通过交换成对的相同定位位，使前N – 1个自然数的乘积最小

给定整数N ，任务是通过任意次数交换任意两个数字的^第i^个位，找到前N个自然数(即[1，(N – 1)])的最小正乘积。

注意： N始终是2的理想幂。由于乘积可能很大，所以以10 ⁹ + 7为模输出答案。

例子：

Input: N = 4
Output: 6
Explanation:
No swapping of bits is required. Therefore, the minimum product is 1*2*3 = 6.

Input: N = 8
Output: 1512
Explanation:
Let the array arr[] stores all the value from 1 to N as {1, 2, 3, 4, 5, 6, 7}
Follow the below steps:
Step 1: In elements 2 = (0010) and 5 = (0101), swap 0th and 1st bit. Therefore, replace 2 with 1 and 5 with 6. arr[] = {1, 1, 3, 4, 6, 6, 7}.
Step 2: In elements 3 = (0011) and 4 = (0100), swap 1th bit. Therefore, replace 3 with 1 and 4 with 6. arr[] = {1, 1, 1, 6, 6, 6, 7}.
Hence, the minimum product = 1*1*1*6*6*6*7 = 1512 % 1e9+7 = 1512.

方法：这个想法是做一些观察。例如，如果N = 8且arr [] = {1,2,3,4,5,6,7} ，请注意，要使乘积最小，必须有三个6，即，必须有一个具有值的元素(N – 2)的出现频率为(1 +(N – 4)/ 2)，并且必须有3个，即，必须有(1 +(N – 4)/ 2)个。最后，将当前乘积乘以(N – 1) 。因此，公式变为：

Minimum product for any value N = ((N – 1) * (N – 2)^{(N – 4)/2 + 1}) % 1e9 + 7 

请按照以下步骤解决问题：

1. 将ans初始化为1 。
2. 在[0，1 +(N – 4)/ 2]范围内迭代。
3. 在每次遍历中，将ans乘以N – 2并将ans更新为ans mod 1e9 + 7 。
4. 完成上述步骤后，打印出ans *(N – 1)mod 1e9 + 7的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
int mod = 1e9 + 7;
 
// Function to find the minimum product
// of 1 to N - 1 after performing the
// given operations
void minProduct(int n)
{
    // Initialize ans with 1
    int ans = 1;
 
    // Multiply ans with N-2
    // ((N - 4)/2) times
    for (int i = 1;
         i <= (n - 4) / 2; i++) {
        ans = (1LL * ans
               * (n - 2))
              % mod;
    }
 
    // Multiply ans with N - 1
    // and N - 2 once
    ans = (1LL * ans
           * (n - 2) * (n - 1))
          % mod;
 
    // Print ans
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 8;
 
    // Function Call
    minProduct(N);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
static int mod = (int)1e9 + 7;
 
// Function to find the
// minimum product of 1
// to N - 1 after performing
// the given operations
static void minProduct(int n)
{
  // Initialize ans with 1
  int ans = 1;
 
  // Multiply ans with N-2
  // ((N - 4)/2) times
  for (int i = 1;
           i <= (n - 4) / 2; i++)
  {
    ans = (int)(1L * ans *
               (n - 2)) % mod;
  }
 
  // Multiply ans with N - 1
  // and N - 2 once
  ans = (int)(1L * ans *
             (n - 2) * (n - 1)) % mod;
 
  // Print ans
  System.out.print(ans + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given Number N
  int N = 8;
 
  // Function Call
  minProduct(N);
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program for the above approach
mod = 1e9 + 7
 
# Function to find the minimum product
# of 1 to N - 1 after performing the
# given operations
def minProduct(n):
     
    # Initialize ans with 1
    ans = 1
 
    # Multiply ans with N-2
    # ((N - 4)/2) times
    for i in range(1, (n - 4) // 2 + 1):
        ans = (ans * (n - 2)) % mod
 
    # Multiply ans with N - 1
    # and N - 2 once
    ans = (ans * (n - 2) * (n - 1)) % mod
 
    # Print ans
    print(int(ans))
 
# Driver Code
if __name__ == '__main__':
     
    # Given number N
    N = 8
 
    # Function call
    minProduct(N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the
// above approach
using System;
class GFG{
 
static int mod = (int)1e9 + 7;
 
// Function to find the
// minimum product of 1
// to N - 1 after performing
// the given operations
static void minProduct(int n)
{
  // Initialize ans with 1
  int ans = 1;
 
  // Multiply ans with N-2
  // ((N - 4)/2) times
  for (int i = 1;
           i <= (n - 4) / 2; i++)
  {
    ans = (int)(1L * ans *
               (n - 2)) % mod;
  }
 
  // Multiply ans with N - 1
  // and N - 2 once
  ans = (int)(1L * ans *
             (n - 2) *
             (n - 1)) % mod;
 
  // Print ans
  Console.Write(ans + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given Number N
  int N = 8;
 
  // Function Call
  minProduct(N);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// JavaScript program for the above approach
 
let mod = 1e9 + 7;
 
// Function to find the minimum product
// of 1 to N - 1 after performing the
// given operations
function minProduct(n)
{
    // Initialize ans with 1
    let ans = 1;
 
    // Multiply ans with N-2
    // ((N - 4)/2) times
    for (let i = 1; i <= Math.floor((n - 4) / 2); i++)
    {
        ans = (1 * ans * (n - 2)) % mod;
    }
 
    // Multiply ans with N - 1
    // and N - 2 once
    ans = (1 * ans * (n - 2) * (n - 1)) % mod;
 
    // Print ans
    document.write(ans + "<br>");
}
 
// Driver Code
 
    // Given Number N
    let N = 8;
 
    // Function Call
    minProduct(N);
 
// This code is contributed by Manoj.

输出

1512

时间复杂度： O(N)，其中N是给定的整数。
辅助空间： O(1)