Verilog-检测输入序列能否被3整除

目录

• 题目
• 编程思路
• 代码
• 仿真波形

题目

输入一个序列，最新输入的数字为最低位，如果当前序列能被3整除，输出1，否则输出0。
例如：输入1010_1111,对应1,2,5,10,21,43,87，175，因此输出为：0000_1010.

编程思路

Last_remainder	In	Remainder	Out
0	0	0	1
1	0	2	0
2	0	1	0
0	1	1	0
1	1	0	1
2	1	2	0

代码

`timescale 1ns / 1ps

module sequence_divide_by_3(
	input clk,
	input rstn,
	input data,
	
	output divide_by_3
    );
	 
parameter Remainder_0 = 2'd0,
			 Remainder_1 = 2'd1,
			 Remainder_2 = 2'd2;
			 
reg [1:0] state,next_state;

always @(posedge clk or negedge rstn) begin
	if(!rstn) state <= Remainder_0;
	else state <= next_state;
end

always @(*) begin
	case(state)
		Remainder_0: next_state = data ? Remainder_1 : Remainder_0;
		Remainder_1: next_state = data ? Remainder_0 : Remainder_2;
		Remainder_2: next_state = data ? Remainder_2 : Remainder_1;
		default : next_state = Remainder_0;
	endcase
end

assign divide_by_3 = (((state == Remainder_0) && (~data)) || ((state == Remainder_1) && (data)))?1'b1:1'b0;

endmodule

仿真波形