JavaScript-Find an algorithm? 0000000 sequence Increment

To raise a chestnut: 0000000 increments, the format is this:

000000100000020000003.....递增到100000010.....递增到1000000100.....递增到1000000100000

Imitating the answer of @g_koala_c, wrote the PHP version:

for ($i = 0; $i < 100; $i++) {     $zero = '';    $k = 7-strlen($i);    for ($j = $k; $j >0; $j--) {         $zero .= 0;    }    echo $zero.$i.'
';}

Reply content:

To raise a chestnut: 0000000 increments, the format is this:

000000100000020000003.....递增到100000010.....递增到1000000100.....递增到1000000100000

Imitating the answer of @g_koala_c, wrote the PHP version:

for ($i = 0; $i < 100; $i++) {     $zero = '';    $k = 7-strlen($i);    for ($j = $k; $j >0; $j--) {         $zero .= 0;    }    echo $zero.$i.'
';}

JS notation, the principle is to first calculate the length of the number, and then 0 in front of the complement. Now it's 100, you can change I to 100000.

for (var i = 0 ; i <= 100; i ++){  var zero = "";  for (var j = 7-i.toString().length; j > 0; j--) {    zero += "0";  }  console.log(zero + i);}

is actually a zero-sum problem, Python 3 writes this:

["{:0>7}".format(i) for i in range(1, 100001)]

And then provide a JS to the wording

Array.from(Array(1000000).keys()).map(function(x){ return "0".repeat(8 - ("" + (x + 1)).length) + (x+1)})

Php

for ($i=0;$i<=9999999;$i++) echo str_pad($i,7,"0",STR_PAD_LEFT);

Provide a method that may be available in the future.

for (let i = 0; i <= 100; i++) console.log(String(i).padStart(7,'0'))

function incrace(){    console.log((n=>(7-n.length)>0?(new Array(7-n.length+1)).join(0)+n:n)(String(i++)));    setTimeout(incrace,500);}var i = 0;incrace();

After writing to find that the problem has been answered ... JS code, for reference

Js

var len = 10;    //长度for (var i = 1; i <= 100; i++) {    console.log((Array(len).join(0) + Math.abs(i)).slice(-len));}

Come and come ... On a set of PHP version of the wild path

// 定义一个位数比结果位数多的初始值$base_num = 10000000;// 开搞for ($i = 0; $i < 100; $i++) {    echo substr($base_num += $i, -7), "\n";}

Look at my C-language:

for (int i = 0; i < 10000000; i++) {    printf("%07d\n", i);}

The length after 0 is actually also available:

// 补零后长度, 注意printf的变化~in len = 7;for (int i = 0; i < 10000000; i++) {    printf("%0*d\n", len, i);}

I find that many of the answers are not real algorithms, and I thought the implementation of the C-language printf function would be similar.

The previous C-language implementation of the printf function, this problem does not have to calculate the length

Please run on a compiler that supports C99. (C89 not supported)

 #include  #include  void Ltoa (long num, int width, char *str) {static char digs[] = "0123456789";//To facilitate extended hex char ac[width+1];//C99 support; +1 is because the string must end with '% ', but this does not count toward the string length. int i = width; This variable can be used directly instead of I, but this function is I wrote before, too lazy to change the following code memset (&ac, ' 0 ', width); Directly put all memory first ' 0 ' ac[i] = ' + ';//The end of the string must be ' \ n '//followed by an integer-to-string code, the idea is that the division and the remainder to get each digit (the conversion is the same idea)//For example 123/10 = 12_3 12/10 = 1_2//1/10 = 0_1//You can see that the remainder of the 3,2,1 is inverted. Copy the 3, 2, 1 upside-down to the memory space. if (num) ac[--i] = digs[num% 10]; num = NUM/10; while (0 < num && 0 < i) {ldiv_t QR = ldiv (num, 10); ldiv_t with structural body, the main convenience is to remove the remainder and quotient num = Qr.quot; Ac[--i] = Digs[qr.rem]; Converting 0,1,3,4 to ' 0 ', ' 1 ', ' 2 ', ' 3 ', can also be used 0+ ' 0 ' method, but it has been explained above, in order to extend the convenience of other systems. } int n = sizeof (AC)-I; memcpy (str, &AC, width + 1);//Copy the string and the \ n to the buffer.}  

    int max = 一个数字;    for (int i = 0; i < max; i++) {        String tem = "";        for (int j = 0; j < len - (i + "").toString().length(); j++) {            tem = tem + "0";        }        System.out.println(tem+i);    }

Somebody change the value to more than 100000?

Why is there no C + + version AH.
I'm relieved to see how complicated you're writing.

Is there any algorithm to write? To a MySQL version of:

CREATE TABLE `test` (  `tid` int(7) UNSIGNED ZEROFILL DEFAULT NULL) ENGINE=InnoDB DEFAULT CHARSET=utf8;

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