在M名学生中分发N份论文集的方法数量

给定N名学生和M题集，其中M≤N 。所有M套都不同，并且每套都有足够的数量。所有的学生都坐在一排。任务是找到分发试卷的方法，以便如果选择了M个连续的学生，则每个学生都有一个唯一的试卷集。答案可能很大，因此以10 ⁹ + 7为模数打印答案。
例子：

Input: N = 2, M = 2 
Output: 2 
(A, B) and (B, A) are the only possible ways.
Input: N = 15, M = 4 
Output: 24 
 

方法：可以观察到，方法的数量与N无关，并且仅取决于M。首先可以给M个学生M套，然后可以重复相同的模式。以这种方式分发试卷的方式为M！ 。例如，

N = 6, M = 3 
A, B, C, A, B, C 
A, C, B, A, C, B 
B, C, A, B, C, A 
B, A, C, B, A, C 
C, A, B, C, A, B 
C, B, A, C, B, A 
 

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
const int MOD = 1000000007;
 
// Function to return n! % 1000000007
int factMod(int n)
{
 
    // To store the factorial
    long fact = 1;
 
    // Find the factorial
    for (int i = 2; i <= n; i++) {
        fact *= (i % MOD);
        fact %= MOD;
    }
 
    return fact;
}
 
// Function to return the
// count of possible ways
int countWays(int n, int m)
{
    return factMod(m);
}
 
// Driver code
int main()
{
    int n = 2, m = 2;
 
    cout << countWays(n, m);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
static int MOD = 1000000007;
 
// Function to return n! % 1000000007
static int factMod(int n)
{
 
    // To store the factorial
    long fact = 1;
 
    // Find the factorial
    for (int i = 2; i <= n; i++)
    {
        fact *= (i % MOD);
        fact %= MOD;
    }
    return (int)fact;
}
 
// Function to return the
// count of possible ways
static int countWays(int n, int m)
{
    return factMod(m);
}
 
// Driver code
public static void main(String args[])
{
    int n = 2, m = 2;
 
    System.out.print(countWays(n, m));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
MOD = 1000000007;
 
# Function to return n! % 1000000007
def factMod(n) :
 
    # To store the factorial
    fact = 1;
 
    # Find the factorial
    for i in range(2, n + 1) :
        fact *= (i % MOD);
        fact %= MOD;
 
    return fact;
 
# Function to return the
# count of possible ways
def countWays(n, m) :
 
    return factMod(m);
 
# Driver code
if __name__ == "__main__" :
 
    n = 2; m = 2;
 
    print(countWays(n, m));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
    static int MOD = 1000000007;
     
    // Function to return n! % 1000000007
    static int factMod(int n)
    {
        // To store the factorial
        int fact = 1;
     
        // Find the factorial
        for (int i = 2; i <= n; i++)
        {
            fact *= (i % MOD);
            fact %= MOD;
        }
        return fact;
    }
 
    // Function to return the
    // count of possible ways
    static int countWays(int n, int m)
    {
        return factMod(m);
    }
     
    // Driver code
    public static void Main()
    {
        int n = 2, m = 2;
        Console.Write(countWays(n, m));
    }
}
 
// This code is contributed by Sanjit Prasad

Javascript

// javascript implementation of the approach
     MOD = 1000000007;
 
    // Function to return n! % 1000000007
    function factMod(n) {
 
        // To store the factorial
        var fact = 1;
 
        // Find the factorial
        for (i = 2; i <= n; i++) {
            fact *= (i % MOD);
            fact %= MOD;
        }
        return parseInt( fact);
    }
 
    // Function to return the
    // count of possible ways
    function countWays(n , m) {
        return factMod(m);
    }
 
    // Driver code
     
        var n = 2, m = 2;
 
        document.write(countWays(n, m));
 
// This code contributed by aashish1995

输出：

2

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