用c语言设计一个程序,编写一个函数，按K1键D1亮，按K2键D2亮，按K3键D3亮，同时按下K1,K2键D4亮，同时按下K1,K3键，四个灯全亮，同时按下K1,K2，K3键,四个灯全灭

为了实现这个功能，我们可以使用C语言的基本输入输出以及一些开关状态变量。这里是一个简单的示例程序，我们假设已经有一个硬件环境，其中K1、K2、K3作为按键，D1、D2、D3、D4作为LED灯。我们将使用位操作来简化表示灯光的状态。

#include <stdio.h>
#include <delay.h> // 假设有个用于延时的库

// 定义键盘和灯的状态
#define KEYBOARD_1 0x01
#define KEYBOARD_2 0x02
#define KEYBOARD_3 0x04
#define LAMP_D1 0x08
#define LAMP_D2 0x10
#define LAMP_D3 0x20
#define LAMP_D4 (LAMP_D1 | LAMP_D2 | LAMP_D3)

int keyboard_state = 0; // 当前按键状态
int lamp_status = 0; // 灯光状态

void handle_key(int key) {
    if (key == KEYBOARD_1) {
        if ((keyboard_state & KEYBOARD_1) && (keyboard_state & KEYBOARD_2)) {
            lamp_status |= LAMP_D4;
        } else if (keyboard_state & KEYBOARD_1) {
            lamp_status |= LAMP_D1;
        }
    } else if (key == KEYBOARD_2) {
        if ((keyboard_state & KEYBOARD_1) && (keyboard_state & KEYBOARD_2)) {
            lamp_status |= LAMP_D4;
        } else if (keyboard_state & KEYBOARD_2) {
            lamp_status |= LAMP_D2;
        }
    } else if (key == KEYBOARD_3) {
        if ((keyboard_state & KEYBOARD_1) && (keyboard_state & KEYBOARD_3)) {
            lamp_status |= LAMP_D4;
        } else if (keyboard_state & KEYBOARD_3) {
            lamp_status |= LAMP_D3;
        }
    }

    // 更新键盘状态并检查是否所有灯都亮着
    keyboard_state |= key;
    if (keyboard_state == KEYBOARD_1 | KEYBOARD_2 | KEYBOARD_3) {
        lamp_status = LAMP_D4; // 如果所有键按下，全部灯亮
    } else {
        lamp_status &= ~LAMP_D4; // 否则清除所有灯
    }
}

void update_lights() {
    printf("Current lighting status: ");
    if (lamp_status & LAMP_D1) printf("D1 ");
    if (lamp_status & LAMP_D2) printf("D2 ");
    if (lamp_status & LAMP_D3) printf("D3 ");
    if (lamp_status & LAMP_D4) printf("D4\n");
    delay_ms(500); // 模拟灯泡点亮和熄灭的效果
}

int main() {
    while (1) {
        char keypress;
        scanf("%c", &keypress);
        handle_key(keypress);
        update_lights();
    }
    return 0;
}