具有给定 AND 值的子集数

给定一个长度为N的数组arr和一个整数X ，任务是找到AND值为X的子集的数量。
例子：

Input: arr[] = {2, 3, 2} X = 2 
Output: 6 
All possible subsets and there AND values are: 
{2} = 2 
{3} = 3 
{2} = 2 
{2, 3} = 2 & 3 = 2 
{3, 2} = 3 & 2 = 2 
{2, 2} = 2 & 2 = 2 
{2, 3, 2} = 2 & 3 & 2 = 2
Input: arr[] = {0, 0, 0}, X = 0 
Output: 7 
 

方法：一个简单的方法是通过生成所有可能的子集，然后通过给定 AND 值计算子集的数量来解决问题。但是，对于较小的数组元素值，可以使用动态规划来解决。
我们先来看递归关系。

dp[i][curr_and] = dp[i + 1][curr_and] + dp[i + 1][curr_and & arr[i]] 
 

上面的递推关系可以定义为子数组arr[i…N-1]的子集数，这样它们与curr_and 的 AND将产生所需的 AND 值。
因为只有路径，所以递归关系是合理的。或者，获取当前元素并将其与curr_and或忽略它并继续前进。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define maxN 20
#define maxM 64
 
// To store states of DP
int dp1[maxN][maxM];
bool v1[maxN][maxM];
 
// Function to return the required count
int findCnt(int* arr, int i, int curr, int n, int m)
{
    // Base case
    if (i == n) {
        return (curr == m);
    }
 
    // If the state has been solved before
    // return the value of the state
    if (v1[i][curr])
        return dp1[i][curr];
 
    // Setting the state as solved
    v1[i][curr] = 1;
 
    // Recurrence relation
    dp1[i][curr]
        = findCnt(arr, i + 1, curr, n, m)
          + findCnt(arr, i + 1, (curr & arr[i]), n, m);
 
    return dp1[i][curr];
}
 
// Driver code
int main()
{
    int arr[] = { 0, 0, 0 };
    int n = sizeof(arr) / sizeof(int);
    int m = 0;
 
    cout << findCnt(arr, 0, ((1 << 6) - 1), n, m);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
static int maxN = 20;
static int maxM = 64;
 
// To store states of DP
static int [][]dp1 = new int[maxN][maxM];
static boolean [][]v1 = new boolean[maxN][maxM];
 
// Function to return the required count
static int findCnt(int []arr, int i,
                   int curr, int n, int m)
{
    // Base case
    if (i == n)
    {
        return (curr == m ? 1 : 0);
    }
 
    // If the state has been solved before
    // return the value of the state
    if (v1[i][curr])
        return dp1[i][curr];
 
    // Setting the state as solved
    v1[i][curr] = true;
 
    // Recurrence relation
    dp1[i][curr] = findCnt(arr, i + 1, curr, n, m) +
                   findCnt(arr, i + 1, (curr & arr[i]), n, m);
 
    return dp1[i][curr];
}
 
// Driver code
public static void main(String []args)
{
    int arr[] = { 0, 0, 0 };
    int n = arr.length;
    int m = 0;
 
    System.out.println(findCnt(arr, 0, ((1 << 6) - 1), n, m));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
import numpy as np
 
maxN = 20
maxM = 64
 
# To store states of DP
dp1 = np.zeros((maxN, maxM));
v1 = np.zeros((maxN, maxM));
 
# Function to return the required count
def findCnt(arr, i, curr, n, m) :
 
    # Base case
    if (i == n) :
        return (curr == m);
 
    # If the state has been solved before
    # return the value of the state
    if (v1[i][curr]) :
        return dp1[i][curr];
 
    # Setting the state as solved
    v1[i][curr] = 1;
 
    # Recurrence relation
    dp1[i][curr] = findCnt(arr, i + 1, curr, n, m) + \
                   findCnt(arr, i + 1, (curr & arr[i]), n, m);
 
    return dp1[i][curr];
 
# Driver code
if __name__ == "__main__" :
     
    arr = [ 0, 0, 0 ];
    n = len(arr);
    m = 0;
 
    print(findCnt(arr, 0, ((1 << 6) - 1), n, m));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
static int maxN = 20;
static int maxM = 64;
 
// To store states of DP
static int [,]dp1 = new int[maxN, maxM];
static bool [,]v1 = new bool[maxN, maxM];
 
// Function to return the required count
static int findCnt(int []arr, int i,
                   int curr, int n, int m)
{
    // Base case
    if (i == n)
    {
        return (curr == m ? 1 : 0);
    }
 
    // If the state has been solved before
    // return the value of the state
    if (v1[i, curr])
        return dp1[i, curr];
 
    // Setting the state as solved
    v1[i, curr] = true;
 
    // Recurrence relation
    dp1[i, curr] = findCnt(arr, i + 1, curr, n, m) +
                   findCnt(arr, i + 1, (curr & arr[i]), n, m);
 
    return dp1[i, curr];
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 0, 0, 0 };
    int n = arr.Length;
    int m = 0;
 
    Console.WriteLine(findCnt(arr, 0, ((1 << 6) - 1), n, m));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the approach
 
var maxN = 20;
var maxM =  64;
 
// To store states of DP
var dp1 = Array.from(Array(maxN), ()=> Array(maxM));
var v1 = Array.from(Array(maxN), ()=> Array(maxM));
 
// Function to return the required count
function findCnt(arr, i, curr, n, m)
{
    // Base case
    if (i == n) {
        return (curr == m);
    }
 
    // If the state has been solved before
    // return the value of the state
    if (v1[i][curr])
        return dp1[i][curr];
 
    // Setting the state as solved
    v1[i][curr] = 1;
 
    // Recurrence relation
    dp1[i][curr]
        = findCnt(arr, i + 1, curr, n, m)
          + findCnt(arr, i + 1, (curr & arr[i]), n, m);
 
    return dp1[i][curr];
}
 
// Driver code
var arr = [0, 0, 0];
var n = arr.length;
var m = 0;
document.write( findCnt(arr, 0, ((1 << 6) - 1), n, m));
 

输出：

7

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。