🌹 喇賽的人 Blog 🌹

🌹 喇低喇賽 🌹

0%

sql 99乘法表

亂寫於 更新於 Disqus：
廢話字數： 5.1k 所需傷眼時間 ≈ 5 分鐘

 
terminal

c#

今天看書時又看到這題，這也是老生常談了，多數看到 c# or java 等程式語言會採用 loop 方式進行兩重迴圈輸出如下

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
//橫式
for(int i = 1; i <= 9; i++) {
    for(int j = 1; j <= 9; j++) {
        string str = $"{i} * {j} = {(i * j).ToString().PadLeft(2,' ')} | ";
        Console.Write(str);
    }
    Console.WriteLine();
    Console.WriteLine();
}

//直式
for (int i = 1; i <= 9; i++)
{
    //繪製 1 to 9
    for(int j = 0; j < 9; j++)
    {
        Console.Write($"   {i}  ");
    }
    Console.WriteLine();


    //繪製乘號及 1 to 9
    for (int j = 1; j <= 9; j++)
    {
        Console.Write($"x  {j}  ");
    }
    Console.WriteLine();


    //繪製填寫答案前的分隔號
    for (int j = 1; j <= 9; j++)
    {
        Console.Write($"----  ");
    }
    Console.WriteLine();

    //繪製解答
    for (int j = 1; j <= 9; j++)
    {
        Console.Write($"{(i * j).ToString().PadLeft(4, ' ')}  ");
    }
    Console.WriteLine();

    //保持多空一行
    Console.WriteLine();
}

其實可以用 linq 寫 , 跟 sql 解法雷同產生笛卡爾積

1
2
3
4
5
var nums = Enumerable.Range( 1, 9 );
var result = from x in nums
			 from y in nums
			 select $"{x} * {y} = {x * y}";
result.ToList().ForEach( x => Console.WriteLine(x));

後來想到這樣寫成直式好像也滿直覺的

1
2
3
4
5
6
7
8
9
var nums = Enumerable.Range( 1, 9 );
var result = from x in nums
			 from y in nums
			 select
			 $@"  {y.ToString( ).PadLeft( 4 )}" + Environment.NewLine +
			 $@"x {x.ToString( ).PadLeft( 4 )}" + Environment.NewLine +
			 $@"------" + Environment.NewLine +
			 $@"  {(x * y).ToString( ).PadLeft( 4 )}" + Environment.NewLine;
result.ToList().ForEach( x => Console.WriteLine(x));

這裡還有很噁爛的 SelectMany 寫法 , 想要詳細知道 linq 可以看這個微軟 mvp 大神的 blog , 應該是講解 linq 最細的 blog

1
2
3
4
5
6
var num = Enumerable.Range(1, 9);
var result = num.SelectMany(
		x => num,
		(x, y) => $"{x} * {y} = {x * y}" 
	);
result.ToList().ForEach(x => Console.WriteLine(x));

sql server

而 SQL 採用集合做為思考，一般都會使用 UNION ALL 搭配遞迴與 CROSS JOIN 進行計算，大概會長得像下面這樣

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
WITH TALLY(N) AS (
    SELECT  1 
    UNION ALL
    SELECT  2
    UNION ALL
    SELECT  3
    UNION ALL
    SELECT  4
    UNION ALL
    SELECT  5
    UNION ALL
    SELECT  6
    UNION ALL
    SELECT  7
    UNION ALL
    SELECT  8
    UNION ALL
    SELECT  9
)
SELECT CAST(A.N AS varchar) + '*'  
    + CAST(B.N AS varchar) + '=' 
    + CAST(A.N * B.N AS varchar)
FROM Tally A 
CROSS JOIN Tally B

--使用遞迴(RECURSIVE)
WITH TALLY(N) AS (
    SELECT  1 N
    UNION ALL
    SELECT 1 + N
    FROM Tally
    WHERE N < 9
)
SELECT CAST(A.N AS varchar) + '*'  
    + CAST(B.N AS varchar) + '=' 
    + CAST(A.N * B.N AS varchar)
FROM Tally A 
CROSS JOIN Tally B

無聊想到好像也搭配 REPLICATE 可以拿來畫個三角形 , REPLICATE 到底還能做啥呢 , 大概統計東西時可以畫個文字版的 chart 之類的吧

1
2
3
4
5
6
7
8
9
WITH TALLY(N) AS (
    SELECT  1 N
    UNION ALL
    SELECT 1 + N
    FROM Tally
    WHERE N < 9
)
SELECT REPLICATE('*',N)
FROM TALLY

偶然看到老外文章
VALUES 也可以當作衍生資料表使用所以又生出一種無腦的寫法

1
2
3
4
5
6
7
8
9
10
11
WITH TALLY AS(
    SELECT *
    FROM (
        VALUES(1),(2),(3),(4),(5),(6),(7),(8),(9)
    ) T(N) 
)
SELECT CAST(A.N AS varchar) + '*'  
    + CAST(B.N AS varchar) + '=' 
    +  CAST(A.N * B.N AS varchar)
FROM TALLY A
CROSS JOIN TALLY B

如果是 sql server 2016 可以用 string_split

1
2
3
4
5
6
7
with cte(n) as (
	select cast(value as int) as n
	from string_split('1,2,3,4,5,6,7,8,9' , ',')
)
select x.n , '*' as '*' , y.n , '=' as '=' , (x.n * y.n) result
from cte x
cross join cte y

postgresql

後來想到 postgresql 其實可以直接用 generate_series 函數產生或是 array達成相同效果，
順便多個機車變化，將計算出的答案以 9 , 8 , 7 , 1 , 2 , 4 , 5 , 6 進行排序，一題完美刁難面試者的題目就誕生了

1
2
3
4
5
6
7
8
9
10
11
12
13
--使用generate_series
select a , b , a || '*' || b || '=' || a * b as result
from generate_series(1,9) a , generate_series(1,9) b
order by 
case when a >= 7 then a else -a end desc 
, b asc;

--使用array
elect a , b , a || '*' || b || '=' || a * b as result
from unnest(array[1,2,3,4,5,6,7,8,9]) a , unnest(array[1,2,3,4,5,6,7,8,9]) b
order by 
case when a >= 7 then a else -a end desc 
, b asc;

oracle

偶然機會下用到 Oracle 想說順便玩玩 , 然後就可以生出有顏色的九九乘法了

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
WITH TALLY AS (
SELECT 1 N FROM DUAL
UNION ALL
SELECT 2 N FROM DUAL
UNION ALL
SELECT 3 N FROM DUAL
UNION ALL
SELECT 4 N FROM DUAL
UNION ALL
SELECT 5 N FROM DUAL
UNION ALL
SELECT 6 N FROM DUAL
UNION ALL
SELECT 7 N FROM DUAL
UNION ALL
SELECT 8 N FROM DUAL
UNION ALL
SELECT 9 N FROM DUAL
)
SELECT CASE WHEN MOD(X.N , 2) = 0 THEN '<html><font color="red">' || X.N || '*' || Y.N || '=' || X.N * Y.N
ELSE X.N || '*' || Y.N || '=' || X.N * Y.N END NINENINE
FROM TALLY X
CROSS JOIN TALLY Y

python

最近搞 python , 來個單迴圈

1
2
3
4
5
6
7
8
9
10
i = 1
j = 1

while j <= 9:
    print(f"{i} * {j} = {i * j}")
    if i == 9:
        j += 1
        i = 1
    else:
        i += 1

反人類遞迴 python

1
2
3
4
5
6
7
8
9
10
11
12
def test(i, j):
    if j <= 9:
        print(f"{i} * {j} = {i * j}")
        if i == 9:
            j += 1
            i = 1
        else:
            i += 1

        test(i, j)

test(1,1)

js

1
2
3
4
5
6
7
8
9
for (var i = 1, j = 1; j <= 9;){
    console.log(`${i} * ${j} = ${i * j}`)
    if (i == 9) {
        i = 1
        j += 1
    } else {
        i += 1
    }
}

雙重遞迴 , 關鍵是在 N 裡面再多包一個 M 的遞迴
首先當數字小於 10 時會不斷印出來 , 當數字執行到 10 的時候會 return
接著再回到 N 開始往下個數字疊加下去 , 此時 m 已經歸零又重新進入 M 的遞迴
有趣的是拿這串 code 去問 chatgpt 問他這串作用是啥 , 他竟然可以回出 99 乘法表 , 真是太噁心了

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
function N(n , m){
	if(n === 10){
		return
	} else {
		M(m + 1, n)
		return N(n + 1 , m)
	}
}

function M(m , n){
	if(m === 10){
		return
	} else {
		console.log(`${n} * ${m} = ${n * m}`)
		return M(m + 1 , n)
	}
}

N(0,0)

rxjs

最近搞 rxjs 開開腦洞

1
2
3
4
5
6
7
8
9
10
let A$ = range(1, 9)
let B$ = range(1, 9)

A$.subscribe(x => {
    console.log(`----${x}-----`)
    B$.subscribe(y =>
        console.log(`${x} * ${y} = ${x * y}`)
    )
    console.log('----------')
})
關閉