of first n natural numbers
bool isDivisible(int n)
{
if (isPrime(n + 1))
return false;
return true;
}
// Driver code
int main()
{
int n = 6;
if (isDivisible(n))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function that returns true if n is prime
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function that return true if the product
// of the first n natural numbers is divisible
// by the sum of first n natural numbers
static boolean isDivisible(int n)
{
if (isPrime(n + 1))
return false;
return true;
}
// Driver code
public static void main(String[] args)
{
int n = 6;
if (isDivisible(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Code_Mech.Python3
# Python 3 implementation of the approach
from math import sqrt
# Function that returns true if n is prime
def isPrime(n):
# Corner cases
if (n <= 1):
return False
if (n <= 3):
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 and n % 3 == 0):
return False
for i in range(5, int(sqrt(n)) + 1, 6):
if (n % i == 0 and n % (i + 2) == 0):
return False
return True
# Function that return true if the product
# of the first n natural numbers is divisible
# by the sum of first n natural numbers
def isDivisible(n):
if (isPrime(n + 1)):
return False
return True
# Driver code
if __name__ == '__main__':
n = 6
if (isDivisible(n)):
print("Yes")
else:
print("No")
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if n is prime
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function that return true if the product
// of the first n natural numbers is divisible
// by the sum of first n natural numbers
static bool isDivisible(int n)
{
if (isPrime(n + 1))
return false;
return true;
}
// Driver code
static void Main()
{
int n = 6;
if (isDivisible(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by mitsPHP
Javascript
输出：
No