计算表达式(N1 *(N – 1)2 *…* 1N)％(109 + 7)

给定整数N ，任务是找到表达式(N ¹ *(N – 1) ² *…* 1 ^N )％(10 ⁹ + 7)的值。

Input: N = 1 
Output: 1 
Explanation: 
1¹ = 1

Input: N = 4 
Output: 288 
Explanation: 
4¹ * (4 – 1)² * (4 – 2)³ * (4-3)⁴ 
= 4 * 9 * 8 * 1 
= 288

天真的方法：解决此问题的最简单方法是在[1，N]范围内进行迭代。对于^第i^次迭代，计算(N – i + 1) ^i的值。最后，打印每次迭代中所有计算值的乘积。
时间复杂度： O(N ² * log ₂ (N))
辅助空间： O(1)

高效的方法：可以基于以下观察来优化上述方法：

F(N) = N¹ * (N – 1)² * … * 1^N 
= N * (N – 1) * (N – 1) * (N – 2) * (N – 2) * (N – 2)* … 1 * 1 * 1 
= N * (N – 1) * (N – 2)*… * 1 * (N -1) * (N – 2) * …* 1 * … 
= N! * (N – 1)! * (N – 2)! * … * 1!

请按照以下步骤解决问题：

• 使用factorial(N)= N * factorial(N – 1)预先计算从1到N的阶乘值。
• 使用以上观察结果，在[1，N]范围内进行迭代，并找到在[1，N]范围内的所有阶乘的乘积
• 最后，打印表达式的值。

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
#define mod 1000000007
 
// Function to find the value of the expression
// ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
int ValOfTheExpression(int n)
{
 
    // factorial[i]: Stores factorial of i
    int factorial[n] = { 0 };
 
    // Base Case for factorial
    factorial[0] = factorial[1] = 1;
 
    // Precompute the factorial
    for (int i = 2; i <= n; i++) {
        factorial[i] = ((factorial[i - 1] % mod)
                        * (i % mod))
                       % mod;
    }
 
    // dp[N]: Stores the value of the expression
    // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    int dp[n] = { 0 };
    dp[1] = 1;
 
    for (int i = 2; i <= n; i++) {
 
        // Update dp[i]
        dp[i] = ((dp[i - 1] % mod)
                 * (factorial[i] % mod))
                % mod;
    }
 
    // Return the answer.
    return dp[n];
}
 
// Driver Code
int main()
{
 
    int n = 4;
    // Function call
    cout << ValOfTheExpression(n) << "\n";
}

Java

// Java program to implement
// the above approach
class GFG
{
  static int mod = 1000000007;
 
  // Function to find the value of the expression
  // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
  static int ValOfTheExpression(int n)
  {
 
    // factorial[i]: Stores factorial of i
    int[] factorial = new int[n + 1];
 
    // Base Case for factorial
    factorial[0] = factorial[1] = 1;
 
    // Precompute the factorial
    for (int i = 2; i <= n; i++)
    {
      factorial[i] = ((factorial[i - 1] % mod)
                      * (i % mod)) % mod;
    }
 
    // dp[N]: Stores the value of the expression
    // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    int[] dp = new int[n + 1];
    dp[1] = 1;
 
    for (int i = 2; i <= n; i++)
    {
 
      // Update dp[i]
      dp[i] = ((dp[i - 1] % mod)
               * (factorial[i] % mod)) % mod;
    }
 
    // Return the answer.
    return dp[n];
  }
 
  // Driver code
  public static void main(String[] args) {
    int n = 4;
 
    // Function call
    System.out.println(ValOfTheExpression(n));
  }
}
 
// This code is contributed by divyesh072019

Python3

# Python 3 program to implement
# the above approach
mod = 1000000007
 
# Function to find the value of the expression
# ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
def ValOfTheExpression(n):
    global mod
     
    # factorial[i]: Stores factorial of i
    factorial = [0 for i in range(n + 1)]
 
    # Base Case for factorial
    factorial[0] = 1
    factorial[1] = 1
 
    # Precompute the factorial
    for i in range(2, n + 1, 1):
        factorial[i] = ((factorial[i - 1] % mod) * (i % mod))%mod
 
    # dp[N]: Stores the value of the expression
    # ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    dp = [0 for i in range(n+1)]
    dp[1] = 1
    for i in range(2, n + 1, 1):
       
        # Update dp[i]
        dp[i] = ((dp[i - 1] % mod)*(factorial[i] % mod)) % mod
 
    # Return the answer.
    return dp[n]
 
# Driver Code
if __name__ == '__main__':
    n = 4
     
    # Function call
    print(ValOfTheExpression(n))
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program to implement
// the above approach
using System;
class GFG
{
 
  static int mod = 1000000007;
 
  // Function to find the value of the expression
  // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
  static int ValOfTheExpression(int n)
  {
 
    // factorial[i]: Stores factorial of i
    int[] factorial = new int[n + 1];
 
    // Base Case for factorial
    factorial[0] = factorial[1] = 1;
 
    // Precompute the factorial
    for (int i = 2; i <= n; i++)
    {
      factorial[i] = ((factorial[i - 1] % mod)
                      * (i % mod))
        % mod;
    }
 
    // dp[N]: Stores the value of the expression
    // ( N^1 * (N – 1)^2 * … * 1^N) % (109 + 7).
    int[] dp = new int[n + 1];
    dp[1] = 1;
 
    for (int i = 2; i <= n; i++)
    {
 
      // Update dp[i]
      dp[i] = ((dp[i - 1] % mod)
               * (factorial[i] % mod))
        % mod;
    }
 
    // Return the answer.
    return dp[n];
  }
 
  // Driver code
  static void Main()
  {
    int n = 4;
 
    // Function call
    Console.WriteLine(ValOfTheExpression(n));
  }
}
 
// This code is contributed by divyeshrabadiya07

输出：

288

时间复杂度： O(N)
辅助空间： O(N)