通过将元素最多增加 K 来最大化数组的按位或
给定一个数组arr[]和一个整数K ，任务是最大化数组arr[]的按位或，其中arr [] 的每个元素几乎可以递增K 。
例子：
Input: arr[]= {1, 3, 7, 0, 6, 1}, K = 2
Output: [1 3 8 0 6 1]
Explanation: Increase the number 7 by 1 ie 8 after that or of the array is maximum that is 15
Input: arr[] ={2, 4, 7, 9}, K = 2
Output: [2, 4, 7, 9]
Explanation: The bitwise OR of arr[] is 15 already so no change is required. 15 is the maximum possible OR.
方法：这个问题可以通过 使用按位运算。请按照以下步骤解决给定的问题。
- 找到所有元素的按位或，这将告诉每个位的状态。
- 从MSB (Most Significant bit)开始，因为 MSB 在最终的按位 OR 中做了很多改变。
- 如果按位或未设置特定位。找到设置该位的最小步骤。
- 设置第 i 位的最小步骤是(1<
- 如果最小步数小于等于K ，则计算最小步数并更新数组。
下面是上述方法的实现。
C++
// C++ code for the above approach
#include
using namespace std;
void MaximizeBitwiseOR(long a[], int k, int n)
{
// For storing initial
// bitwise OR of array arr[]
long oris = 0;
long bitwiseOr = 0;
for (int i = 0; i < n; ++i)
{
bitwiseOr |= a[i];
}
for (int i = 60; i >= 0; i--)
{
if (((1L << i) & bitwiseOr) == 0)
{
long minSteps = LONG_MAX;
int mini = -1;
for (int j = 0; j < n; j++)
{
long y = ((1L << i) - 1) & a[j];
long steps = (1L << i) - y;
if (steps <= k && steps < minSteps)
{
minSteps = steps;
mini = j;
}
}
if (mini != -1)
{
k -= minSteps;
a[mini] += minSteps;
long orr = 0;
for (int j = 0; j < n; j++)
{
orr |= a[j];
}
bitwiseOr = orr;
}
}
}
// Print the required result
for (long elements = 0; elements < n; elements++)
{
if(a[elements]>0)
cout << a[elements] << " ";
else
cout<<0<<" ";
}
}
// Driver code
int main()
{
int N = 6;
int K = 2;
long arr[] = {1, 3, 7, 0, 6, 1};
MaximizeBitwiseOR(arr, K, N);
}
// This code is contributed by Potta Lokesh Java
import java.io.*;
class Main {
static void MaximizeBitwiseOR(long[] a, int k, int n)
{
// For storing initial
// bitwise OR of array arr[]
long or = 0;
long bitwiseOr = 0;
for (int i = 0; i < a.length; ++i) {
bitwiseOr |= a[i];
}
for (int i = 60; i >= 0; i--) {
if (((1L << i) & bitwiseOr) == 0) {
long minSteps = Long.MAX_VALUE;
int mini = -1;
for (int j = 0; j < n; j++) {
long y = ((1L << i) - 1) & a[j];
long steps = (1L << i) - y;
if (steps <= k
&& steps < minSteps) {
minSteps = steps;
mini = j;
}
}
if (mini != -1) {
k -= minSteps;
a[mini] += minSteps;
long orr = 0;
for (int j = 0; j < n; j++) {
orr |= a[j];
}
bitwiseOr = orr;
}
}
}
// Print the required result
for (long elements : a) {
System.out.print(elements + " ");
}
}
// Driver code
public static void main(String[] args)
{
int N = 6;
int K = 2;
long arr[] = { 1, 3, 7, 0, 6, 1 };
MaximizeBitwiseOR(arr, K, N);
}
}Python3
# Python3 code for the above approach
import sys
def MaximizeBitwiseOR(a, k, n):
# For storing initial
# bitwise OR of array arr[]
oris = 0
bitwiseOr = 0
for i in range(n):
bitwiseOr |= a[i]
for i in range(60, -1, -1):
if (((1 << i) & bitwiseOr) == 0):
minSteps = sys.maxsize
mini = -1
for j in range(n):
y = ((1 << i) - 1) & a[j]
steps = (1 << i) - y
if (steps <= k and steps < minSteps):
minSteps = steps
mini = j
if (mini != -1):
k -= minSteps
a[mini] += minSteps
orr = 0
for j in range(n):
orr |= a[j]
bitwiseOr = orr
# Print the required result
for elements in range(n):
if(a[elements] > 0):
print(a[elements], end=" ")
else:
print(0, end=" ")
# Driver code
if __name__ == "__main__":
N = 6
K = 2
arr = [1, 3, 7, 0, 6, 1]
MaximizeBitwiseOR(arr, K, N)
# This code is contributed by ukasp.C#
using System;
class GFG {
static void MaximizeBitwiseOR(long[] a, int k, int n)
{
// For storing initial
// bitwise OR of array arr[]
long or = 0;
long bitwiseOr = 0;
for (int i = 0; i < a.Length; ++i) {
bitwiseOr |= a[i];
}
for (int i = 60; i >= 0; i--) {
if (((1L << i) & bitwiseOr) == 0) {
long minSteps = Int32.MaxValue;
int mini = -1;
for (int j = 0; j < n; j++) {
long y = ((1L << i) - 1) & a[j];
long steps = (1L << i) - y;
if (steps <= k
&& steps < minSteps) {
minSteps = steps;
mini = j;
}
}
if (mini != -1) {
k -= (int)minSteps;
a[mini] += minSteps;
long orr = 0;
for (int j = 0; j < n; j++) {
orr |= a[j];
}
bitwiseOr = orr;
}
}
}
// Print the required result
foreach (long elements in a) {
Console.Write(elements + " ");
}
}
// Driver code
public static void Main()
{
int N = 6;
int K = 2;
long []arr = { 1, 3, 7, 0, 6, 1 };
MaximizeBitwiseOR(arr, K, N);
}
}
// This code is contributed by Samim Hossain Mondal.输出
1 3 8 0 6 1 时间复杂度： O(N*60)
辅助空间： O(1)