quartus编写一段代码实现按一下key0数码管上显示的数字加一，连续按两下key0数码管上显示的数字减一

在Quartus中，要实现这样的功能，我们通常会用到FPGA或CPLD的GPIO模块配合状态机的设计。这里假设你正在使用Verilog HDL语言编写代码。首先，我们需要定义按键输入、数码管输出以及相应的状态变量。

module key_counter(
    input [7:0] keypad_input,   // Keypad inputs (key0 is assumed to be bit 0)
    output reg [3:0] display,    // Seven-segment display outputs
    input clk,                  // System clock
    input reset                // Reset signal
);

// States for the state machine
typedef enum {IDLE, KEY0_PRESSED, KEY0_RELEASED} KeyState;

reg [2:0] currentState = IDLE;
reg prev_key_state = IDLE;

always @(posedge clk or posedge reset) begin
    if(reset) begin
        currentState <= IDLE;
        prev_key_state <= IDLE;
    end else if(keypad_input[0]) begin
        if(currentState == KEY0_PRESSED && prev_key_state != KEY0_PRESSED) begin
            currentState <= KEY0_RELEASED;   // Subtract one when double press detected
        end else if(currentState == KEY0_RELEASED) begin
            currentState <= KEY0_PRESSED;     // Add one on single press
        end
    end else begin
        currentState <= prev_key_state;
    end
    
    // Update display based on current state
    case(currentState)
        IDLE: display <= 0;           // Display '0' initially
        KEY0_PRESSED:
            display <= display + 1;     // Increment display value
        KEY0_RELEASED:
            display <= display - 1;     // Decrement display value
    endcase

    prev_key_state <= currentState;
end

endmodule

这个例子中，`currentState`表示当前的状态，`prev_key_state`用于记录前一次按键状态，以判断是否为双击。当按下并释放`key0`时，如果状态从按下变为释放，那么执行减一操作；单次按下则执行加一操作。