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Reverse int without conversion but it returns 0. Why?

I am trying to reverse integer values without converting to string and then back to int. Correct me if I'm wrong but that's costly & unnecessary. I tried this function and a console.log within the while and just after it returns, for example, 5 (if n = 5) --- but the return value is 0. Why?
function reverseInt(n) {
const ten = 10;
const zero = 0;
// No need to convert, which is expensive
let rev_num = zero;
while (n > zero) {
rev_num = rev_num * ten + (n % ten);
n = Math.floor(n / ten);
}
return rev_num * Math.sign(n);
}
I was expecting 5, not 0.

Math.sign returns 0 since n is 0 at the end of the loop. Extract the sign before iterating instead.
function reverseInt(n) {
const ten = 10;
const zero = 0;
const sign = Math.sign(n);
n = Math.abs(n);
let rev_num = zero;
while (n > zero) {
rev_num = rev_num * ten + (n % ten);
n = Math.floor(n / ten);
}
return rev_num * sign;
}
console.log(reverseInt(123));

Different results for the same problem solution in node.js and python

I solved the following leetCode problem with some code :
You have d dice, and each die has f faces numbered 1, 2, ..., f.
Return the number of possible ways modulo 10^9 + 7 to roll the dice so the sum of the face up numbers equals t.
I made two versions of the solution code, one in node.js using mathjs, and one in python using the math module .
In node.js
const { combinations: comb, bignumber: Big } = require("mathjs");
function dice(d, f, t) {
if (t > d * f || t < d) return 0;
var result = Big(0);
var i = 0;
var sign = 1;
var n = t - 1;
var k = t - d;
while (k >= 0 && i <= d) {
result = result.add(
comb(Big(d), Big(i))
.times(comb(Big(n), Big(k)))
.times(sign)
);
i++;
n -= f;
k -= f;
sign *= -1;
}
return result;
}
console.log(
dice(30, 30, 500).mod(
Big(10)
.pow(9)
.add(7)
)
);
In python :
import math
def dice(d, f, t):
if t > d * f or t < d:
return 0
result = 0
i = 0
sign = 1
n = t - 1
k = t - d
while k >= 0 and i <= d:
result += math.comb(d, i) * math.comb(n, k) * sign
i += 1
n -= f
k -= f
sign *= -1
return result
print(dice(30, 30, 500) % (math.pow(10, 9) + 7))
Now when i run the code with these parameters : d=30 f=30 t=500 (the last line of each version of the code), i expect the result to be 222616187 .
In the node.js version , that's exactly what i get .
But in the python version , i'm getting 811448245.0 i can't figure out why is that happening .
So why is there a difference in the results ?

The math module usesfloat, not arbitrary precision int.
math - Mathematical functions
[...]
The following functions are provided by this module. Except when
explicitly noted otherwise, all return values are floats.
Since math.pow returns a float, the leading argument to % is converted to a float as well.
The result of dice(30, 30, 500) is too large to be accurately represented as a float. Its float representation is off by -14999044413600247749080617.
The ** operator and its function version operator.pow do not force float conversion and provide an integer if all parameters are integers.
>>> print(dice(30, 30, 500) % (10 ** 9 + 7))
222616187

Solved, in a weird way. It turns out, math.pow returns a float instead of int and somehow bugged. I think int % float has a different cast operation in it and treated differently by the compiler. It can be investigated further. If you cast it to int, that would be your answer.
import math
def dice(d, f, t):
if t > d * f or t < d:
return 0
result = 0
i = 0
sign = 1
n = t - 1
k = t - d
while k >= 0 and i <= d:
result += math.comb(d, i) * math.comb(n, k) * sign
i += 1
n -= f
k -= f
sign *= -1
return result
print(dice(30, 30, 500) % int((math.pow(10, 9) + 7)))

How to do popcount or count bits on arbitrarily long bit sequence in JavaScript

This looks like it works well on integers that are within the max integer size in JavaScript:
function bitCount (n) {
var bits = 0
while (n !== 0) {
bits += bitCount32(n | 0)
n /= 0x100000000
}
return bits
}
function bitCount32 (n) {
n = n - ((n >> 1) & 0x55555555)
n = (n & 0x33333333) + ((n >> 2) & 0x33333333)
return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24
}
I'm wondering though how to count bits generally, on any sized bit stream, efficiently, ideally without converting to string.

The comment with the lookup table, roughly:
var lookup=new Uint8Array(256);
for(var i=0;i<256;i++){
var c=0;
for(var j=i;j;j>>=1)
if(j&1)c++;
lookup[i]=c;
}
function count(arr){
var arr8=new Uint8Array(arr);
return arr8.reduce((a,e)=>a+lookup[e],0);
}
console.log(count(new Uint8Array([0x12,0x34,0x56,0x78,0x9A,0xBC,0xDE,0xF])));
//11 21 22 31 22 32 33 4 = 32
Of course the table-generation could use your magic too.

If you know your buffer will have length divisible by 4,
let array32 = new Uint32Array(buffer);
let numBits = array32.reduce((a, e) => a + bitCount32(e), 0);
Otherwise probably tevemadar's suggestion is better, use Uint8Array and count bits in a byte, not in a dword.

which are alternative of tofixed() in javascript [duplicate]

Suppose I have a value of 15.7784514, I want to display it 15.77 with no rounding.
var num = parseFloat(15.7784514);
document.write(num.toFixed(1)+"<br />");
document.write(num.toFixed(2)+"<br />");
document.write(num.toFixed(3)+"<br />");
document.write(num.toFixed(10));
Results in -
15.8
15.78
15.778
15.7784514000
How do I display 15.77?

Convert the number into a string, match the number up to the second decimal place:
function calc(theform) {
var num = theform.original.value, rounded = theform.rounded
var with2Decimals = num.toString().match(/^-?\d+(?:\.\d{0,2})?/)[0]
rounded.value = with2Decimals
}
<form onsubmit="return calc(this)">
Original number: <input name="original" type="text" onkeyup="calc(form)" onchange="calc(form)" />
<br />"Rounded" number: <input name="rounded" type="text" placeholder="readonly" readonly>
</form>
The toFixed method fails in some cases unlike toString, so be very careful with it.

Update 5 Nov 2016
New answer, always accurate
function toFixed(num, fixed) {
var re = new RegExp('^-?\\d+(?:\.\\d{0,' + (fixed || -1) + '})?');
return num.toString().match(re)[0];
}
As floating point math in javascript will always have edge cases, the previous solution will be accurate most of the time which is not good enough.
There are some solutions to this like num.toPrecision, BigDecimal.js, and accounting.js.
Yet, I believe that merely parsing the string will be the simplest and always accurate.
Basing the update on the well written regex from the accepted answer by #Gumbo, this new toFixed function will always work as expected.
Old answer, not always accurate.
Roll your own toFixed function:
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}

Another single-line solution :
number = Math.trunc(number*100)/100
I used 100 because you want to truncate to the second digit, but a more flexible solution would be :
number = Math.trunc(number*Math.pow(10, digits))/Math.pow(10, digits)
where digits is the amount of decimal digits to keep.
See Math.trunc specs for details and browser compatibility.

I opted to write this instead to manually remove the remainder with strings so I don't have to deal with the math issues that come with numbers:
num = num.toString(); //If it's not already a String
num = num.slice(0, (num.indexOf("."))+3); //With 3 exposing the hundredths place
Number(num); //If you need it back as a Number
This will give you "15.77" with num = 15.7784514;

Update (Jan 2021)
Depending on its range, a number in javascript may be shown in scientific notation. For example, if you type 0.0000001 in the console, you may see it as 1e-7, whereas 0.000001 appears unchanged (0.000001).
If your application works on a range of numbers for which scientific notation is not involved, you can just ignore this update and use the original answer below.
This update is about adding a function that checks if the number is in scientific format and, if so, converts it into decimal format. Here I'm proposing this one, but you can use any other function that achieves the same goal, according to your application's needs:
function toFixed(x) {
if (Math.abs(x) < 1.0) {
let e = parseInt(x.toString().split('e-')[1]);
if (e) {
x *= Math.pow(10,e-1);
x = '0.' + (new Array(e)).join('0') + x.toString().substring(2);
}
} else {
let e = parseInt(x.toString().split('+')[1]);
if (e > 20) {
e -= 20;
x /= Math.pow(10,e);
x += (new Array(e+1)).join('0');
}
}
return x;
}
Now just apply that function to the parameter (that's the only change with respect to the original answer):
function toFixedTrunc(x, n) {
x = toFixed(x)
// From here on the code is the same than the original answer
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
This updated version addresses also a case mentioned in a comment:
toFixedTrunc(0.000000199, 2) => "0.00"
Again, choose what fits your application needs at best.
Original answer (October 2017)
General solution to truncate (no rounding) a number to the n-th decimal digit and convert it to a string with exactly n decimal digits, for any n≥0.
function toFixedTrunc(x, n) {
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
where x can be either a number (which gets converted into a string) or a string.
Here are some tests for n=2 (including the one requested by OP):
0 => 0.00
0.01 => 0.01
0.5839 => 0.58
0.999 => 0.99
1.01 => 1.01
2 => 2.00
2.551 => 2.55
2.99999 => 2.99
4.27 => 4.27
15.7784514 => 15.77
123.5999 => 123.59
And for some other values of n:
15.001097 => 15.0010 (n=4)
0.000003298 => 0.0000032 (n=7)
0.000003298257899 => 0.000003298257 (n=12)

parseInt is faster then Math.floor
function floorFigure(figure, decimals){
if (!decimals) decimals = 2;
var d = Math.pow(10,decimals);
return (parseInt(figure*d)/d).toFixed(decimals);
};
floorFigure(123.5999) => "123.59"
floorFigure(123.5999, 3) => "123.599"

num = 19.66752
f = num.toFixed(3).slice(0,-1)
alert(f)
This will return 19.66

Simple do this
number = parseInt(number * 100)/100;

Just truncate the digits:
function truncDigits(inputNumber, digits) {
const fact = 10 ** digits;
return Math.floor(inputNumber * fact) / fact;
}

This is not a safe alternative, as many others commented examples with numbers that turn into exponential notation, that scenery is not covered by this function
// typescript
// function formatLimitDecimals(value: number, decimals: number): number {
function formatLimitDecimals(value, decimals) {
const stringValue = value.toString();
if(stringValue.includes('e')) {
// TODO: remove exponential notation
throw 'invald number';
} else {
const [integerPart, decimalPart] = stringValue.split('.');
if(decimalPart) {
return +[integerPart, decimalPart.slice(0, decimals)].join('.')
} else {
return integerPart;
}
}
}
console.log(formatLimitDecimals(4.156, 2)); // 4.15
console.log(formatLimitDecimals(4.156, 8)); // 4.156
console.log(formatLimitDecimals(4.156, 0)); // 4
console.log(formatLimitDecimals(0, 4)); // 0
// not covered
console.log(formatLimitDecimals(0.000000199, 2)); // 0.00

These solutions do work, but to me seem unnecessarily complicated. I personally like to use the modulus operator to obtain the remainder of a division operation, and remove that. Assuming that num = 15.7784514:
num-=num%.01;
This is equivalent to saying num = num - (num % .01).

I fixed using following simple way-
var num = 15.7784514;
Math.floor(num*100)/100;
Results will be 15.77

My version for positive numbers:
function toFixed_norounding(n,p)
{
var result = n.toFixed(p);
return result <= n ? result: (result - Math.pow(0.1,p)).toFixed(p);
}
Fast, pretty, obvious. (version for positive numbers)

The answers here didn't help me, it kept rounding up or giving me the wrong decimal.
my solution converts your decimal to a string, extracts the characters and then returns the whole thing as a number.
function Dec2(num) {
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length == 1) {
return Number(num);
}
else {
return Number(numarr[0]+"."+numarr[1].charAt(0)+numarr[1].charAt(1));
}
}
else {
return Number(num);
}
}
Dec2(99); // 99
Dec2(99.9999999); // 99.99
Dec2(99.35154); // 99.35
Dec2(99.8); // 99.8
Dec2(10265.985475); // 10265.98

The following code works very good for me:
num.toString().match(/.\*\\..{0,2}|.\*/)[0];

This worked well for me. I hope it will fix your issues too.
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
// 5.56789 ----> $5.56
// 0.342 ----> $0.34
// -10.3484534 ----> $-10.34
// 600 ----> $600.00
function convertNumber(){
var result = toFixedNumber(document.getElementById("valueText").value);
document.getElementById("resultText").value = result;
}
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
<div>
<input type="text" id="valueText" placeholder="Input value here..">
<br>
<button onclick="convertNumber()" >Convert</button>
<br><hr>
<input type="text" id="resultText" placeholder="result" readonly="true">
</div>

An Easy way to do it is the next but is necessary ensure that the amount parameter is given as a string.
function truncate(amountAsString, decimals = 2){
var dotIndex = amountAsString.indexOf('.');
var toTruncate = dotIndex !== -1 && ( amountAsString.length > dotIndex + decimals + 1);
var approach = Math.pow(10, decimals);
var amountToTruncate = toTruncate ? amountAsString.slice(0, dotIndex + decimals +1) : amountAsString;
return toTruncate
? Math.floor(parseFloat(amountToTruncate) * approach ) / approach
: parseFloat(amountAsString);
}
console.log(truncate("7.99999")); //OUTPUT ==> 7.99
console.log(truncate("7.99999", 3)); //OUTPUT ==> 7.999
console.log(truncate("12.799999999999999")); //OUTPUT ==> 7.99

Here you are. An answer that shows yet another way to solve the problem:
// For the sake of simplicity, here is a complete function:
function truncate(numToBeTruncated, numOfDecimals) {
var theNumber = numToBeTruncated.toString();
var pointIndex = theNumber.indexOf('.');
return +(theNumber.slice(0, pointIndex > -1 ? ++numOfDecimals + pointIndex : undefined));
}
Note the use of + before the final expression. That is to convert our truncated, sliced string back to number type.
Hope it helps!

truncate without zeroes
function toTrunc(value,n){
return Math.floor(value*Math.pow(10,n))/(Math.pow(10,n));
}
or
function toTrunc(value,n){
x=(value.toString()+".0").split(".");
return parseFloat(x[0]+"."+x[1].substr(0,n));
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.4
truncate with zeroes
function toTruncFixed(value,n){
return toTrunc(value,n).toFixed(n);
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.40

If you exactly wanted to truncate to 2 digits of precision, you can go with a simple logic:
function myFunction(number) {
var roundedNumber = number.toFixed(2);
if (roundedNumber > number)
{
roundedNumber = roundedNumber - 0.01;
}
return roundedNumber;
}

I used (num-0.05).toFixed(1) to get the second decimal floored.

It's more reliable to get two floating points without rounding.
Reference Answer
var number = 10.5859;
var fixed2FloatPoints = parseInt(number * 100) / 100;
console.log(fixed2FloatPoints);
Thank You !

My solution in typescript (can easily be ported to JS):
/**
* Returns the price with correct precision as a string
*
* #param price The price in decimal to be formatted.
* #param decimalPlaces The number of decimal places to use
* #return string The price in Decimal formatting.
*/
type toDecimal = (price: number, decimalPlaces?: number) => string;
const toDecimalOdds: toDecimal = (
price: number,
decimalPlaces: number = 2,
): string => {
const priceString: string = price.toString();
const pointIndex: number = priceString.indexOf('.');
// Return the integer part if decimalPlaces is 0
if (decimalPlaces === 0) {
return priceString.substr(0, pointIndex);
}
// Return value with 0s appended after decimal if the price is an integer
if (pointIndex === -1) {
const padZeroString: string = '0'.repeat(decimalPlaces);
return `${priceString}.${padZeroString}`;
}
// If numbers after decimal are less than decimalPlaces, append with 0s
const padZeroLen: number = priceString.length - pointIndex - 1;
if (padZeroLen > 0 && padZeroLen < decimalPlaces) {
const padZeroString: string = '0'.repeat(padZeroLen);
return `${priceString}${padZeroString}`;
}
return priceString.substr(0, pointIndex + decimalPlaces + 1);
};
Test cases:
expect(filters.toDecimalOdds(3.14159)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 2)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 0)).toBe('3');
expect(filters.toDecimalOdds(3.14159, 10)).toBe('3.1415900000');
expect(filters.toDecimalOdds(8.2)).toBe('8.20');
Any improvements?

Another solution, that truncates and round:
function round (number, decimals, truncate) {
if (truncate) {
number = number.toFixed(decimals + 1);
return parseFloat(number.slice(0, -1));
}
var n = Math.pow(10.0, decimals);
return Math.round(number * n) / n;
};

function limitDecimalsWithoutRounding(val, decimals){
let parts = val.toString().split(".");
return parseFloat(parts[0] + "." + parts[1].substring(0, decimals));
}
var num = parseFloat(15.7784514);
var new_num = limitDecimalsWithoutRounding(num, 2);

Roll your own toFixed function: for positive values Math.floor works fine.
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
For negative values Math.floor is round of the values. So you can use Math.ceil instead.
Example,
Math.ceil(-15.778665 * 10000) / 10000 = -15.7786
Math.floor(-15.778665 * 10000) / 10000 = -15.7787 // wrong.

Gumbo's second solution, with the regular expression, does work but is slow because of the regular expression. Gumbo's first solution fails in certain situations due to imprecision in floating points numbers. See the JSFiddle for a demonstration and a benchmark. The second solution takes about 1636 nanoseconds per call on my current system, Intel Core i5-2500 CPU at 3.30 GHz.
The solution I've written involves adding a small compensation to take care of floating point imprecision. It is basically instantaneous, i.e. on the order of nanoseconds. I clocked 2 nanoseconds per call but the JavaScript timers are not very precise or granular. Here is the JS Fiddle and the code.
function toFixedWithoutRounding (value, precision)
{
var factorError = Math.pow(10, 14);
var factorTruncate = Math.pow(10, 14 - precision);
var factorDecimal = Math.pow(10, precision);
return Math.floor(Math.floor(value * factorError + 1) / factorTruncate) / factorDecimal;
}
var values = [1.1299999999, 1.13, 1.139999999, 1.14, 1.14000000001, 1.13 * 100];
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 2));
}
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 4));
}
console.log("type of result is " + typeof toFixedWithoutRounding(1.13 * 100 / 100, 2));
// Benchmark
var value = 1.13 * 100;
var startTime = new Date();
var numRun = 1000000;
var nanosecondsPerMilliseconds = 1000000;
for (var run = 0; run < numRun; run++)
toFixedWithoutRounding(value, 2);
var endTime = new Date();
var timeDiffNs = nanosecondsPerMilliseconds * (endTime - startTime);
var timePerCallNs = timeDiffNs / numRun;
console.log("Time per call (nanoseconds): " + timePerCallNs);

Building on David D's answer:
function NumberFormat(num,n) {
var num = (arguments[0] != null) ? arguments[0] : 0;
var n = (arguments[1] != null) ? arguments[1] : 2;
if(num > 0){
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length > 1) {
if(n > 0){
var temp = numarr[0] + ".";
for(var i = 0; i < n; i++){
if(i < numarr[1].length){
temp += numarr[1].charAt(i);
}
}
num = Number(temp);
}
}
}
}
return Number(num);
}
console.log('NumberFormat(123.85,2)',NumberFormat(123.85,2));
console.log('NumberFormat(123.851,2)',NumberFormat(123.851,2));
console.log('NumberFormat(0.85,2)',NumberFormat(0.85,2));
console.log('NumberFormat(0.851,2)',NumberFormat(0.851,2));
console.log('NumberFormat(0.85156,2)',NumberFormat(0.85156,2));
console.log('NumberFormat(0.85156,4)',NumberFormat(0.85156,4));
console.log('NumberFormat(0.85156,8)',NumberFormat(0.85156,8));
console.log('NumberFormat(".85156",2)',NumberFormat(".85156",2));
console.log('NumberFormat("0.85156",2)',NumberFormat("0.85156",2));
console.log('NumberFormat("1005.85156",2)',NumberFormat("1005.85156",2));
console.log('NumberFormat("0",2)',NumberFormat("0",2));
console.log('NumberFormat("",2)',NumberFormat("",2));
console.log('NumberFormat(85156,8)',NumberFormat(85156,8));
console.log('NumberFormat("85156",2)',NumberFormat("85156",2));
console.log('NumberFormat("85156.",2)',NumberFormat("85156.",2));
// NumberFormat(123.85,2) 123.85
// NumberFormat(123.851,2) 123.85
// NumberFormat(0.85,2) 0.85
// NumberFormat(0.851,2) 0.85
// NumberFormat(0.85156,2) 0.85
// NumberFormat(0.85156,4) 0.8515
// NumberFormat(0.85156,8) 0.85156
// NumberFormat(".85156",2) 0.85
// NumberFormat("0.85156",2) 0.85
// NumberFormat("1005.85156",2) 1005.85
// NumberFormat("0",2) 0
// NumberFormat("",2) 0
// NumberFormat(85156,8) 85156
// NumberFormat("85156",2) 85156
// NumberFormat("85156.",2) 85156

Already there are some suitable answer with regular expression and arithmetic calculation, you can also try this
function myFunction() {
var str = 12.234556;
str = str.toString().split('.');
var res = str[1].slice(0, 2);
document.getElementById("demo").innerHTML = str[0]+'.'+res;
}
// output: 12.23

Here is what is did it with string
export function withoutRange(number) {
const str = String(number);
const dotPosition = str.indexOf('.');
if (dotPosition > 0) {
const length = str.substring().length;
const end = length > 3 ? 3 : length;
return str.substring(0, dotPosition + end);
}
return str;
}

Javascript intelligent rounding

I currently need to round numbers up to their nearest major number. (Not sure what the right term is here)
But see an example of what I'm trying to achieve
IE:
13 // 20
349 // 400
5645 // 6000
9892 // 10000
13988 // 20000
93456 // 100000
231516 // 300000
etc. etc.
I have implemented a way of doing this but its so painful and only handles numbers up to a million and if I want it to go higher I need to add more if statements (yeah see how i implmented it :P im not very proud, but brain is stuck)
There must be something out there already but google is not helping me very much probably due to me not knowing the correct term for the kind of rounding i want to do

<script type="text/javascript">
function intelliRound(num) {
var len=(num+'').length;
var fac=Math.pow(10,len-1);
return Math.ceil(num/fac)*fac;
}
alert(intelliRound(13));
alert(intelliRound(349));
alert(intelliRound(5645));
// ...
</script>
See http://jsfiddle.net/fCLjp/

One way;
var a = [13, // 20
349, // 400
5645, // 6000
9892, // 10000
13988, // 20000
93456, // 100000
231516 // 300000
]
for (var i in a) {
var num = a[i];
var scale = Math.pow(10, Math.floor(Math.log(num) / Math.LN10));
print([ num, Math.ceil(num / scale) * scale ])
}
13,20
349,400
5645,6000
9892,10000
13988,20000
93456,100000
231516,300000

The answer from #rabudde works well, but for those that need to handle negative numbers, here's an updated version:
function intelliRound(num) {
var len = (num + '').length;
var result = 0;
if (num < 0) {
var fac = Math.pow(10, len - 2);
result = Math.floor(num / fac) * fac;
}
else {
var fac = Math.pow(10, len - 1);
result = Math.ceil(num / fac) * fac;
}
return result;
}
alert(intelliRound(13));
alert(intelliRound(349));
alert(intelliRound(5645));
alert(intelliRound(-13));
alert(intelliRound(-349));
alert(intelliRound(-5645));

you can use Math.ceil function, as described here:
javascript - ceiling of a dollar amount
to get your numbers right you'll have to divide them by 10 (if they have 2 digits), 100 (if they have 3 digits), and so on...

The intelliRound function from the other answers works well, but break with negative numbers. Here I have extended these solutions to support decimals (e.g. 0.123, -0.987) and non-numbers:
/**
* Function that returns the floor/ceil of a number, to an appropriate magnitude
* #param {number} num - the number you want to round
*
* e.g.
* magnitudeRound(0.13) => 1
* magnitudeRound(13) => 20
* magnitudeRound(349) => 400
* magnitudeRound(9645) => 10000
* magnitudeRound(-3645) => -4000
* magnitudeRound(-149) => -200
*/
function magnitudeRound(num) {
const isValidNumber = typeof num === 'number' && !Number.isNaN(num);
const result = 0;
if (!isValidNumber || num === 0) return result;
const abs = Math.abs(num);
const sign = Math.sign(num);
if (abs > 0 && abs <= 1) return 1 * sign; // percentages on a scale -1 to 1
if (abs > 1 && abs <= 10) return 10 * sign;
const zeroes = `${Math.round(abs)}`.length - 1; // e.g 123 => 2, 4567 => 3
const exponent = 10 ** zeroes; // math floor and ceil only work on integer
const roundingDirection = sign < 0 ? 'floor' : 'ceil';
return Math[roundingDirection](num / exponent) * exponent;
}