在 M 名学生中分发 N 套试卷的方式数量

原文:https://www . geeksforgeeks . org/在 m 学生中分发 n 套纸张的方式数/

给定 N 个学生，共 M 套试卷，其中 M ≤ N 。所有的 M 套都不一样，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，这样如果有任何 M 连续的学生被选中，那么每个学生都有一个独特的试卷集。答案可能很大，所以以 10 9 + 7 为模打印答案。 例:

输入: N = 2，M = 2 T3】输出: 2 (A，B)和(B，A)是唯一可能的方式。 输入: N = 15，M = 4 输出: 24

进路:可以观察到进路数与 N 无关，只依赖于 M 。先给 M 的学生发 M 套，然后重复同样的花样。这样分发试题纸的方式数量是 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

以下是上述方法的实现:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)

// 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

Python 3

# 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

java 描述语言

<script>
// 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
</script>

Output: 

2

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