给定数组arr [] ，任务是找到具有给定AND值M的最长子序列。如果没有这样的子序列，则打印0 。
例子：
Input: arr[] = {3, 7, 2, 3}, M = 3
Output: 3
{3, 7, 3} is the required subsequence.
3 & 7 & 3 = 3
Input: arr[] = {2, 2}, M = 3
Output: 0
天真的方法：解决此问题的一种简单方法是生成所有可能的子序列，然后在其中找到具有所需AND值的最大子序列。
高效方法：一项主要观察结果是，所需子序列中的所有数字与M进行“与”运算时，都应产生值M。所以，过滤掉所有那些并与M等于M使得元素。
现在，任务是在此过滤后的子集中找到最长的子序列。很明显，所有这些数字都将与在一起。如果此AND的结果为M，则答案将等于此过滤后的集合的大小。否则答案将为0 。这是因为“与”操作仅会取消设置位。因此，集合中的数字越大，它越优化。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the required length
int findLen(int* arr, int n, int m)
{
// To store the filtered numbers
vector filter;
// Filtering the numbers
for (int i = 0; i < n; i++)
if ((arr[i] & m) == m)
filter.push_back(arr[i]);
// If there are no elements to check
if (filter.size() == 0)
return 0;
// Find the AND of all the
// filtered elements
int c_and = filter[0];
for (int i = 1; i < filter.size(); i++)
c_and &= filter[i];
// Check if the AND is equal to m
if (c_and == m)
return filter.size();
return 0;
}
// Driver code
int main()
{
int arr[] = { 7, 3, 3, 1, 3 };
int n = sizeof(arr) / sizeof(int);
int m = 3;
cout << findLen(arr, n, m);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the required length
static int findLen(int []arr, int n, int m)
{
// To store the filtered numbers
Vector filter = new Vector<>();
// Filtering the numbers
for (int i = 0; i < n; i++)
if ((arr[i] & m) == m)
filter.add(arr[i]);
// If there are no elements to check
if (filter.size() == 0)
return 0;
// Find the AND of all the
// filtered elements
int c_and = filter.get(0);
for (int i = 1; i < filter.size(); i++)
c_and &= filter.get(i);
// Check if the AND is equal to m
if (c_and == m)
return filter.size();
return 0;
}
// Driver code
public static void main(String []args)
{
int arr[] = { 7, 3, 3, 1, 3 };
int n = arr.length;
int m = 3;
System.out.println(findLen(arr, n, m));
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 implementation of the approach
# Function to return the required length
def findLen(arr, n, m) :
# To store the filtered numbers
filter = [];
# Filtering the numbers
for i in range(n) :
if ((arr[i] & m) == m) :
filter.append(arr[i]);
# If there are no elements to check
if (len(filter) == 0) :
return 0;
# Find the OR of all the
# filtered elements
c_and = filter[0];
for i in range(1, len(filter)) :
c_and &= filter[i];
# Check if the OR is equal to m
if (c_and == m) :
return len(filter);
# Driver code
if __name__ == "__main__" :
arr = [ 7, 3, 3, 1, 3 ];
n = len(arr);
m = 3;
print(findLen(arr, n, m));
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the required length
static int findLen(int []arr, int n, int m)
{
// To store the filtered numbers
List filter = new List();
// Filtering the numbers
for (int i = 0; i < n; i++)
if ((arr[i] & m) == m)
filter.Add(arr[i]);
// If there are no elements to check
if (filter.Count == 0)
return 0;
// Find the AND of all the
// filtered elements
int c_and = filter[0];
for (int i = 1; i < filter.Count; i++)
c_and &= filter[i];
// Check if the AND is equal to m
if (c_and == m)
return filter.Count;
return 0;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 7, 3, 3, 1, 3 };
int n = arr.Length;
int m = 3;
Console.WriteLine(findLen(arr, n, m));
}
}
// This code is contributed by 29AjayKumar 输出：
4
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