题解 | #游戏机计费程序#
游戏机计费程序
http://www.nowcoder.com/practice/50188fb7e23b4eee86f8c463c8284f5e
题意整理
money是预付费，根据boost的值递减，boost为0时，每个时钟减一，boost为1时，每个时钟减2。同时把每个时刻变化之后的值作为remain的值输出。当set为1时，表示玩家续费，将money的值加到remain上。当remain小于10时，将yellow的值置为1。当费用不足时，将red的值置为1，yellow的值置为0。
值得注意的是：由于在boost情况下，每个时钟remain的数值减二，所以remain不一定可以达到0，费用不足应包括两个情况：remain=0或remain =1 &&boost=1。题解主体
首先明确remain的取值，当费用不足时（remain == 0 || (remain==1&&boost)），remain的值保持为0。在boost的值为0时，remain每个时钟减一；boost为1时，remain的值每个时钟减二。当set信号有效，应把money的数值加到remain之中。always @(posedge clk or negedge rst_n)
begin
if(!rst_n)
remain <= 0;
else if (set)
remain <= money + remain;
else if (remain == 0 || (remain==1&&boost))
remain <= remain;
else
begin
if (!boost)
remain <= remain -1;
else if (boost)
remain <= remain -2;
else remain <= remain;
end
end 然后明确指示灯的逻辑：当remain小于10时，将yellow的值置为1。remain=0或remain =1 &&boost=1时，将red的值置为1，yellow的值置为0。
always @(posedge clk or negedge rst_n)
if(!rst_n)
begin
yellow <= 1'd0;
red <= 1'd0;
end
else if (remain <10 && remain !== 0)
begin
yellow <= 1'd1;
red <= 1'd0;
end
else if (remain == 0 || (remain==1&&boost))
begin
yellow <= 1'd0;
red <= 1'd1;
end
参考答案
`timescale 1ns/1ns
module game_count
(
input rst_n, //异位复位信号，低电平有效
input clk, //时钟信号
input [9:0]money,
input set,
input boost,
output reg[9:0]remain,
output reg yellow,
output reg red
);
always @(posedge clk or negedge rst_n)
begin
if(!rst_n)
remain <= 0;
else if (set&&!boost)
remain <= money + remain -1;
else if (set&&!boost)
remain <= money + remain -2;
else if (remain == 0 || (remain==1&&boost))
remain <= remain;
else
begin
if (!boost)
remain <= remain -1;
else if (boost)
remain <= remain -2;
else remain <= remain;
end
end
always @(posedge clk or negedge rst_n)
if(!rst_n)
begin
yellow <= 1'd0;
red <= 1'd0;
end
else if (remain <12 && remain !== 0)
begin
yellow <= 1'd1;
red <= 1'd0;
end
else if (remain == 0 || (remain==1&&boost))
begin
yellow <= 1'd0;
red <= 1'd1;
end
else
begin
yellow <= 1'd0;
red <= 1'd0;
end
endmodule
查看25道真题和解析
