问题给定N边的方阵，我们必须找到一个子矩阵，使其元素的按位XOR最大，并且必须打印最大的按位XOR。
例子：
Input :
matrix is
{ {1, 2, 3, 4}
{5, 6, 7, 8}
{9, 10, 11, 12}
{13, 14, 15, 16} }
Output : 31
We get the value 31 by doing XOR of submatrix [15, 16}
朴素方法蛮力方法是通过运行四个循环(两个用于开始行和列，两个用于结束行和列)来找到数组的所有子矩阵，然后找到子矩阵所有元素的异或，并找出异或。检查此子矩阵的xor是否最大。
时间复杂度：O(n ^ 6)
高效的方法在这种方法中，我们将计算从1，1到i，j的每个子矩阵的异或，这可以通过使用以下公式在O(N * N)中完成
(子矩阵从1，1到i，j的异或)=(子矩阵从1，1到i-1，j的异或)^(子矩阵从1，1到i，j-1的异或)^(子矩阵的异或从1，1到i-1，j-1)^ arr [i] [j]。
现在，为了计算从i，j到i1，j1的子矩阵的xor，我们可以使用以下公式
(子矩阵从i，j到i1，j1的异或)=(子矩阵从1，1到i1，j1的异或)^(子矩阵从1，1到i-1，j-1的异或)^(子矩阵的异或从1，1到i1，j-1)^(子矩阵从1，1到i-1，j1的异或)。
我们必须运行四个循环以找出数组的所有子矩阵，并且该子矩阵的xor可以在O(1)时间内计算出来。
因此时间复杂度降低为O(N ^ 4)
C++
// C++ program to implement the above approach
#include
using namespace std;
#define N 101
// Compute the xor of elements from (1, 1) to
// (i, j) and store it in prefix_xor[i][j]
void prefix(int arr[N][N], int prefix_xor[N][N], int n)
{
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
// xor of submatrix from 1, 1 to i, j is
// (xor of submatrix from 1, 1 to i-1,
// j )^(xor of submatrix from 1, 1 to i, j-1)
// ^(xor of submatrix from 1, 1 to i-1, j-1) ^
// arr[i][j]
prefix_xor[i][j] = arr[i][j] ^
prefix_xor[i - 1][j] ^
prefix_xor[i][j - 1] ^
prefix_xor[i - 1][j - 1];
}
}
}
// find the submatrix with maximum xor value
void Max_xor(int prefix_xor[N][N], int n)
{
int max_value = 0;
// we need four loops to find all the submatrix
// of a matrix
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
for (int i1 = i; i1 <= n; i1++) {
for (int j1 = j; j1 <= n; j1++) {
// xor of submatrix from i, j to i1, j1 is
// (xor of submatrix from 1, 1 to i1, j1 )
// ^(xor of submatrix from 1, 1 to i-1, j-1)
// ^(xor of submatrix from 1, 1 to i1, j-1)
// ^(xor of submatrix from 1, 1 to i-1, j1)
int x = 0;
x ^= prefix_xor[i1][j1];
x ^= prefix_xor[i - 1][j - 1];
x ^= prefix_xor[i1][j - 1];
x ^= prefix_xor[i - 1][j1];
// if the xor is greater than maximum value
// substitute it
max_value = max(max_value, x);
}
}
}
}
cout << max_value << endl;
}
// Driver code
int main()
{
int arr[N][N] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
int n = 4;
int prefix_xor[N][N] = { 0 };
// Find the prefix_xor
prefix(arr, prefix_xor, n);
// Find submatrix with maximum bitwise xor
Max_xor(prefix_xor, n);
return 0;
} Java
// Java program to implement the above approach
public class GFG
{
static int N =101;
// Compute the xor of elements from (1, 1) to
// (i, j) and store it in prefix_xor[i][j]
static void prefix(int arr[][], int prefix_xor[][], int n)
{
for (int i = 1; i < n; i++)
{
for (int j = 1; j < n; j++)
{
// xor of submatrix from 1, 1 to i, j is
// (xor of submatrix from 1, 1 to i-1,
// j )^(xor of submatrix from 1, 1 to i, j-1)
// ^(xor of submatrix from 1, 1 to i-1, j-1) ^
// arr[i][j]
prefix_xor[i][j] = arr[i][j] ^
prefix_xor[i - 1][j] ^
prefix_xor[i][j - 1] ^
prefix_xor[i - 1][j - 1];
}
}
}
// find the submatrix with maximum xor value
static void Max_xor(int prefix_xor[][], int n)
{
int max_value = 0;
// we need four loops to find all the submatrix
// of a matrix
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
for (int i1 = i; i1 <= n; i1++)
{
for (int j1 = j; j1 <= n; j1++)
{
// xor of submatrix from i, j to i1, j1 is
// (xor of submatrix from 1, 1 to i1, j1 )
// ^(xor of submatrix from 1, 1 to i-1, j-1)
// ^(xor of submatrix from 1, 1 to i1, j-1)
// ^(xor of submatrix from 1, 1 to i-1, j1)
int x = 0;
x ^= prefix_xor[i1][j1];
x ^= prefix_xor[i - 1][j - 1];
x ^= prefix_xor[i1][j - 1];
x ^= prefix_xor[i - 1][j1];
// if the xor is greater than maximum value
// substitute it
max_value = Math.max(max_value, x);
}
}
}
}
System.out.println(max_value);
}
// Driver code
public static void main(String[] args)
{
int arr[][] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
int n = 4;
int prefix_xor[][] = new int[N][N];
// Find the prefix_xor
prefix(arr, prefix_xor, n);
// Find submatrix with maximum bitwise xor
Max_xor(prefix_xor, n);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to implement the above approach
N = 101
# Compute the xor of elements from (1, 1) to
# (i, j) and store it in prefix_xor[i][j]
def prefix(arr, prefix_xor, n):
for i in range(1, n):
for j in range(1, n):
# xor of submatrix from 1, 1 to i, j is
# (xor of submatrix from 1, 1 to i-1,
# j )^(xor of submatrix from 1, 1 to i, j-1)
# ^(xor of submatrix from 1, 1 to i-1, j-1) ^
# arr[i][j]
prefix_xor[i][j] = (arr[i][j] ^
prefix_xor[i - 1][j] ^
prefix_xor[i][j - 1] ^
prefix_xor[i - 1][j - 1])
# find the submatrix with maximum xor value
def Max_xor(prefix_xor, n):
max_value = 0
# we need four loops to find
# all the submatrix of a matrix
for i in range(1, n + 1):
for j in range(1, n + 1):
for i1 in range(i, n + 1):
for j1 in range(j, n + 1):
# xor of submatrix from i, j to i1, j1 is
# (xor of submatrix from 1, 1 to i1, j1 )
# ^(xor of submatrix from 1, 1 to i-1, j-1)
# ^(xor of submatrix from 1, 1 to i1, j-1)
# ^(xor of submatrix from 1, 1 to i-1, j1)
x = 0
x ^= prefix_xor[i1][j1]
x ^= prefix_xor[i - 1][j - 1]
x ^= prefix_xor[i1][j - 1]
x ^= prefix_xor[i - 1][j1]
# if the xor is greater than maximum value
# substitute it
max_value = max(max_value, x)
print(max_value)
# Driver code
arr = [[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9, 10, 11, 12 ],
[ 13, 14, 15, 16 ]]
n = 4
prefix_xor = [[ 0 for i in range(N)]
for i in range(N)]
# Find the prefix_xor
prefix(arr, prefix_xor, n)
# Find submatrix with maximum bitwise xor
Max_xor(prefix_xor, n)
# This code is contributed by Mohit KumarC#
// C# program to implement the above approach
using System;
class GFG
{
static int N =101;
// Compute the xor of elements from (1, 1) to
// (i, j) and store it in prefix_xor[i][j]
static void prefix(int [ , ] arr, int [ , ] prefix_xor, int n)
{
for (int i = 1; i < n; i++)
{
for (int j = 1; j < n; j++)
{
// xor of submatrix from 1, 1 to i, j is
// (xor of submatrix from 1, 1 to i-1,
// j )^(xor of submatrix from 1, 1 to i, j-1)
// ^(xor of submatrix from 1, 1 to i-1, j-1) ^
// arr[i][j]
prefix_xor[i, j] = arr[i, j] ^
prefix_xor[i - 1, j] ^
prefix_xor[i, j - 1] ^
prefix_xor[i - 1, j - 1];
}
}
}
// find the submatrix with maximum xor value
static void Max_xor(int[ , ] prefix_xor, int n)
{
int max_value = 0;
// we need four loops to find all the submatrix
// of a matrix
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
for (int i1 = i; i1 <= n; i1++)
{
for (int j1 = j; j1 <= n; j1++)
{
// xor of submatrix from i, j to i1, j1 is
// (xor of submatrix from 1, 1 to i1, j1 )
// ^(xor of submatrix from 1, 1 to i-1, j-1)
// ^(xor of submatrix from 1, 1 to i1, j-1)
// ^(xor of submatrix from 1, 1 to i-1, j1)
int x = 0;
x ^= prefix_xor[i1, j1];
x ^= prefix_xor[i - 1, j - 1];
x ^= prefix_xor[i1, j - 1];
x ^= prefix_xor[i - 1, j1];
// if the xor is greater than maximum value
// substitute it
max_value = Math.Max(max_value, x);
}
}
}
}
Console.WriteLine(max_value);
}
// Driver code
public static void Main()
{
int [ , ] arr = new int [ , ] { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
int n = 4;
int [ , ] prefix_xor = new int[N, N];
// Find the prefix_xor
prefix(arr, prefix_xor, n);
// Find submatrix with maximum bitwise xor
Max_xor(prefix_xor, n);
}
}
// This code is contributed by ihritik输出：
31