原文:
给你一个有序矩阵 n*n 。任务是通过将给定矩阵的镜像与矩阵本身相加来找到结果矩阵。 示例 :
input : mat[][] = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
output : 4 4 4
10 10 10
16 16 16
explanation:
resultant matrix = {{1, 2, 3}, {{3, 2, 1},
{4, 5, 6}, {6, 5, 4},
{7, 8, 9}} {9, 8, 7}}
input : mat[][] = {{1, 2},
{3, 4}}
output : 3 3
7 7
当找到矩阵的镜像时，每个元素的行将保持不变，但其列的值将重新排列。对于任何元素 a ij ，其在镜像中的新位置将是 a i(n-j) 。获得矩阵的镜像后，将其添加到原始矩阵并打印结果。 积分照顾:
- 矩阵的索引将从 0，0 开始，在 n-1，n-1 结束，因此任何元素 a ij 的位置将是 a i(n-1-j)。
- 打印结果时，请注意正确的输出格式
以下是上述方法的实现:
c
// c program to find sum of matrix and
// its mirror image
#include
#define n 4
using namespace std;
// function to print the resultant matrix
void printsum(int mat[][n])
{
for (int i = 0; i < n; i ) {
for (int j = 0; j < n; j ) {
cout << setw(3) << mat[i][n - 1 - j] mat[i][j] << " ";
}
cout << "\n";
}
}
// driver code
int main()
{
int mat[n][n] = { { 2, 4, 6, 8 },
{ 1, 3, 5, 7 },
{ 8, 6, 4, 2 },
{ 7, 5, 3, 1 } };
printsum(mat);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to find sum of
// matrix and its mirror image
import java.io.*;
class gfg
{
static int n = 4;
// function to print the
// resultant matrix
static void printsum(int mat[][])
{
for (int i = 0; i < n; i )
{
for (int j = 0; j < n; j )
{
system.out.print((mat[i][n - 1 - j]
mat[i][j]) " ");
}
system.out.println();
}
}
// driver code
public static void main (string[] args)
{
int mat[][] = { { 2, 4, 6, 8 },
{ 1, 3, 5, 7 },
{ 8, 6, 4, 2 },
{ 7, 5, 3, 1 } };
printsum(mat);
}
}
// this code is contributed by anuj_67
python 3
# python 3 program to find sum of matrix
# and its mirror image
n = 4
# function to print the resultant matrix
def printsum(mat):
for i in range(n):
for j in range(n):
print('{:>3}'.format(mat[i][n - 1 - j]
mat[i][j]), end =" ")
print("\n", end = "")
# driver code
if __name__ == '__main__':
mat = [[2, 4, 6, 8],
[1, 3, 5, 7],
[8, 6, 4, 2],
[7, 5, 3, 1]]
printsum(mat)
# this code is contributed by
# surendra_gangwar
c
// c# program to find sum of
// matrix and its mirror image
using system;
class gfg
{
static int n = 4;
// function to print the
// resultant matrix
static void printsum(int [,]mat)
{
for (int i = 0; i < n; i )
{
for (int j = 0; j < n; j )
{
console.write((mat[i, n - 1 - j]
mat[i, j]) " ");
}
console.writeline();
}
}
// driver code
public static void main ()
{
int [,]mat = { { 2, 4, 6, 8 },
{ 1, 3, 5, 7 },
{ 8, 6, 4, 2 },
{ 7, 5, 3, 1 } };
printsum(mat);
}
}
// this code is contributed by shs..
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
output:
10 10 10 10
8 8 8 8
10 10 10 10
8 8 8 8
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