为行中的所有框上色，以便每M个连续的框都是唯一的

给定N个保持直线的盒子和M个颜色，使得M≤N 。盒子的位置无法更改。任务是找到给盒子上色的方法的数量，这样，如果考虑了M个连续的盒子集，那么每个盒子的颜色都是唯一的。由于答案可能很大，因此以10 ⁹ + 7为模数打印答案。
例子：

Input: N = 3, M = 2 
Output: 2 
If colours are c1 and c2 the only possible 
ways are {c1, c2, c1} and {c2, c1, c2}.
Input: N = 13, M = 10 
Output: 3628800 
 

方法：方法的数量与N无关，并且仅取决于M。可以用给定的M个颜色对前M个盒子进行着色而无需重复，然后可以对下一组M个盒子重复相同的图案。可以对颜色的每个排列进行此操作。因此，为盒子上色的方式将为M！ 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
#define MOD 1000000007
 
// Function to return (m! % MOD)
int modFact(int n, int m)
{
    int result = 1;
    for (int i = 1; i <= m; i++)
        result = (result * i) % MOD;
 
    return result;
}
 
// Driver code
int main()
{
    int n = 3, m = 2;
 
    cout << modFact(n, m);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
    static final int MOD = 1000000007;
     
    // Function to return (m! % MOD)
    static int modFact(int n, int m)
    {
        int result = 1;
        for (int i = 1; i <= m; i++)
            result = (result * i) % MOD;
     
        return result;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 3, m = 2;
     
        System.out.println(modFact(n, m));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
MOD = 1000000007
 
# Function to return (m! % MOD)
def modFact(n, m) :
     
    result = 1
    for i in range(1, m + 1) :
        result = (result * i) % MOD
 
    return result
 
# Driver code
n = 3
m = 2
 
print(modFact(n, m))
 
# This code is contributed by
# divyamohan123

C#

// C# implementation of the above approach
using System;
class GFG
{
    const int MOD = 1000000007;
     
    // Function to return (m! % MOD)
    static int modFact(int n, int m)
    {
        int result = 1;
        for (int i = 1; i <= m; i++)
            result = (result * i) % MOD;
     
        return result;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 3, m = 2;
     
        Console.WriteLine(modFact(n, m));
    }
}
 
// This code is contributed by Nidhi_biet

Javascript

// Javascript implementation of the approach
 
const MOD = 1000000007;
 
// Function to return (m! % MOD)
function modFact(n, m)
{
    let result = 1;
    for (let i = 1; i <= m; i++)
        result = (result * i) % MOD;
 
    return result;
}
 
// Driver code
    let n = 3, m = 2;
 
    document.write(modFact(n, m));
 

输出：

2

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