N个数字，可从M个数字中除以5

给定M个唯一数字和一个数字N。任务是查找可以由给定M个数字形成的N个数字的数目，该数字可以被5整除，并且没有重复的数字。
注意：如果无法从给定的数字中形成N位数字，请打印-1。
例子：

Input : N = 3, M = 6, digits[] = {2, 3, 5, 6, 7, 9} 
Output : 20
Input : N = 5, M = 6, digits[] = {0, 3, 5, 6, 7, 9} 
Output : 240

对于要被5整除的数字，唯一的条件是数字中单位位置的数字必须为0或5 。
因此，要查找可以被给定数字形成的可被5整除的数字计数，请执行以下操作：

• 检查给定的数字是否同时包含0和5。
• 如果给定的数字同时包含0和5，则可以用2种方式填充单位位置，否则可以用1种方式填充单位位置。
• 现在，十位数可以用剩余的M-1个数字填充。因此，有(M-1)种填充十位的方法。
• 同样，现在可以用其余(M-2)个数字中的任意一个填充百位，依此类推。

因此，如果给定的数字同时具有0和5：

Required number of numbers = 2 * (M-1)* (M-2)...N-times. 

否则，如果给定的数字具有0和5之一而不是两者都为：

Required number of numbers = 1 * (M-1)* (M-2)...N-times. 

下面是上述方法的实现。

C++

// CPP program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
#include 
using namespace std;
 
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
int numbers(int n, int arr[], int m)
{
    int isZero = 0, isFive = 0;
    int result = 0;
 
    // If it is not possible to form
    // n digit number from the given
    // m digits without repetition
    if (m < n) {
        return -1;
    }
 
    for (int i = 0; i < m; i++) {
        if (arr[i] == 0)
            isZero = 1;
 
        if (arr[i] == 5)
            isFive = 1;
    }
 
    // If both zero and five exists
    if (isZero && isFive) {
        result = 2;
 
        // Remaining N-1 iterations
        for (int i = 0; i < n - 1; i++) {
            result = result * (--m);
        }
    }
    else if (isZero || isFive) {
        result = 1;
 
        // Remaining N-1 iterations
        for (int i = 0; i < n - 1; i++) {
            result = result * (--m);
        }
    }
    else
        result = -1;
 
    return result;
}
 
// Driver code
int main()
{
    int n = 3, m = 6;
 
    int arr[] = { 2, 3, 5, 6, 7, 9 };
 
    cout << numbers(n, arr, m);
 
    return 0;
}

Java

// Java program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
 
class GFG {
 
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
    static int numbers(int n, int arr[], int m) {
        int isZero = 0, isFive = 0;
        int result = 0;
 
        // If it is not possible to form
        // n digit number from the given
        // m digits without repetition
        if (m < n) {
            return -1;
        }
 
        for (int i = 0; i < m; i++) {
            if (arr[i] == 0) {
                isZero = 1;
            }
 
            if (arr[i] == 5) {
                isFive = 1;
            }
        }
 
        // If both zero and five exists
        if (isZero == 1 && isFive == 1) {
            result = 2;
 
            // Remaining N-1 iterations
            for (int i = 0; i < n - 1; i++) {
                result = result * (--m);
            }
        } else if (isZero == 1 || isFive == 1) {
            result = 1;
 
            // Remaining N-1 iterations
            for (int i = 0; i < n - 1; i++) {
                result = result * (--m);
            }
        } else {
            result = -1;
        }
 
        return result;
    }
 
// Driver code
    public static void main(String[] args) {
        int n = 3, m = 6;
 
        int arr[] = {2, 3, 5, 6, 7, 9};
        System.out.println(numbers(n, arr, m));
 
    }
}
// This code is contributed by RAJPUT-JI

Python 3

# Python 3 program to find the count
# of all possible N digit numbers which
# are divisible by 5 formed from M digits
 
# Function to find the count of all
# possible N digit numbers which are
# divisible by 5 formed from M digits
def numbers(n, arr, m):
 
    isZero = 0
    isFive = 0
    result = 0
 
    # If it is not possible to form
    # n digit number from the given
    # m digits without repetition
    if (m < n) :
        return -1
 
    for i in range(m) :
        if (arr[i] == 0):
            isZero = 1
 
        if (arr[i] == 5):
            isFive = 1
 
    # If both zero and five exists
    if (isZero and isFive) :
        result = 2
 
        # Remaining N-1 iterations
        for i in range( n - 1):
            m -= 1
            result = result * (m)
 
    elif (isZero or isFive) :
        result = 1
 
        # Remaining N-1 iterations
        for i in range(n - 1) :
            m -= 1
            result = result * (m)
    else:
        result = -1
 
    return result
 
# Driver code
if __name__ == "__main__":
    n = 3
    m = 6
 
    arr = [ 2, 3, 5, 6, 7, 9]
 
    print(numbers(n, arr, m))
 
# This code is contributed by ChitraNayal

C#

// C# program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
using System;
public class GFG {
 
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
    static int numbers(int n, int []arr, int m) {
        int isZero = 0, isFive = 0;
        int result = 0;
 
        // If it is not possible to form
        // n digit number from the given
        // m digits without repetition
        if (m < n) {
            return -1;
        }
 
        for (int i = 0; i < m; i++) {
            if (arr[i] == 0) {
                isZero = 1;
            }
 
            if (arr[i] == 5) {
                isFive = 1;
            }
        }
 
        // If both zero and five exists
        if (isZero == 1 && isFive == 1) {
            result = 2;
 
            // Remaining N-1 iterations
            for (int i = 0; i < n - 1; i++) {
                result = result * (--m);
            }
        } else if (isZero == 1 || isFive == 1) {
            result = 1;
 
            // Remaining N-1 iterations
            for (int i = 0; i < n - 1; i++) {
                result = result * (--m);
            }
        } else {
            result = -1;
        }
 
        return result;
    }
 
// Driver code
    public static void Main() {
        int n = 3, m = 6;
 
        int []arr = {2, 3, 5, 6, 7, 9};
        Console.WriteLine(numbers(n, arr, m));
 
    }
}
// This code is contributed by RAJPUT-JI

Javascript

    // Javascript program to find the count of all
    // possible N digit numbers which are
    // divisible by 5 formed from M digits
     
    // Function to find the count of all
    // possible N digit numbers which are
    // divisible by 5 formed from M digits
    function numbers(n, arr, m) {
        let isZero = 0, isFive = 0;
        let result = 0;
   
        // If it is not possible to form
        // n digit number from the given
        // m digits without repetition
        if (m < n) {
            return -1;
        }
   
        for (let i = 0; i < m; i++) {
            if (arr[i] == 0) {
                isZero = 1;
            }
   
            if (arr[i] == 5) {
                isFive = 1;
            }
        }
   
        // If both zero and five exists
        if (isZero == 1 && isFive == 1) {
            result = 2;
   
            // Remaining N-1 iterations
            for (let i = 0; i < n - 1; i++) {
                result = result * (--m);
            }
        } else if (isZero == 1 || isFive == 1) {
            result = 1;
   
            // Remaining N-1 iterations
            for (let i = 0; i < n - 1; i++) {
                result = result * (--m);
            }
        } else {
            result = -1;
        }
   
        return result;
    }
     
    let n = 3, m = 6;
   
    let arr = [2, 3, 5, 6, 7, 9];
    document.write(numbers(n, arr, m));
     

输出：

20