4*3
要实现上下左右OK五个按键来操作一个4x3矩阵，您可以按照类似的思路进行编程。以下是一个示例代码：
#include <stdio.h>
#define ROWS 4
#define COLS 3
int matrix[ROWS][COLS] = {
{
1, 2, 3},
{
4, 5, 6},
{
7, 8, 9},
{
10, 11, 12}
};
void printMatrix() {
for (int i = 0; i < ROWS; i++) {
for (int j = 0; j < COLS; j++) {
printf("%-2d ", matrix[i][j]);
}
printf("\n");
}
printf("\n");
}
void moveUp() {
int temp[COLS];
for (int j = 0; j < COLS; j++) {
temp[j] = matrix[0][j];
}
for (int i = 0; i < ROWS - 1; i++) {
for (int j = 0; j < COLS; j++) {
matrix[i][j] = matrix[i + 1][j];
}
}
for (int j = 0; j < COLS; j++) {
matrix[ROWS - 1][j] = temp[j];
}
}
void moveDown() {
int temp[COLS];
for (int j = 0; j < COLS; j++) {
temp[j] = matrix[ROWS - 1][j];
}
for (int i = ROWS - 1; i > 0; i--) {
for (int j = 0; j < COLS; j++) {
matrix[i][j] = matrix[i - 1][j];
}
}
for (int j = 0; j < COLS; j++) {
matrix[0][j] = temp[j];
}
}
void moveLeft() {
int temp[ROWS];
for (int i = 0; i < ROWS; i++) {
temp[i] = matrix[i][0];
}
for (int i = 0; i < ROWS; i++) {
for (int j = 0; j < COLS - 1; j++) {
matrix[i][j] = matrix[i][j + 1];
}
}
for (int i = 0; i < ROWS; i++) {
matrix[i][COLS - 1] = temp[i];
}
}
void moveRight() {
int temp[ROWS];
for (int i = 0; i < ROWS; i++) {
temp[i] = matrix[i][COLS - 1];
}
for (int i = 0; i < ROWS; i++) {
for (int j = COLS - 1; j > 0; j--) {
matrix[i][j] = matrix[i][j - 1];
}
}
for (int i = 0; i < ROWS; i++) {
matrix[i][0] = temp[i];
}
}
void pressOK() {
int sum = 0;
for (int i = 0; i < ROWS; i++) {
for (int j = 0; j < COLS; j++) {
sum += matrix[i][j];
}
}
printf("Sum: %d\n", sum);
}
int main() {
int choice;
while (1) {
printMatrix();
printf("Press 1: Up, 2: Down, 3: Left, 4: Right, 5: OK\n");
printf("Enter your choice: ");
scanf("%d", &choice);
switch (choice) {
case 1:
moveUp();
break;
case 2:
moveDown();
break;
case 3:
moveLeft();
break;
case 4:
moveRight();
break;
case 5:
pressOK();
break;
default:
printf("Invalid choice!\n");
}
printf("\n");
}
return 0;
}
这个示例与之前的示例非常相似，只是调整了矩阵的行和列的顺序，以适应4x3的矩阵。移动操作的实现也有一些微调。
3*4
要实现上下左右OK五个按键来操作一个3x4矩阵，您可以使用一个二维数组来表示该矩阵，并利用按键输入来进行矩阵操作。以下是一个示例代码：
#include <stdio.h>
#define ROWS 3
#define COLS 4
int matrix[ROWS][COLS] = {
{
1, 2, 3, 4},
{
5, 6, 7, 8},
{
9, 10, 11, 12}
};
void printMatrix() {
for (int i = 0; i < ROWS; i++) {
for (int j = 0; j < COLS; j++) {
printf("%-2d ", matrix[i][j]);
}
printf("\n");
}
printf("\n");
}
void moveUp() {
int temp = matrix[0][0];
for (int i = 0; i < ROWS - 1; i++) {
for (int j = 0; j < COLS; j++) {
matrix[i][j] = matrix[i + 1][j];
}
}
for (int j = 0; j < COLS; j++) {
matrix[ROWS - 1][j] = temp;
}
}
void moveDown() {
int temp = matrix[ROWS - 1][0];
for (int i = ROWS - 1; i > 0; i--) {
for (int j = 0; j < COLS; j++) {
matrix[i][j] = matrix[i - 1][j];
}
}
for (int j = 0; j < COLS; j++) {
matrix[0][j] = temp;
}
}
void moveLeft() {
int temp = matrix[0][0];
for (int i = 0; i < ROWS; i++) {
for (int j = 0; j < COLS - 1; j++) {
matrix[i][j] = matrix[i][j + 1];
}
}
for (int i = 0; i < ROWS; i++) {
matrix[i][COLS - 1] = temp;
}
}
void moveRight() {
int temp = matrix[0][COLS - 1];
for (int i = 0; i < ROWS; i++) {
for (int j = COLS - 1; j > 0; j--) {
matrix[i][j] = matrix[i][j - 1];
}
}
for (int i = 0; i < ROWS; i++) {
matrix[i][0] = temp;
}
}
void pressOK() {
int sum = 0;
for (int i = 0; i < ROWS; i++) {
for (int j = 0; j < COLS; j++) {
sum += matrix[i][j];
}
}
printf("Sum: %d\n", sum);
}
int main() {
int choice;
while (1) {
printMatrix();
printf("Press 1: Up, 2: Down, 3: Left, 4: Right, 5: OK\n");
printf("Enter your choice: ");
scanf("%d", &choice);
switch (choice) {
case 1:
moveUp();
break;
case 2:
moveDown();
break;
case 3:
moveLeft();
break;
case 4:
moveRight();
break;
case 5:
pressOK();
break;
default:
printf("Invalid choice!\n");
}
printf("\n");
}
return 0;
}
在这个示例中，我们定义了一个3x4矩阵matrix，以及相应的按键操作函数。printMatrix函数用于打印矩阵。moveUp、moveDown、moveLeft、moveRight函数分别实现了上、下、左、右按键的移动操作，将矩阵按指定方向进行移动。pressOK函数用于计算矩阵中所有元素的和。
在main函数中，我们通过一个循环来接受用户的按键输入，并根据选择调用相应的按键操作函数或打印矩阵的和。
