题解 | #求最小公倍数#

//
//	最小公倍数，最大公约数
module lcm#(
parameter DATA_W = 8)
(
input [DATA_W-1:0] 	A,	//	输入2个无符号数
input [DATA_W-1:0] 	B,
input 				vld_in,	//	有效输入
input				rst_n,
input 				clk,

output	wire	[DATA_W*2-1:0] 	lcm_out,	//	最小公倍数
output	wire 	[DATA_W-1:0]	mcd_out,	//	最大公约数

output	reg						vld_out
);

reg		[DATA_W-1:0]	mcd, a_buf, b_buf;
reg		[DATA_W*2-1:0]	mul_buf;
//	reg						mcd_vld;
reg		[1:0]			cur_st, nx_st;

parameter		idle	= 2'b00,
				s1		= 2'b01,
				s2		= 2'b10;
//				s2		= 2'b11;

//	两段式状态机
always@(posedge clk or negedge rst_n)
	if(!rst_n)
		cur_st	<= idle;
	else
		cur_st	<= nx_st;

always@(posedge clk or negedge rst_n)		begin
	if(!rst_n)		begin
		a_buf		<= 8'd0;
		b_buf		<= 8'd0;
		mul_buf		<= 16'd0;

		mcd			<= 8'd0;
		vld_out		<= 1'b0;
		nx_st		<= idle;		
		end
	else 
		case(cur_st)
			idle:	begin
				if(vld_in)	begin
					a_buf		<= A;
					b_buf		<= B;
					mul_buf		<= A*B;
					nx_st		<= s1;
					end
				else	begin
					nx_st		<= idle;
					end
				vld_out	<= 1'b0;	
				end
			s1:
				if(a_buf != b_buf)	begin
					if(a_buf > b_buf)	begin
						a_buf	<= a_buf - b_buf;
						b_buf	<= b_buf;
						end
					else	begin
						a_buf	<= a_buf;
						b_buf	<= b_buf - a_buf;
						end
					nx_st	<= s1;
					end
					
				else	
					nx_st	<= s2;
			
			s2:	
				begin
					vld_out	<= 1'b1;
					mcd		<= a_buf;	//	最大公约数
					nx_st	<= idle;
				end
				
			default:	nx_st	<= idle;
			
		endcase
end

//	注意：这里也是对的，但是在牛客网的测试上没通过.
//	assign	mcd_out		= (vld_out == 1'b1) ? mcd : 8'd0;
//	assign	lcm_out		= (vld_out == 1'b1)	? mul_buf/mcd_out : 16'd0;

assign	mcd_out		= mcd;
assign	lcm_out		= mul_buf/mcd_out;

endmodule