题解 | #求最小公倍数#
求最小公倍数
https://www.nowcoder.com/practice/ce067d6beee2413c8a26d37ca1a9431f
`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] minusA_reg, minusB_reg;
wire [DATA_W-1:0] minusA, minusB;
always@(posedge clk or negedge rst_n) begin
if(!rst_n) begin
minusA_reg <= 'd0;
minusB_reg <= 'd0;
end
else begin
minusA_reg <= vld_in?A:(minusA - minusB);
minusB_reg <= vld_in?B:minusB;
end
end
assign minusA = (minusA_reg>minusB_reg)?minusA_reg:minusB_reg;
assign minusB = (minusA_reg>minusB_reg)?minusB_reg:minusA_reg;
reg [DATA_W*2-1:0] lcm_out_reg;
reg [DATA_W-1:0] mcd_out_reg;
always@(posedge clk or negedge rst_n) begin
if(!rst_n) begin
vld_out <= 1'b0;
{lcm_out_reg,mcd_out_reg} <= 'd0;
end
else begin
vld_out <= (minusA_reg==minusB_reg)&(minusA_reg!='d0);
lcm_out_reg <= ((minusA_reg==minusB_reg)&(minusA_reg!='d0))?(A*B/minusB):'d0;
mcd_out_reg <= ((minusA_reg==minusB_reg)&(minusA_reg!='d0))?minusB:'d0;
end
end
assign mcd_out = mcd_out_reg;
assign lcm_out = lcm_out_reg;
endmodule
