给定一个“ N * N”矩阵,任务是找到所有可能子矩阵的XOR的XOR。
例子:
Input :arr = {{3, 1},
{1, 3}}
Output : 0
Explanation: All the elements lie in 4 submatrices each.
4 being even, there total contribution towards
final answer becomes 0. Thus, ans = 0.
Input : arr = {{6, 7, 13},
{8, 3, 4},
{9, 7, 6}};
Output : 4
一种简单的方法是生成所有可能的子矩阵,唯一地找到每个子矩阵的XOR,然后对它们进行XOR。该方法的时间复杂度将为O(n 6 )。
更好的解决方案:对于每个索引(R,C),我们将尝试查找该索引所在的子矩阵的数量。如果子矩阵的数量为奇数,则最终答案将更新为ans =(ans ^ arr [R] [C]) 。如果是偶数,我们不需要更新答案。之所以可行,是因为与自身进行异或的数字为零,并且运算顺序不会影响最终的异或值。
假设基于0的索引,则索引(R,C)所在的子矩阵的数量等于
(R + 1)*(C + 1)*(N - R)*(N - C)
下面是上述方法的实现:
C++
// C++ program to find the XOR of XOR's of
// all submatrices
#include
using namespace std;
#define n 3
// Function to find to required
// XOR value
int submatrixXor(int arr[][n])
{
int ans = 0;
// Nested loop to find the
// number of sub-matrix each
// index belongs to
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Number of ways to choose
// from top-left elements
int top_left = (i + 1) * (j + 1);
// Number of ways to choose
// from bottom-right elements
int bottom_right = (n - i) * (n - j);
if ((top_left % 2 == 1) && (bottom_right % 2 == 1))
ans = (ans ^ arr[i][j]);
}
}
return ans;
}
// Driver Code
int main()
{
int arr[][n] = { { 6, 7, 13 },
{ 8, 3, 4 },
{ 9, 7, 6 } };
cout << submatrixXor(arr);
return 0;
}
Java
//Java program to find the XOR of XOR's
// of all submatrices
class GFG
{
// Function to find to required
// XOR value
static int submatrixXor(int[][]arr)
{
int n = 3;
int ans = 0;
// Nested loop to find the
// number of sub-matrix each
// index belongs to
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// Number of ways to choose
// from top-left elements
int top_left = (i + 1) * (j + 1);
// Number of ways to choose
// from bottom-right elements
int bottom_right = (n - i) * (n - j);
if ((top_left % 2 == 1) &&
(bottom_right % 2 == 1))
ans = (ans ^ arr[i][j]);
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int[][] arr = {{ 6, 7, 13},
{ 8, 3, 4 },
{ 9, 7, 6 }};
System.out.println(submatrixXor(arr));
}
}
// This code is contributed
// by Code_Mech.
Python3
# Python3 program to find the XOR of
# XOR's of all submatrices
# Function to find to required
# XOR value
def submatrixXor(arr, n):
ans = 0
# Nested loop to find the
# number of sub-matrix each
# index belongs to
for i in range(0, n):
for j in range(0, n):
# Number of ways to choose
# from top-left elements
top_left = (i + 1) * (j + 1)
# Number of ways to choose
# from bottom-right elements
bottom_right = (n - i) * (n - j)
if (top_left % 2 == 1 and
bottom_right % 2 == 1):
ans = (ans ^ arr[i][j])
return ans
# Driver code
n = 3
arr = [[6, 7, 13],
[8, 3, 4],
[9, 7, 6]]
print(submatrixXor(arr, n))
# This code is contributed by Shrikant13
C#
// C# program to find the XOR of XOR's
// of all submatrices
using System;
class GFG
{
// Function to find to required
// XOR value
static int submatrixXor(int [,]arr)
{
int n = 3;
int ans = 0;
// Nested loop to find the
// number of sub-matrix each
// index belongs to
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// Number of ways to choose
// from top-left elements
int top_left = (i + 1) * (j + 1);
// Number of ways to choose
// from bottom-right elements
int bottom_right = (n - i) * (n - j);
if ((top_left % 2 == 1) &&
(bottom_right % 2 == 1))
ans = (ans ^ arr[i, j]);
}
}
return ans;
}
// Driver Code
public static void Main()
{
int [, ]arr = {{ 6, 7, 13},
{ 8, 3, 4 },
{ 9, 7, 6 }};
Console.Write(submatrixXor(arr));
}
}
// This code is contributed
// by Akanksha Rai
PHP
Javascript
输出:
4
时间复杂度:O(N 2 )
辅助空间: O(1)
如果您希望与行业专家一起参加现场课程,请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。