题解 | #求最小公倍数#

辗转相除法，不需要使用状态机，而是引入了一个flag_r标志正在进行辗转相除，比参考答案节约了7个cycle

`timescale 1ns/1ns

module lcm#(
parameter DATA_W = 8)
(
input [DATA_W-1:0] A,
input [DATA_W-1:0] B,
input 			vld_in,
input			rst_n,
input 			clk,
output	wire	[DATA_W*2-1:0] 	lcm_out,
output	wire 	[DATA_W-1:0]	mcd_out,
output	reg					vld_out
);
    
    reg [DATA_W-1:0] a_r;
    reg [DATA_W-1:0] b_r;
    wire [DATA_W-1:0] a_w;
    wire [DATA_W-1:0] b_w;
    wire [DATA_W-1:0] res_w;
    reg flag_r;
    reg [DATA_W*2-1:0] lcm_out_r;
    
    assign vld_out = flag_r && (a_r == b_r);
    assign res_w = a_r - b_r;
    assign {a_w, b_w} = res_w > b_r ? {res_w, b_r} : {b_r, res_w};
    assign mcd_out = vld_out ? a_r : 'd0;
    assign lcm_out = lcm_out_r/ mcd_out;
    
    always @(posedge clk or negedge rst_n) begin
        if (!rst_n) begin
            a_r <= 'd0;
            b_r <= 'd0;
            flag_r <= 'd0;
        end
        else if (vld_in) begin
            {a_r, b_r} <= A > B ? {A, B} : {B, A};
            lcm_out_r <= A * B;
            flag_r <= 'd1;
        end
        else if (vld_out) begin
            flag_r <= 'd0;
        end
        else if (flag_r) begin
            a_r <= a_w;
            b_r <= b_w;
        end
    end
endmodule