可以使用给定的一组数字生成的不同 N 位奇数整数的计数

给定一个大小为N的数组arr[]表示从0到9的数字，任务是计算可以使用数组中给定数字形成的不同奇数N位整数的数量。

例子：

Input: arr[] = {1, 0, 2, 4}
Output: 4
Explaination: The possible 4-digit odd integers that can be formed using the given digits are 2041, 2401, 4021 and 4201.

Input: arr[] = {2, 3, 4, 1, 2, 3}
Output: 90

方法：给定的问题可以通过以下观察来解决：

• 对于奇数，它的个位（即第 1^位）应该有一个奇数，即 {1, 3, 5, 7, 9}
• 由于整数应该有N位，因此最高有效位（^第N 位的数字）不能等于0 。
• 除了第 1^位和第 N^位的数字之外的所有数字都可以有任何其他数字。
• X位数的排列方式总数是X！ / ( freq[0]! * freq[1]! *…* freq[9]! ) ，其中freq[i]表示给定X位中第i位的频率。

为了解决这个问题，请跟踪变量odd中奇数位数和变量zero中等于 0 的位数。所以根据上面的观察，如果i代表第^N位， j代表第¹位，迭代i和j的所有可能值，对于每个有效的(i, j) ，计算排列方式的数量剩余的(N-2) 个数字。

下面是上述方法的实现：

C++14

// C++ Program for the above approach
#include 
using namespace std;
 
// Function to find the count of distinct
// odd integers with N digits using the
// given digits in the array arr[]
int countOddIntegers(int arr[], int N)
{
    // Stores the factorial of a number
    int Fact[N] = {};
 
    // Calculate the factorial of all
    // numbers from 1 to N
    Fact[0] = 1;
    for (int i = 1; i < N; i++) {
        Fact[i] = i * Fact[i - 1];
    }
 
    // Stores the frequency of each digit
    int freq[10] = {};
    for (int i = 0; i < N; i++) {
        freq[arr[i]]++;
    }
 
    // Stores the final answer
    int ans = 0;
 
    // Loop to iterate over all values of
    // Nth digit i and 1st digit j
    for (int i = 1; i <= 9; i += 2) {
 
        // If digit i does not exist in
        // the given array move to next i
        if (!freq[i])
            continue;
 
        // Fixing i as Nth digit
        freq[i]--;
 
        for (int j = 1; j <= 9; j++) {
 
            // Stores the answer of a specific
            // value of i and j
            int cur_ans = 0;
 
            // If digit j does not exist
            // move to the next j
            if (freq[j] == 0) {
                continue;
            }
 
            // Fixing j as 1st digit
            freq[j]--;
 
            // Calculate number of ways to
            // arrange remaining N-2 digits
            cur_ans = Fact[N - 2];
            for (int k = 0; k <= 9; k++) {
                cur_ans = cur_ans / Fact[freq[k]];
            }
            ans += cur_ans;
 
            // Including j back into
            // the set of digits
            freq[j]++;
        }
        // Including i back into the
        // set of the digits
        freq[i]++;
    }
 
    // Return Answer
    return ans;
}
 
// Driver Code
int main()
{
    int A[] = { 2, 3, 4, 1, 2, 3 };
    int N = sizeof(A) / sizeof(int);
 
    // Function Call
    cout << countOddIntegers(A, N);
 
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
 
class GFG{
 
// Function to find the count of distinct
// odd integers with N digits using the
// given digits in the array arr[]
static int countOddIntegers(int arr[], int N)
{
     
    // Stores the factorial of a number
    int Fact[] = new int[N];
 
    // Calculate the factorial of all
    // numbers from 1 to N
    Fact[0] = 1;
    for(int i = 1; i < N; i++)
    {
        Fact[i] = i * Fact[i - 1];
    }
 
    // Stores the frequency of each digit
    int freq[] = new int[10];
    for(int i = 0; i < N; i++)
    {
        freq[arr[i]]++;
    }
 
    // Stores the final answer
    int ans = 0;
 
    // Loop to iterate over all values of
    // Nth digit i and 1st digit j
    for(int i = 1; i <= 9; i += 2)
    {
         
        // If digit i does not exist in
        // the given array move to next i
        if (freq[i] == 0)
            continue;
 
        // Fixing i as Nth digit
        freq[i]--;
 
        for(int j = 1; j <= 9; j++)
        {
             
            // Stores the answer of a specific
            // value of i and j
            int cur_ans = 0;
 
            // If digit j does not exist
            // move to the next j
            if (freq[j] == 0)
            {
                continue;
            }
 
            // Fixing j as 1st digit
            freq[j]--;
 
            // Calculate number of ways to
            // arrange remaining N-2 digits
            cur_ans = Fact[N - 2];
            for(int k = 0; k <= 9; k++)
            {
                cur_ans = cur_ans / Fact[freq[k]];
            }
            ans += cur_ans;
 
            // Including j back into
            // the set of digits
            freq[j]++;
        }
         
        // Including i back into the
        // set of the digits
        freq[i]++;
    }
 
    // Return Answer
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 2, 3, 4, 1, 2, 3 };
    int N = A.length;
 
    // Function Call
    System.out.print(countOddIntegers(A, N));
}
}
 
// This code is contributed by code_hunt

Python3

# Python Program for the above approach
from array import *
from math import *
 
# Function to find the count of distinct
# odd integers with N digits using the
# given digits in the array arr[]
def countOddIntegers(arr, N):
   
    # Stores the factorial of a number
    # Calculate the factorial of all
    # numbers from 1 to N
    Fact = [0] * N
    Fact[0] = 1
    for i in range(1,N):
        Fact[i] = i * Fact[i - 1]
     
    # Stores the frequency of each digit
    freq= [0] * 10
    for i in range(len(freq)):
        freq[i] = 0;
    for i in range(N):
        freq[arr[i]] = freq[arr[i]] + 1;
     
 
    # Stores the final answer
    ans = 0
 
    # Loop to iterate over all values of
    # Nth digit i and 1st digit j
    for i in range(1, 10, 2) :
 
        # If digit i does not exist in
        # the given array move to next i
        if (freq[i] == 0):
            continue
 
        # Fixing i as Nth digit
        freq[i] = freq[i] - 1;
 
        for j in range(1, 10, 1) :
 
            # Stores the answer of a specific
            # value of i and j
            cur_ans = 0
 
            # If digit j does not exist
            # move to the next j
            if (freq[j] == 0) :
                continue
             
            # Fixing j as 1st digit
            freq[j]=freq[j]-1;
 
            # Calculate number of ways to
            # arrange remaining N-2 digits
            cur_ans = Fact[N - 2]
            for k in range(10):
                cur_ans = cur_ans / Fact[freq[k]]
             
            ans += cur_ans
 
            # Including j back into
            # the set of digits
            freq[j] = freq[j] + 1;
         
        # Including i back into the
        # set of the digits
        freq[i] = freq[i] + 1;
 
    # Return Answer
    return ceil(ans)
 
# Driver Code
if __name__ ==  "__main__":
    A = [ 2, 3, 4, 1, 2, 3 ]
    N = len(A)
 
    # Function Call
    print(countOddIntegers(A, N))
 
    # This code is contributed by anudeep23042002.

C#

// C# program of the above approach
using System;
 
class GFG{
 
// Function to find the count of distinct
// odd integers with N digits using the
// given digits in the array arr[]
static int countOddIntegers(int []arr, int N)
{
     
    // Stores the factorial of a number
    int []Fact = new int[N];
 
    // Calculate the factorial of all
    // numbers from 1 to N
    Fact[0] = 1;
    for(int i = 1; i < N; i++)
    {
        Fact[i] = i * Fact[i - 1];
    }
 
    // Stores the frequency of each digit
    int []freq = new int[10];
    for(int i = 0; i < N; i++)
    {
        freq[arr[i]]++;
    }
 
    // Stores the final answer
    int ans = 0;
 
    // Loop to iterate over all values of
    // Nth digit i and 1st digit j
    for(int i = 1; i <= 9; i += 2)
    {
         
        // If digit i does not exist in
        // the given array move to next i
        if (freq[i] == 0)
            continue;
 
        // Fixing i as Nth digit
        freq[i]--;
 
        for(int j = 1; j <= 9; j++)
        {
             
            // Stores the answer of a specific
            // value of i and j
            int cur_ans = 0;
 
            // If digit j does not exist
            // move to the next j
            if (freq[j] == 0)
            {
                continue;
            }
 
            // Fixing j as 1st digit
            freq[j]--;
 
            // Calculate number of ways to
            // arrange remaining N-2 digits
            cur_ans = Fact[N - 2];
            for(int k = 0; k <= 9; k++)
            {
                cur_ans = cur_ans / Fact[freq[k]];
            }
            ans += cur_ans;
 
            // Including j back into
            // the set of digits
            freq[j]++;
        }
         
        // Including i back into the
        // set of the digits
        freq[i]++;
    }
 
    // Return Answer
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []A = { 2, 3, 4, 1, 2, 3 };
    int N = A.Length;
 
    // Function Call
    Console.Write(countOddIntegers(A, N));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find the count of distinct
        // odd integers with N digits using the
        // given digits in the array arr[]
        function countOddIntegers(arr, N)
        {
            // Stores the factorial of a number
            let Fact = new Array(N);
 
            // Calculate the factorial of all
            // numbers from 1 to N
            Fact[0] = 1;
            for (let i = 1; i < N; i++) {
                Fact[i] = i * Fact[i - 1];
            }
 
            // Stores the frequency of each digit
            let freq = new Array(10).fill(0);
            for (let i = 0; i < N; i++) {
                freq[arr[i]]++;
            }
 
            // Stores the final answer
            let ans = 0;
 
            // Loop to iterate over all values of
            // Nth digit i and 1st digit j
            for (let i = 1; i <= 9; i += 2) {
 
                // If digit i does not exist in
                // the given array move to next i
                if (!freq[i]) {
                    continue;
                }
 
                // Fixing i as Nth digit
                freq[i]--;
 
                for (let j = 1; j <= 9; j++) {
 
                    // Stores the answer of a specific
                    // value of i and j
                    let cur_ans = 0;
 
                    // If digit j does not exist
                    // move to the next j
                    if (freq[j] == 0) {
                        continue;
                    }
 
                    // Fixing j as 1st digit
                    freq[j]--;
 
                    // Calculate number of ways to
                    // arrange remaining N-2 digits
                    cur_ans = Fact[N - 2];
                    for (let k = 0; k <= 9; k++) {
                        cur_ans = Math.floor(cur_ans / Fact[freq[k]]);
                    }
                    ans = ans + cur_ans;
 
                    // Including j back into
                    // the set of digits
                    freq[j]++;
                }
                // Including i back into the
                // set of the digits
                freq[i]++;
            }
 
            // Return Answer
            return ans;
        }
 
        // Driver Code
 
        let A = [2, 3, 4, 1, 2, 3];
        let N = A.length;
 
        // Function Call
        document.write(countOddIntegers(A, N));
 
// This code is contributed by Potta Lokesh
    

输出：

90

时间复杂度： O(N * 50)
辅助空间： O(1)