制作 3*3 矩阵幻方所需的最小更改
给定一个 3*3 矩阵，找出需要对其进行的最少更改次数，以便将其变成幻方。幻方是一个方阵，其所有行的总和相同，所有列的总和相同，并且两条对角线的总和相同。
例子：
Input : 8 5 6
3 8 7
4 9 2
Output : 2
5 in row 1 col 2 should be changed to 1
8 in row 2 col 2 should be changed to 5
Total 2 changes are required.
Input : 8 9 4
5 9 3
6 1 8
Output : 3
8 in row 1 col 1 should be changed to 2
5 in row 2 col 1 should be changed to 7
9 in row 2 col 2 should be changed to 5
Total 3 changes are required.有8 个可能的 3*3 维幻方，它们是
8 1 6
3 5 7
4 9 2
6 1 8
7 5 3
2 9 4
4 9 2
3 5 7
8 1 6
2 9 4
7 5 3
6 1 8
8 3 4
1 5 9
6 7 2
4 3 8
9 5 1
2 7 6
6 7 2
1 5 9
8 3 4
2 7 6
9 5 1
4 3 8
我们创建了一个 3D 数组来存储这 8 个矩阵。现在我们检查这 8 个矩阵中的每一个的输入矩阵，并找出可以通过最少数量的变化获得的矩阵。
C++
#include
using namespace std;
// This program takes in two 2D arrays as
// input and compares them to find out the
// minimum number of changes that needs to
// be made to convert arr to ms.
int findMinimumFromMS(int arr[][3], int ms[][3])
{
int count = 0;
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
if (arr[i][j] != ms[i][j])
count++;
}
}
return count;
}
int findMinimum(int arr[][3])
{
int ms[][3][3] = {
{ { 8, 1, 6 }, { 3, 5, 7 }, { 4, 9, 2 } },
{ { 6, 1, 8 }, { 7, 5, 3 }, { 2, 9, 4 } },
{ { 4, 9, 2 }, { 3, 5, 7 }, { 8, 1, 6 } },
{ { 2, 9, 4 }, { 7, 5, 3 }, { 6, 1, 8 } },
{ { 8, 3, 4 }, { 1, 5, 9 }, { 6, 7, 2 } },
{ { 4, 3, 8 }, { 9, 5, 1 }, { 2, 7, 6 } },
{ { 6, 7, 2 }, { 1, 5, 9 }, { 8, 3, 4 } },
{ { 2, 7, 6 }, { 9, 5, 1 }, { 4, 3, 8 } }, };
// If all the elements need to be changed,
// there would be 9 changes, so we take the
// max as 9
int min = 9;
for(int i = 0; i < 8; i++)
{
int x = findMinimumFromMS(arr, ms[i]);
if (x < min)
min = x;
}
return min;
}
// Driver code
int main()
{
int arr[][3] = { { 8, 5, 6 },
{ 3, 8, 7 },
{ 4, 9, 2 } };
cout << findMinimum(arr) << endl;
return 0;
}
// This code is contributed by divyeshrabadiya07 Java
class GfG {
// this program takes in two 2D arrays as
// input and compares them to find out the
// minimum number of changes that needs to
// be made to convert arr to ms.
public static int findMinimumFromMS(int[][] arr,
int[][] ms)
{
int count = 0;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (arr[i][j] != ms[i][j])
count++;
}
}
return count;
}
public static int findMinimum(int[][] arr)
{
int[][][] ms = {
{ { 8, 1, 6 }, { 3, 5, 7 }, { 4, 9, 2 } },
{ { 6, 1, 8 }, { 7, 5, 3 }, { 2, 9, 4 } },
{ { 4, 9, 2 }, { 3, 5, 7 }, { 8, 1, 6 } },
{ { 2, 9, 4 }, { 7, 5, 3 }, { 6, 1, 8 } },
{ { 8, 3, 4 }, { 1, 5, 9 }, { 6, 7, 2 } },
{ { 4, 3, 8 }, { 9, 5, 1 }, { 2, 7, 6 } },
{ { 6, 7, 2 }, { 1, 5, 9 }, { 8, 3, 4 } },
{ { 2, 7, 6 }, { 9, 5, 1 }, { 4, 3, 8 } },
};
// If all the elements need to be changed,
// there would be 9 changes, so we take the
// max as 9
int min = 9;
for (int i = 0; i < 8; i++) {
int x = findMinimumFromMS(arr, ms[i]);
if (x < min)
min = x;
}
return min;
}
public static void main(String[] args)
{
int[][] arr = { { 8, 5, 6 }, { 3, 8, 7 },
{ 4, 9, 2 } };
System.out.println(findMinimum(arr));
}
}Python3
# This program takes in two 2D arrays as
# input and compares them to find out the
# minimum number of changes that needs to
# be made to convert arr to ms.
def findMinimumFromMS(arr, ms):
count = 0
for i in range(0, 3):
for j in range(0, 3):
if arr[i][j] != ms[i][j]:
count += 1
return count
def findMinimum(arr):
ms = [
[ [ 8, 1, 6 ], [ 3, 5, 7 ], [ 4, 9, 2 ] ],
[ [ 6, 1, 8 ], [ 7, 5, 3 ], [ 2, 9, 4 ] ],
[ [ 4, 9, 2 ], [ 3, 5, 7 ], [ 8, 1, 6 ] ],
[ [ 2, 9, 4 ], [ 7, 5, 3 ], [ 6, 1, 8 ] ],
[ [ 8, 3, 4 ], [ 1, 5, 9 ], [ 6, 7, 2 ] ],
[ [ 4, 3, 8 ], [ 9, 5, 1 ], [ 2, 7, 6 ] ],
[ [ 6, 7, 2 ], [ 1, 5, 9 ], [ 8, 3, 4 ] ],
[ [ 2, 7, 6 ], [ 9, 5, 1 ], [ 4, 3, 8 ] ],
]
# If all the elements need to be changed, there
# would be 9 changes, so we take the max as 9
Min = 9
for i in range(0, 8):
x = findMinimumFromMS(arr, ms[i])
if x < Min:
Min = x
return Min
if __name__ == "__main__":
arr = [ [ 8, 5, 6 ], [ 3, 8, 7 ], [ 4, 9, 2 ] ]
print(findMinimum(arr))
# This code is contributed by Rituraj JainC#
using System;
class GFG{
// This program takes in two 2D arrays as
// input and compares them to find out the
// minimum number of changes that needs to
// be made to convert arr to ms.
static int findMinimumFromMS(int[,] arr, int[,,] ms,
int row)
{
int count = 0;
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
if (arr[i, j] != ms[row, i, j])
{
count++;
}
}
}
return count;
}
static int findMinimum(int[,] arr)
{
int[,,] ms = {
{ { 8, 1, 6 }, { 3, 5, 7 }, { 4, 9, 2 } },
{ { 6, 1, 8 }, { 7, 5, 3 }, { 2, 9, 4 } },
{ { 4, 9, 2 }, { 3, 5, 7 }, { 8, 1, 6 } },
{ { 2, 9, 4 }, { 7, 5, 3 }, { 6, 1, 8 } },
{ { 8, 3, 4 }, { 1, 5, 9 }, { 6, 7, 2 } },
{ { 4, 3, 8 }, { 9, 5, 1 }, { 2, 7, 6 } },
{ { 6, 7, 2 }, { 1, 5, 9 }, { 8, 3, 4 } },
{ { 2, 7, 6 }, { 9, 5, 1 }, { 4, 3, 8 } },
};
// If all the elements need to be changed,
// there would be 9 changes, so we take the
// max as 9
int min = 9;
for(int i = 0; i < 8; i++)
{
int x = findMinimumFromMS(arr, ms, i);
if (x < min)
{
min = x;
}
}
return min;
}
// Driver Code
static public void Main()
{
int[,] arr = { { 8, 5, 6 },
{ 3, 8, 7 },
{ 4, 9, 2 } };
Console.WriteLine(findMinimum(arr));
}
}
// This code is contributed by avanitrachhadiya2155Javascript
输出：
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