将数组划分为 K 个段，使得单个段总和的按位 AND 最大化

给定一个大小为N的数组和一个整数K 。任务是将数组划分为K 个段，使得单个段总和的按位AND最大化。找出可以得到的按位与的最大值。

例子：

Input : a[] = { 2, 5, 6, 9, 1, 3, 10, 12 }, K = 3 
Output : 8 
Explanation : 
Maximum bitwise AND can be obtained by making cut at 3rd element and 5th element(1 based indexing) 
(2+5+6)&(9+1)&(3+10+12) = 8
Input : a[] = { 1, 2, 7, 10, 23, 21, 6, 8, 7, 3 } , K = 2 
Output : 41 
Explanation: 
Maximum bitwise AND can be obtained by making cut at 5th element(1 based indexing) 
(1+2+7+10+23)&(21+6+8+7+3) = 41

方法：
首先，尝试回答一个更简单的问题：给定一个整数x并确定是否可以将给定数组划分为K 个段，使得段总和的按位AND具有x 的所有设置位?
让我们将第 i段中元素的总和表示为$S\textsubscript{i}$ .另外，让我们称一个段为好，如果$S\textsubscript{i}$具有x 的所有设置位。现在很清楚，对于一个好的段i ， AND ( $S\textsubscript{i}$ , x )= x 。
此外，所有K段都应该有利于获得x 的按位与。现在检查我们是否可以将数组划分为k 个好的段。我们可以在这里使用动态规划。
让dp[i][j]= true ，表示可以将前i 个元素划分为j 个段，使得所有j都是好的段，否则为false 。
以上 dp 的循环是：

dp[i][j] is 1, if for some index k

贪婪地从可能答案的最高位开始构建dp表。
下面是上述方法的实现：

C++

// CPP program to find maximum possible AND
#include 
using namespace std;
 
// Function to check whether a k segment partition
// is possible such that bitwise AND is 'mask'
bool checkpossible(int mask, int* arr, int* prefix, int n,
                                                  int k)
{
    // dp[i][j] stores whether it is possible to partition
    // first i elements into j segments such that all j
    // segments are 'good'
    bool dp[n + 1][k + 1];
 
    // Initialising dp
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
 
    // Filling dp in bottom-up manner
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= k; j++) {
            // Finding a cut such that first l elements
            // can be partitioned into j-1 'good' segments
            // and arr[l+1]+...+arr[i] is a 'good' segment
            for (int l = i - 1; l >= 0; --l) {
                if (dp[l][j - 1] && (((prefix[i] - prefix[l])
                           & mask) == mask)) {
                    dp[i][j] = 1;
                    break;
                }
            }
        }
    }
 
    return dp[n][k];
}
 
// Function to find maximum possible AND
int Partition(int arr[], int n, int k)
{
    // Array to store prfix sums
    int prefix[n+1];
 
    for (int i = 1; i <= n; i++) {
        prefix[i] = prefix[i - 1] + arr[i];
    }
 
    // Maximum no of bits in the possible answer
    int LOGS = 20;
 
    // This will store the final answer
    int ans = 0;
 
    // Constructing answer greedily selecting
    // from the higher most bit
    for (int i = LOGS; i >= 0; --i) {
        // Checking if array can be partitioned
        // such that the bitwise AND is ans|(1<

---

Java

// Java program to find maximum possible AND
 
class GFG
{
     
    // Function to check whether a k segment partition
    // is possible such that bitwise AND is 'mask'
    static boolean checkpossible(int mask, int arr[],
                                int prefix[], int n, int k)
    {
        int i,j;
         
        // dp[i][j] stores whether it is possible to partition
        // first i elements into j segments such that all j
        // segments are 'good'
        boolean dp[][] = new boolean[n + 1][k + 1];
     
        // Initialising dp
        for(i = 0; i < n + 1; i++)
        {
            for(j = 0; j < k + 1; j++)
            {
                dp[i][j] = false ;
            }
        }
         
        dp[0][0] = true;
     
        // Filling dp in bottom-up manner
        for ( i = 1; i <= n; i++)
        {
            for (j = 1; j <= k; j++)
            {
                // Finding a cut such that first l elements
                // can be partitioned into j-1 'good' segments
                // and arr[l+1]+...+arr[i] is a 'good' segment
                for (int l = i - 1; l >= 0; --l)
                {
                    if (dp[l][j - 1] && (((prefix[i] - prefix[l])
                            & mask) == mask))
                    {
                        dp[i][j] = true;
                        break;
                    }
                }
            }
        }
     
        return dp[n][k];
    }
     
    // Function to find maximum possible AND
    static int Partition(int arr[], int n, int k)
    {
        // Array to store prfix sums
        int prefix[] = new int[n+1];
     
        for (int i = 1; i <= n; i++)
        {
            prefix[i] = prefix[i - 1] + arr[i];
        }
     
        // Maximum no of bits in the possible answer
        int LOGS = 20;
     
        // This will store the final answer
        int ans = 0;
     
        // Constructing answer greedily selecting
        // from the higher most bit
        for (int i = LOGS; i >= 0; --i)
        {
            // Checking if array can be partitioned
            // such that the bitwise AND is ans|(1<

---

Python3

# Python3 program to find maximum possible AND
 
# Function to check whether a k segment partition
# is possible such that bitwise AND is 'mask'
def checkpossible(mask,arr,prefix, n,k):
     
    # dp[i][j] stores whether it is possible to partition
    # first i elements into j segments such that all j
    # segments are 'good'
    dp=[[0 for i in range(k+1)] for i in range(n + 1)]
 
    # Initialising dp
    dp[0][0] = 1
 
    # Filling dp in bottom-up manner
    for i in range(1, n+1):
        for j in range(1, k+1):
             
            # Finding a cut such that first l elements
            # can be partitioned into j-1 'good' segments
            # and arr[l+1]+...+arr[i] is a 'good' segment
            for l in range(i-1,-1,-1):
                if (dp[l][j - 1] and (((prefix[i] - prefix[l]) & mask) == mask)):
                    dp[i][j] = 1
                    break
 
    return dp[n][k]
 
 
# Function to find maximum possible AND
def Partition(arr, n, k):
    # Array to store prfix sums
    prefix=[0 for i in range(n+1)]
 
    for i in range(1,n+1):
        prefix[i] = prefix[i - 1] + arr[i]
 
    # Maximum no of bits in the possible answer
    LOGS = 20
 
    # This will store the final answer
    ans = 0
 
    # Constructing answer greedily selecting
    # from the higher most bit
    for i in range(LOGS,-1,-1):
        # Checking if array can be partitioned
        # such that the bitwise AND is ans|(1<

---

C#

// C# program to find maximum possible AND
using System;
     
class GFG
{
     
    // Function to check whether a
    // k-segment partition is possible
    // such that bitwise AND is 'mask'
    static Boolean checkpossible(int mask, int []arr,
                                 int []prefix,
                                 int n, int k)
    {
        int i, j;
         
        // dp[i,j] stores whether it is possible
        // to partition first i elements into
        // j-segments such that all j-segments are 'good'
        Boolean[,] dp = new Boolean[n + 1, k + 1];
     
        // Initialising dp
        for(i = 0; i < n + 1; i++)
        {
            for(j = 0; j < k + 1; j++)
            {
                dp[i, j] = false;
            }
        }
         
        dp[0, 0] = true;
     
        // Filling dp in bottom-up manner
        for ( i = 1; i <= n; i++)
        {
            for (j = 1; j <= k; j++)
            {
                // Finding a cut such that first l elements
                // can be partitioned into j-1 'good' segments
                // and arr[l+1]+...+arr[i] is a 'good' segment
                for (int l = i - 1; l >= 0; --l)
                {
                    if (dp[l, j - 1] &&
                     (((prefix[i] - prefix[l]) &
                        mask) == mask))
                    {
                        dp[i, j] = true;
                        break;
                    }
                }
            }
        }
        return dp[n, k];
    }
     
    // Function to find maximum possible AND
    static int Partition(int []arr, int n, int k)
    {
        // Array to store prfix sums
        int []prefix = new int[n + 1];
     
        for (int i = 1; i <= n; i++)
        {
            prefix[i] = prefix[i - 1] + arr[i];
        }
     
        // Maximum no of bits in the possible answer
        int LOGS = 20;
     
        // This will store the final answer
        int ans = 0;
     
        // Constructing answer greedily selecting
        // from the higher most bit
        for (int i = LOGS; i >= 0; --i)
        {
            // Checking if array can be partitioned
            // such that the bitwise AND is ans|(1<

---

Javascript

    // Javascript program to find maximum possible AND
     
    // Function to check whether a k segment partition
    // is possible such that bitwise AND is 'mask'
    function checkpossible(mask, arr, prefix, n, k)
    {
        let i,j;
           
        // dp[i][j] stores whether it is possible to partition
        // first i elements into j segments such that all j
        // segments are 'good'
        let dp = new Array(n + 1);
       
        // Initialising dp
        for(i = 0; i < n + 1; i++)
        {
            dp[i] = new Array(k + 1);
            for(j = 0; j < k + 1; j++)
            {
                dp[i][j] = false ;
            }
        }
           
        dp[0][0] = true;
       
        // Filling dp in bottom-up manner
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= k; j++)
            {
                // Finding a cut such that first l elements
                // can be partitioned into j-1 'good' segments
                // and arr[l+1]+...+arr[i] is a 'good' segment
                for (let l = i - 1; l >= 0; --l)
                {
                    if (dp[l][j - 1] && (((prefix[i] - prefix[l])
                            & mask) == mask))
                    {
                        dp[i][j] = true;
                        break;
                    }
                }
            }
        }
       
        return dp[n][k];
    }
       
    // Function to find maximum possible AND
    function Partition(arr, n, k)
    {
        // Array to store prfix sums
        let prefix = new Array(n+1);
        prefix.fill(0);
       
        for (let i = 1; i <= n; i++)
        {
            prefix[i] = prefix[i - 1] + arr[i];
        }
       
        // Maximum no of bits in the possible answer
        let LOGS = 20;
       
        // This will store the final answer
        let ans = 0;
       
        // Constructing answer greedily selecting
        // from the higher most bit
        for (let i = LOGS; i >= 0; --i)
        {
            // Checking if array can be partitioned
            // such that the bitwise AND is ans|(1<<i)
            if (checkpossible(ans | (1 << i), arr, prefix, n, k))
            {
                // if possible, update the answer
                ans = ans | (1 << i);
            }
        }
       
        // Return the final answer
        return ans;
    }
     
    let arr = [0, 1, 2, 7, 10, 23, 21, 6, 8, 7, 3], k = 2;
       
    // n = 11 , first element is zero
    // to make array 1 based indexing. So, number of
    // elements are 10
    let n = arr.length - 1 ;
 
    // Function call
    document.write(Partition(arr, n, k));
 
// This code is contributed by divyeshrabadiya07.

输出：

41

时间复杂度： O( $n^2$ klog(S))，其中 S = $S\textsubscript{1}$ + $S\textsubscript{2}$ +….+ $S\textsubscript{k}$ .

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