五进制转换十三进制
 
朋友面试遇见的一个面试题，就敲了下试试，新手上路记录下。
 
这个是让写一个函数
 
function qiuzhi($wu){
	if(substr($wu,-1,1)>4){
		return '不是五进制';
	}	
	$len = strlen($wu);
	$n=1;
	$shi = 0;
	for($i=1;$i<=$len;$i++){
	 $zhi = substr($wu,-$i,1);
	 $shi += $zhi*$n;
	 $n *= 5;
	}
	$right = intval($shi/13);
	$left = $shi % 13;
	$units = '0';
		switch ($left) {
			case 10:
				$units = 'A';
				break;
			case 11:
				$units = 'B';
				break;
			case 12:
				$units = 'C';
				break;
			default:
				$units = $left;
	    		break;
		}
	return $right.$units;
}
echo qiuzhi(444);
 
还有一个php自带的函数：
 
$hexadecimal = '444';//十进制
echo base_convert($hexadecimal, 5, 13);

num=input('请输入要判断的数')
new_num=num.split(".")
def shiwu(num):
    list1 = []
    maolist = {0: '0', 1: '1', 2: '2', 3: '3', 4: '4', 5: '5', 6: '6', 7: '7', 8: '8', 9: "9", 10: 'A', 11: 'B',
               12: 'C', 13: 'D', 14: 'E'}
    new_num = int(num)
    flag = True
    while flag:
        shang = new_num // 15
        yushu = new_num % 15
        if shang >= 15:
            new_num = shang
            list1.append(yushu)
        else:
            list1.append(yushu)
            list1.append(shang)
            flag = False
    last_list = reversed(list1)
    new_last_list=[]
    for i in last_list:
        new_last_list.append(i)
    for i in range(len(new_last_list)):
        new_last_list[i]=maolist[new_last_list[i]]
    return new_last_list

if len(new_num) != 1:
    # 小数
    zhengshu=new_num[0]
    zhengshu_list=shiwu(zhengshu)
    zhengshu_last_list="".join(zhengshu_list)
    xiaoshu=new_num[1]
    xiaoshu_list=shiwu(xiaoshu)
    xiaoshu_last_list="".join(xiaoshu_list)
    print(zhengshu_last_list+'.'+xiaoshu_last_list)
else:
    # 整数
    shu_list=shiwu(new_num[0])
    new_last_num="".join(shu_list)
    print(new_last_num)

一个简单的五进制加法计数器
 
1.1 verilog代码
 
 
 
module counter_fsm(clk,rst_n,Z);
 input clk,rst_n;
 output reg Z;
 reg [2:0] pre_state,next_state;
 parameter s0=3’b000,s1=3’b001,s2=3’b010,s3=3’b011,s4=3’b100;
 always@(posedge clk or negedge rst_n)
 if(!rst_n) pre_state<=s0;
 else pre_state<=next_state;
 always@(pre_state) begin
 next_state=3’bxxx;
 Z=1’b0;
 case(pre_state)
 s0:begin next_state=s1;Z=1’b0;end
 s1:begin next_state=s2;Z=1’b0;end
 s2:begin next_state=s3;Z=1’b0;end
 s3:begin next_state=s4;Z=1’b0;end
 s4:begin next_state=s0;Z=1’b1;end
 default:begin next_state=s0; end
 endcase
 end
 endmodule
 
 
1.2 testbench
 
 
 
`timescale 1ns/1ns
 module counter_fsm_tb;
 reg clk,rst_n;
 wire Z;
 counter_fsm U1(.clk(clk),.rst_n(rst_n),.Z(Z));
 always #10 clk=~clk;
 initial begin
 clk=1’b0;
 rst_n=1’b1;
 #10 rst_n=1’b0;
 #10 rst_n=1’b1;
 end
 endmodule
 
 
1.3 wave
 

题目要求
 
输入一个十进制整数，将这个数转化成对应的十五进制数（在十五进制中，A表示10，B表示11，C表示12，D表示13，E表示14），请写出转换程序。
 
例如：235表示为10A
 
代码实现
 
package test;

import java.util.ArrayList;
import java.util.Scanner;

public class Convert {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			int nextInt = sc.nextInt();
			System.out.println(getResult(nextInt));
		}
	}
	
	
	public static String getResult(int x) {
		ArrayList<Integer> remainder = new ArrayList<>();
		StringBuilder sb = new StringBuilder();
		// 找到终点
		while(x != 0) {
			
			remainder.add(x % 15);
			x /= 15;
			
		}
		// 反向输出
		for(int i = remainder.size() -1; i>=0; i--) {
			if(remainder.get(i) == 10) {
				sb.append('A');
			}else if(remainder.get(i) == 11){
				sb.append('B');
			}else if(remainder.get(i) == 12){
				sb.append('C');
			}else if(remainder.get(i) == 13){
				sb.append('D');
			}else if(remainder.get(i) == 14){
				sb.append('E');
			}else {
				sb.append(remainder.get(i));
			}
		}
		return sb.toString();
	}
	
	
}