I got asked a question and now I am kicking myself for not being able to come up with the exact/correct result.
Imagine we have a function that splits a string into multiple lines but each line has to have x number of characters before we "split" to the new line:
private string[] GetPagedMessages(string input, int maxCharsPerLine) { ... }
For each line, we need to incorporate, at the end of the line "x/y" which is basically 1/4, 2/4 etc...
Now, the paging mechanism must also be part of the length restriction per line.
I have been overworked and overthinking and tripping up on things and this seems pretty straight forward but for the life of me, I cannot figure it out! What am I not "getting"?
What am I interested in? The calculation and some part of the logic but mainly the calculation of how many lines are required to split the input based on the max chars per line which also needs to include the x/y.
Remember: we can have more than a single digit for the x/y (i.e: not just 1/4 but also 10/17 or 99/200)
Samples:
input = "This is a long message"
maxCharsPerLine = 10
output:
This i 1/4 // << Max 10 chars
s a lo 2/4 // << Max 10 chars
ng mes 3/4 // << Max 10 chars
sage 4/4 // << Max 10 chars
Overall the logic is simple but its just the calculation that is throwing me off.
The idea: First, find how many digits is the number of lines:
(n = input.Length, maxCharsPerLine = 10)
if n <= 9*(10-4) ==> 1 digit
if n <= 9*(10-5) + 90*(10-6) ==> 2 digits
if n <= 9*(10-6) + 90*(10-7) + 900*(10-8) ==> 3 digits
if n <= 9*(10-7) + 90*(10-8) + 900*(10-9) + 9000*(10-10) ==> No solution
Then, subtract the spare number of lines. The solution:
private static int GetNumberOfLines(string input, int maxCharsPerLine)
{
int n = input.Length;
int x = maxCharsPerLine;
for (int i = 4; i < x; i++)
{
int j, sum = 0, d = 9, numberOfLines = 0;
for (j = i; j <= i + i - 4; j++)
{
if (x - j <= 0)
return -1; // No solution
sum += d * (x - j);
numberOfLines += d;
d *= 10;
}
if (n <= sum)
return numberOfLines - (sum - n) / (x - j + 1);
}
return -2; // Invalid
}
Usage:
private static string[] GetPagedMessages(string input, int maxCharsPerLine)
{
int numberOfLines = GetNumberOfLines(input, maxCharsPerLine);
if (numberOfLines < 0)
return null;
string[] result = new string[numberOfLines];
int spaceLeftForLine = maxCharsPerLine - numberOfLines.ToString().Length - 2; // Remove the chars of " x/y" except the incremental 'x'
int inputPosition = 0;
for (int line = 1; line < numberOfLines; line++)
{
int charsInLine = spaceLeftForLine - line.ToString().Length;
result[line - 1] = input.Substring(inputPosition, charsInLine) + $" {line}/{numberOfLines}";
inputPosition += charsInLine;
}
result[numberOfLines-1] = input.Substring(inputPosition) + $" {numberOfLines}/{numberOfLines}";
return result;
}
A naive approach is to start counting the line lengths minus the "pager"'s size, until the line count changes in size ("1/9" is shorter than "1/10", which is shorter than "11/20", and so on):
private static int[] GetLineLengths(string input, int maxCharsPerLine)
{
/* The "pager" (x/y) is at least 4 characters (including the preceding space) and at most ... 8?
* 7/9 (4)
* 1/10 (5)
* 42/69 (6)
* 3/123 (6)
* 42/420 (7)
* 999/999 (8)
*/
int charsRemaining = input.Length;
var lineLengths = new List<int>();
// Start with " 1/2", (1 + 1 + 2) = 4 length
var highestLineNumberLength = 1;
var lineNumber = 0;
do
{
lineNumber++;
var currentLineNumberLength = lineNumber.ToString().Length; // 1 = 1, 99 = 2, ...
if (currentLineNumberLength > highestLineNumberLength)
{
// Pager size changed, reset
highestLineNumberLength = currentLineNumberLength;
lineLengths.Clear();
lineNumber = 0;
charsRemaining = input.Length;
continue;
}
var pagerSize = currentLineNumberLength + highestLineNumberLength + 2;
var lineLength = maxCharsPerLine - pagerSize;
if (lineLength <= 0)
{
throw new ArgumentException($"Can't split input of size {input.Length} into chunks of size {maxCharsPerLine}");
}
lineLengths.Add(lineLength);
charsRemaining -= lineLength;
}
while (charsRemaining > 0);
return lineLengths.ToArray();
}
Usage:
private static string[] GetPagedMessages(string input, int maxCharsPerLine)
{
if (input.Length <= maxCharsPerLine)
{
// Assumption: no pager required for a message that takes one line
return new[] { input };
}
var lineLengths = GetLineLengths(input, maxCharsPerLine);
var result = new string[lineLengths.Length];
// Cut the input and append the pager
var previousIndex = 0;
for (var i = 0; i < lineLengths.Length; i++)
{
var lineLength = Math.Min(lineLengths[i], input.Length - previousIndex); // To cater for final line being shorter
result[i] = input.Substring(previousIndex, lineLength) + " " + (i + 1) + "/" + lineLengths.Length;
previousIndex += lineLength;
}
return result;
}
Prints, for example:
This 1/20
is a 2/20
long 3/20
strin 4/20
g tha 5/20
t wil 6/20
l spa 7/20
n mor 8/20
e tha 9/20
n te 10/20
n li 11/20
nes 12/20
beca 13/20
use 14/20
of i 15/20
ts e 16/20
norm 17/20
ous 18/20
leng 19/20
th 20/20
My codes are like below.
I just want to try to convert number to binary and summarize binary numbers as if they are decimal ones, e.g. the desired outcome for 3 is 22:
1 -> 1
2 -> 10
3 -> 11
-------
22 == 1 + 10 + 11
But number array is growing and the code blowing :)
static void Main(string[] args)
{
long deger = 1;
for (int i = 0; i < 1000000; i++)
{
deger *= (deger + 1);
int result = solve(deger);
Console.WriteLine(result);
}
}
public static int solve(long a)
{
ulong[] lastValue = new ulong[a];
for (int i = 1; i < a; i++)
{
var binary = Convert.ToString(i, 2);
lastValue[i] = Convert.ToUInt64(binary);// this part gives an error
}
var result = lastValue.Aggregate((t, c) => t + c);
return (int)result;
}
Well, UInt64 is not large enough; you can try either BigInteger or you may sum up strings:
private string MyBinarySum(string left, string right) {
var x = left
.PadLeft(Math.Max(left.Length, right.Length) + 1, '0')
.Reverse()
.Select(c => c - '0')
.ToArray();
var y = right
.PadLeft(Math.Max(left.Length, right.Length) + 1, '0')
.Reverse().Select(c => c - '0')
.ToArray();
StringBuilder sb = new StringBuilder(left.Length);
int shift = 0;
for (int i = 0; i < x.Length; ++i) {
int v = x[i] + y[i] + shift;
shift = v / 2;
v = v % 2;
sb.Append((char)('0' + v));
}
return String.Concat(sb.ToString().TrimEnd('0').Reverse());
}
Then, with a help of Linq
var result = Enumerable
.Range(0, 1000000)
.Select(item => Convert.ToString(item, 2))
.Aggregate((sum, item) => MyBinarySum(sum, item));
Console.Write(result);
Outcome:
111010001101010010010101110011011100000
which is beyond UInt64.MaxValue == 18446744073709551615
Look at UInt64.MaxValue and UInt64.MinValue.
They defined as 18446744073709551615 and 0 corresponding.
Is there a built in function that would allow me to calculate the modular inverse of a(mod n)?
e.g. 19^-1 = 11 (mod 30), in this case the 19^-1 == -11==19;
Since .Net 4.0+ implements BigInteger with a special modular arithmetics function ModPow (which produces “X power Y modulo Z”), you don't need a third-party library to emulate ModInverse. If n is a prime, all you need to do is to compute:
a_inverse = BigInteger.ModPow(a, n - 2, n)
For more details, look in Wikipedia: Modular multiplicative inverse, section Using Euler's theorem, the special case “when m is a prime”. By the way, there is a more recent SO topic on this: 1/BigInteger in c#, with the same approach suggested by CodesInChaos.
int modInverse(int a, int n)
{
int i = n, v = 0, d = 1;
while (a>0) {
int t = i/a, x = a;
a = i % x;
i = x;
x = d;
d = v - t*x;
v = x;
}
v %= n;
if (v<0) v = (v+n)%n;
return v;
}
The BouncyCastle Crypto library has a BigInteger implementation that has most of the modular arithmetic functions. It's in the Org.BouncyCastle.Math namespace.
Here is a slightly more polished version of Samuel Allan's algorithm. The TryModInverse method returns a bool value, that indicates whether a modular multiplicative inverse exists for this number and modulo.
public static bool TryModInverse(int number, int modulo, out int result)
{
if (number < 1) throw new ArgumentOutOfRangeException(nameof(number));
if (modulo < 2) throw new ArgumentOutOfRangeException(nameof(modulo));
int n = number;
int m = modulo, v = 0, d = 1;
while (n > 0)
{
int t = m / n, x = n;
n = m % x;
m = x;
x = d;
d = checked(v - t * x); // Just in case
v = x;
}
result = v % modulo;
if (result < 0) result += modulo;
if ((long)number * result % modulo == 1L) return true;
result = default;
return false;
}
There is no library for getting inverse mod, but the following code can be used to get it.
// Given a and b->ax+by=d
long[] u = { a, 1, 0 };
long[] v = { b, 0, 1 };
long[] w = { 0, 0, 0 };
long temp = 0;
while (v[0] > 0)
{
double t = (u[0] / v[0]);
for (int i = 0; i < 3; i++)
{
w[i] = u[i] - ((int)(Math.Floor(t)) * v[i]);
u[i] = v[i];
v[i] = w[i];
}
}
// u[0] is gcd while u[1] gives x and u[2] gives y.
// if u[1] gives the inverse mod value and if it is negative then the following gives the first positive value
if (u[1] < 0)
{
while (u[1] < 0)
{
temp = u[1] + b;
u[1] = temp;
}
}
In C#, what's the best way to get the 1st digit in an int? The method I came up with is to turn the int into a string, find the 1st char of the string, then turn it back to an int.
int start = Convert.ToInt32(curr.ToString().Substring(0, 1));
While this does the job, it feels like there is probably a good, simple, math-based solution to such a problem. String manipulation feels clunky.
Edit: irrespective of speed differences, mystring[0] instead of Substring() is still just string manipulation
Benchmarks
Firstly, you must decide on what you mean by "best" solution, of course that takes into account the efficiency of the algorithm, its readability/maintainability, and the likelihood of bugs creeping up in the future. Careful unit tests can generally avoid those problems, however.
I ran each of these examples 10 million times, and the results value is the number of ElapsedTicks that have passed.
Without further ado, from slowest to quickest, the algorithms are:
Converting to a string, take first character
int firstDigit = (int)(Value.ToString()[0]) - 48;
Results:
12,552,893 ticks
Using a logarithm
int firstDigit = (int)(Value / Math.Pow(10, (int)Math.Floor(Math.Log10(Value))));
Results:
9,165,089 ticks
Looping
while (number >= 10)
number /= 10;
Results:
6,001,570 ticks
Conditionals
int firstdigit;
if (Value < 10)
firstdigit = Value;
else if (Value < 100)
firstdigit = Value / 10;
else if (Value < 1000)
firstdigit = Value / 100;
else if (Value < 10000)
firstdigit = Value / 1000;
else if (Value < 100000)
firstdigit = Value / 10000;
else if (Value < 1000000)
firstdigit = Value / 100000;
else if (Value < 10000000)
firstdigit = Value / 1000000;
else if (Value < 100000000)
firstdigit = Value / 10000000;
else if (Value < 1000000000)
firstdigit = Value / 100000000;
else
firstdigit = Value / 1000000000;
Results:
1,421,659 ticks
Unrolled & optimized loop
if (i >= 100000000) i /= 100000000;
if (i >= 10000) i /= 10000;
if (i >= 100) i /= 100;
if (i >= 10) i /= 10;
Results:
1,399,788 ticks
Note:
each test calls Random.Next() to get the next int
Here's how
int i = Math.Abs(386792);
while(i >= 10)
i /= 10;
and i will contain what you need
Try this
public int GetFirstDigit(int number) {
if ( number < 10 ) {
return number;
}
return GetFirstDigit ( (number - (number % 10)) / 10);
}
EDIT
Several people have requested the loop version
public static int GetFirstDigitLoop(int number)
{
while (number >= 10)
{
number = (number - (number % 10)) / 10;
}
return number;
}
The best I can come up with is:
int numberOfDigits = Convert.ToInt32(Math.Floor( Math.Log10( value ) ) );
int firstDigit = value / Math.Pow( 10, numberOfDigits );
variation on Anton's answer:
// cut down the number of divisions (assuming i is positive & 32 bits)
if (i >= 100000000) i /= 100000000;
if (i >= 10000) i /= 10000;
if (i >= 100) i /= 100;
if (i >= 10) i /= 10;
int myNumber = 8383;
char firstDigit = myNumber.ToString()[0];
// char = '8'
Had the same idea as Lennaert
int start = number == 0 ? 0 : number / (int) Math.Pow(10,Math.Floor(Math.Log10(Math.Abs(number))));
This also works with negative numbers.
If you think Keltex's answer is ugly, try this one, it's REALLY ugly, and even faster.
It does unrolled binary search to determine the length.
... leading code along the same lines
/* i<10000 */
if (i >= 100){
if (i >= 1000){
return i/1000;
}
else /* i<1000 */{
return i/100;
}
}
else /* i<100*/ {
if (i >= 10){
return i/10;
}
else /* i<10 */{
return i;
}
}
P.S. MartinStettner had the same idea.
Very simple (and probably quite fast because it only involves comparisons and one division):
if(i<10)
firstdigit = i;
else if (i<100)
firstdigit = i/10;
else if (i<1000)
firstdigit = i/100;
else if (i<10000)
firstdigit = i/1000;
else if (i<100000)
firstdigit = i/10000;
else (etc... all the way up to 1000000000)
An obvious, but slow, mathematical approach is:
int firstDigit = (int)(i / Math.Pow(10, (int)Math.Log10(i))));
int temp = i;
while (temp >= 10)
{
temp /= 10;
}
Result in temp
I know it's not C#, but it's surprising curious that in python the "get the first char of the string representation of the number" is the faster!
EDIT: no, I made a mistake, I forgot to construct again the int, sorry. The unrolled version it's the fastest.
$ cat first_digit.py
def loop(n):
while n >= 10:
n /= 10
return n
def unrolled(n):
while n >= 100000000: # yea... unlimited size int supported :)
n /= 100000000
if n >= 10000:
n /= 10000
if n >= 100:
n /= 100
if n >= 10:
n /= 10
return n
def string(n):
return int(str(n)[0])
$ python -mtimeit -s 'from first_digit import loop as test' \
'for n in xrange(0, 100000000, 1000): test(n)'
10 loops, best of 3: 275 msec per loop
$ python -mtimeit -s 'from first_digit import unrolled as test' \
'for n in xrange(0, 100000000, 1000): test(n)'
10 loops, best of 3: 149 msec per loop
$ python -mtimeit -s 'from first_digit import string as test' \
'for n in xrange(0, 100000000, 1000): test(n)'
10 loops, best of 3: 284 msec per loop
$
I just stumbled upon this old question and felt inclined to propose another suggestion since none of the other answers so far returns the correct result for all possible input values and it can still be made faster:
public static int GetFirstDigit( int i )
{
if( i < 0 && ( i = -i ) < 0 ) return 2;
return ( i < 100 ) ? ( i < 1 ) ? 0 : ( i < 10 )
? i : i / 10 : ( i < 1000000 ) ? ( i < 10000 )
? ( i < 1000 ) ? i / 100 : i / 1000 : ( i < 100000 )
? i / 10000 : i / 100000 : ( i < 100000000 )
? ( i < 10000000 ) ? i / 1000000 : i / 10000000
: ( i < 1000000000 ) ? i / 100000000 : i / 1000000000;
}
This works for all signed integer values inclusive -2147483648 which is the smallest signed integer and doesn't have a positive counterpart. Math.Abs( -2147483648 ) triggers a System.OverflowException and - -2147483648 computes to -2147483648.
The implementation can be seen as a combination of the advantages of the two fastest implementations so far. It uses a binary search and avoids superfluous divisions. A quick benchmark with the index of a loop with 100,000,000 iterations shows that it is twice as fast as the currently fastest implementation.
It finishes after 2,829,581 ticks.
For comparison I also measured a corrected variant of the currently fastest implementation which took 5,664,627 ticks.
public static int GetFirstDigitX( int i )
{
if( i < 0 && ( i = -i ) < 0 ) return 2;
if( i >= 100000000 ) i /= 100000000;
if( i >= 10000 ) i /= 10000;
if( i >= 100 ) i /= 100;
if( i >= 10 ) i /= 10;
return i;
}
The accepted answer with the same correction needed 16,561,929 ticks for this test on my computer.
public static int GetFirstDigitY( int i )
{
if( i < 0 && ( i = -i ) < 0 ) return 2;
while( i >= 10 )
i /= 10;
return i;
}
Simple functions like these can easily be proven for correctness since iterating all possible integer values takes not much more than a few seconds on current hardware. This means that it is less important to implement them in a exceptionally readable fashion as there simply won't ever be the need to fix a bug inside them later on.
Did some tests with one of my co-workers here, and found out most of the solutions don't work for numbers under 0.
public int GetFirstDigit(int number)
{
number = Math.Abs(number); <- makes sure you really get the digit!
if (number < 10)
{
return number;
}
return GetFirstDigit((number - (number % 10)) / 10);
}
Using all the examples below to get this code:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
namespace Benfords
{
class Program
{
static int FirstDigit1(int value)
{
return Convert.ToInt32(value.ToString().Substring(0, 1));
}
static int FirstDigit2(int value)
{
while (value >= 10) value /= 10;
return value;
}
static int FirstDigit3(int value)
{
return (int)(value.ToString()[0]) - 48;
}
static int FirstDigit4(int value)
{
return (int)(value / Math.Pow(10, (int)Math.Floor(Math.Log10(value))));
}
static int FirstDigit5(int value)
{
if (value < 10) return value;
if (value < 100) return value / 10;
if (value < 1000) return value / 100;
if (value < 10000) return value / 1000;
if (value < 100000) return value / 10000;
if (value < 1000000) return value / 100000;
if (value < 10000000) return value / 1000000;
if (value < 100000000) return value / 10000000;
if (value < 1000000000) return value / 100000000;
return value / 1000000000;
}
static int FirstDigit6(int value)
{
if (value >= 100000000) value /= 100000000;
if (value >= 10000) value /= 10000;
if (value >= 100) value /= 100;
if (value >= 10) value /= 10;
return value;
}
const int mcTests = 1000000;
static void Main(string[] args)
{
Stopwatch lswWatch = new Stopwatch();
Random lrRandom = new Random();
int liCounter;
lswWatch.Start();
for (liCounter = 0; liCounter < mcTests; liCounter++)
FirstDigit1(lrRandom.Next());
lswWatch.Stop();
Console.WriteLine("Test {0} = {1} ticks", 1, lswWatch.ElapsedTicks);
lswWatch.Reset();
lswWatch.Start();
for (liCounter = 0; liCounter < mcTests; liCounter++)
FirstDigit2(lrRandom.Next());
lswWatch.Stop();
Console.WriteLine("Test {0} = {1} ticks", 2, lswWatch.ElapsedTicks);
lswWatch.Reset();
lswWatch.Start();
for (liCounter = 0; liCounter < mcTests; liCounter++)
FirstDigit3(lrRandom.Next());
lswWatch.Stop();
Console.WriteLine("Test {0} = {1} ticks", 3, lswWatch.ElapsedTicks);
lswWatch.Reset();
lswWatch.Start();
for (liCounter = 0; liCounter < mcTests; liCounter++)
FirstDigit4(lrRandom.Next());
lswWatch.Stop();
Console.WriteLine("Test {0} = {1} ticks", 4, lswWatch.ElapsedTicks);
lswWatch.Reset();
lswWatch.Start();
for (liCounter = 0; liCounter < mcTests; liCounter++)
FirstDigit5(lrRandom.Next());
lswWatch.Stop();
Console.WriteLine("Test {0} = {1} ticks", 5, lswWatch.ElapsedTicks);
lswWatch.Reset();
lswWatch.Start();
for (liCounter = 0; liCounter < mcTests; liCounter++)
FirstDigit6(lrRandom.Next());
lswWatch.Stop();
Console.WriteLine("Test {0} = {1} ticks", 6, lswWatch.ElapsedTicks);
Console.ReadLine();
}
}
}
I get these results on an AMD Ahtlon 64 X2 Dual Core 4200+ (2.2 GHz):
Test 1 = 2352048 ticks
Test 2 = 614550 ticks
Test 3 = 1354784 ticks
Test 4 = 844519 ticks
Test 5 = 150021 ticks
Test 6 = 192303 ticks
But get these on a AMD FX 8350 Eight Core (4.00 GHz)
Test 1 = 3917354 ticks
Test 2 = 811727 ticks
Test 3 = 2187388 ticks
Test 4 = 1790292 ticks
Test 5 = 241150 ticks
Test 6 = 227738 ticks
So whether or not method 5 or 6 is faster depends on the CPU, I can only surmise this is because the branch prediction in the command processor of the CPU is smarter on the new processor, but I'm not really sure.
I dont have any Intel CPUs, maybe someone could test it for us?
Check this one too:
int get1digit(Int64 myVal)
{
string q12 = myVal.ToString()[0].ToString();
int i = int.Parse(q12);
return i;
}
Also good if you want multiple numbers:
int get3digit(Int64 myVal) //Int64 or whatever numerical data you have
{
char mg1 = myVal.ToString()[0];
char mg2 = myVal.ToString()[1];
char mg3 = myVal.ToString()[2];
char[] chars = { mg1, mg2, mg3 };
string q12= new string(chars);
int i = int.Parse(q12);
return i;
}
while (i > 10)
{
i = (Int32)Math.Floor((Decimal)i / 10);
}
// i is now the first int
Non iterative formula:
public static int GetHighestDigit(int num)
{
if (num <= 0)
return 0;
return (int)((double)num / Math.Pow(10f, Math.Floor(Math.Log10(num))));
}
Just to give you an alternative, you could repeatedly divide the integer by 10, and then rollback one value once you reach zero. Since string operations are generally slow, this may be faster than string manipulation, but is by no means elegant.
Something like this:
while(curr>=10)
curr /= 10;
start = getFirstDigit(start);
public int getFirstDigit(final int start){
int number = Math.abs(start);
while(number > 10){
number /= 10;
}
return number;
}
or
public int getFirstDigit(final int start){
return getFirstDigit(Math.abs(start), true);
}
private int getFirstDigit(final int start, final boolean recurse){
if(start < 10){
return start;
}
return getFirstDigit(start / 10, recurse);
}
int start = curr;
while (start >= 10)
start /= 10;
This is more efficient than a ToString() approach which internally must implement a similar loop and has to construct (and parse) a string object on the way ...
Very easy method to get the Last digit:
int myInt = 1821;
int lastDigit = myInt - ((myInt/10)*10); // 1821 - 1820 = 1
int i = 4567789;
int digit1 = int.Parse(i.ToString()[0].ToString());
This is what I usually do ,please refer my function below :
This function can extract first number occurance from any string you can modify and use this function according to your usage
public static int GetFirstNumber(this string strInsput)
{
int number = 0;
string strNumber = "";
bool bIsContNo = true;
bool bNoOccued = false;
try
{
var arry = strInsput.ToCharArray(0, strInsput.Length - 1);
foreach (char item in arry)
{
if (char.IsNumber(item))
{
strNumber = strNumber + item.ToString();
bIsContNo = true;
bNoOccued = true;
}
else
{
bIsContNo = false;
}
if (bNoOccued && !bIsContNo)
{
break;
}
}
number = Convert.ToInt32(strNumber);
}
catch (Exception ex)
{
return 0;
}
return number;
}
public static int GetFirstDigit(int n, bool removeSign = true)
{
if (removeSign)
return n <= -10 || n >= 10 ? Math.Abs(n) % 10 : Math.Abs(n);
else
return n <= -10 || n >= 10 ? n % 10 : n;
}
//Your code goes here
int[] test = new int[] { -1574, -221, 1246, -4, 8, 38546};
foreach(int n in test)
Console.WriteLine(string.Format("{0} : {1}", n, GetFirstDigit(n)));
Output:
-1574 : 4
-221 : 1
1246 : 6
-4 : 4
8 : 8
38546 : 6
Here is a simpler way that does not involve looping
int number = 1234
int firstDigit = Math.Floor(number/(Math.Pow(10, number.ToString().length - 1))
That would give us 1234/Math.Pow(10, 4 - 1) = 1234/1000 = 1