给定或值的子集数量
给定一个长度为 N 的数组 arr[] ，任务是找到给定 OR 值 M 的子集数量。 示例:
输入: arr[] = {2，3，2} M = 3 输出: 4 所有可能的子集和 OR 值为: { 2 } = 2 { 3 } =3 { 2 } = 2 { 2，3} = 2 | 3 = 3 {3，2} = 3 | 2 = 3 {2
输入: arr[] = {1，3，2，2}，M = 5 T3】输出: 0
方法:一个简单的方法是通过生成所有可能的子集，然后计算给定 OR 值的子集数量来解决问题。但是，对于较小的数组元素值，可以使用动态编程解决。 我们先看递归关系。
DP[I][curr _ or]= DP[I+1][curr _ or]+DP[I+1][curr _ or | arr[I]]
上述递归关系可以定义为子阵列arr【I…N-1】的子集的数量，使得用 curr_or 对它们进行 or 运算将产生所需的 OR 值。 递归关系是合理的，因为只有路径。要么取当前元素，用 curr_or 进行 OR 运算，要么忽略，继续前进。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define maxN 20
#define maxM 64
// To store states of DP
int dp[maxN][maxM];
bool v[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 (v[i][curr])
return dp[i][curr];
// Setting the state as solved
v[i][curr] = 1;
// Recurrence relation
dp[i][curr]
= findCnt(arr, i + 1, curr, n, m)
+ findCnt(arr, i + 1, (curr | arr[i]), n, m);
return dp[i][curr];
}
// Driver code
int main()
{
int arr[] = { 2, 3, 2 };
int n = sizeof(arr) / sizeof(int);
int m = 3;
cout << findCnt(arr, 0, 0, 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 maxN = 20;
static int maxM = 64;
// To store states of DP
static int [][]dp = new int[maxN][maxM];
static boolean [][]v = 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 (v[i][curr])
return dp[i][curr];
// Setting the state as solved
v[i][curr] = true;
// Recurrence relation
dp[i][curr] = findCnt(arr, i + 1, curr, n, m) +
findCnt(arr, i + 1, (curr | arr[i]), n, m);
return dp[i][curr];
}
// Driver code
public static void main(String []args)
{
int arr[] = { 2, 3, 2 };
int n = arr.length;
int m = 3;
System.out.println(findCnt(arr, 0, 0, n, m));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
import numpy as np
maxN = 20
maxM = 64
# To store states of DP
dp = np.zeros((maxN, maxM));
v = 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 (v[i][curr]) :
return dp[i][curr];
# Setting the state as solved
v[i][curr] = 1;
# Recurrence relation
dp[i][curr] = findCnt(arr, i + 1, curr, n, m) + \
findCnt(arr, i + 1, (curr | arr[i]), n, m);
return dp[i][curr];
# Driver code
if __name__ == "__main__" :
arr = [ 2, 3, 2 ];
n = len(arr);
m = 3;
print(findCnt(arr, 0, 0, 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 [,]dp = new int[maxN, maxM];
static Boolean [,]v = 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 (v[i, curr])
return dp[i, curr];
// Setting the state as solved
v[i, curr] = true;
// Recurrence relation
dp[i, curr] = findCnt(arr, i + 1, curr, n, m) +
findCnt(arr, i + 1, (curr | arr[i]), n, m);
return dp[i, curr];
}
// Driver code
public static void Main(String []args)
{
int []arr = { 2, 3, 2 };
int n = arr.Length;
int m = 3;
Console.WriteLine(findCnt(arr, 0, 0, n, m));
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript implementation of the approach
var maxN = 20
var maxM = 64
// To store states of DP
var dp = Array.from(Array(maxN), ()=> Array(maxM));
var v = 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 (v[i][curr])
return dp[i][curr];
// Setting the state as solved
v[i][curr] = 1;
// Recurrence relation
dp[i][curr]
= findCnt(arr, i + 1, curr, n, m)
+ findCnt(arr, i + 1, (curr | arr[i]), n, m);
return dp[i][curr];
}
// Driver code
var arr = [2, 3, 2 ];
var n = arr.length;
var m = 3;
document.write( findCnt(arr, 0, 0, n, m));
</script>
Output
4
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