I'm trying to get the factorial value of number 66, but my method resulting me an output 0. But whenever I try to get the factorial of 5, it is resulting me an output 120. Could anyone please tell me why?
public static int factorial(int n)
{
if (n == 1)
return n;
return n * factorial(n - 1);
}
Sure - factorials get very big, very fast. You're overflowing the bounds of int very quickly... and at some point you'll have multiplied by enough factors to get an overflow to 0, which will then keep the value at 0 forever.
According to a quick Google search, 66 factorial is 5.44344939 × 1092 - which is considerably more than int can handle, or even long or decimal. You could get double to handle it - you'd lose a huge amount of precision, and that would accumulate really quickly too, but at least it wouldn't overflow...
Your method overflows. See the following example:
static void Main(string[] args)
{
Console.WriteLine(factorial(66));
}
public static int factorial(int n)
{
if (n == 1)
return n;
var result = n * factorial(n - 1);
Console.WriteLine("{0} : {1}", n, result);
return result;
}
With this example, the results of each iteration is printed.
You see that at one point, result becomes 0 and this means that every iteration from that point on becomes n * 0.
You can try using BigInteger. This will give the correct result. Calculate factorials in C# contains more information on this.
66! does not fit into an int. Use BigInteger.
The problem is that the factorial of 66 is way to large to fit into an int. I think it will also we way to large to fit into a long.
As an example, factorial(20) will return 2432902008176640000
The factorial of 50 is 3.0414093202×1064 which exeeds already what a int can contain.
Use long or BigInteger for this.
You get numeric overflow, 66! ~= 5e92 which is way larger than an int can handle. Also, factorials are better calculated using a for loop.
Approximately 13 or 14 is the largest number whose factorial fits in an int... If you switch to long, it'll be aroung 18 or 19 if I recall correctly. If you wish arbitraty big numbers you'd have to write your own big arithmetic library or use an existing one :)
You need to use the appropriate data type.
In this case the Big Integer data type is probably best seeing as how fast the numbers get real big.
Here is the way you use this data type.
Right click on your project, choose the add reference menu.
Look for the system.numerics library and add it.
You then add the using clause in your code.
And then you can initialise variable with the keyword as usual.
Related
It'sa me, with another problem.
I need to calculate a factorial of a really huge number, lets assume it is 95.
So as we all know 95! equals to:
10329978488239062144133688859495761720042551046933218543167809699858950620982142410696539365993509132394773015016946331626553858953528454377577119744
I have used a simple method calculating factorials using BigIntegers that I found somewhere around here few months ago:
public static BigInteger FactorialTest(BigInteger x)
{
if (x == 0)
return 1;
BigInteger res = x;
x--;
while (x > 1)
{
res *= x;
x--;
}
return res;
}
And only got a rounded up number:
10329978488239059262599702099394727095397746340117372869212250571234293987594703124871765375385424468563282236864226607350415360000000000000000000000
Next step was using the builtin BigInteger methods for addition, multiplication etc, hoping it will fix the problem - no still did not work.
Last thing I tried was using code of someone smarter, so reached for SolverFoundation, unfortunately
Microsoft.SolverFoundation.Common.BigInteger.Factorial(95)
still returns the rounded up number.
Is there anything I am missing that could get me the proper result? I really hoped BigIntegers would not lose precision like that.
According to http://2000clicks.com/MathHelp/BasicFactorialTable.aspx and other sites the number you are getting from your calculation is correct.
Let say an event have the probability P to succeed. (0 < P < 1 )
and I have to make N tests to see if this happens and I want the total number of successes:
I could go
int countSuccesses = 0;
while(N-- > 0)
{
if(Random.NextDouble()<P) countSuccesses++; // NextDouble is from 0.0 to 1.0
}
But is there not a more efficient way to do this? I want to have a single formula so I just can use ONE draw random number to determine the total number of successes. (EDIT The idea of using only one draw was to get below O(n))
I want to be able call a method
GetSuccesses( n, P)
and it to be O(1)
UPDATE
I will try to go with the
MathNet.Numerics.Distributions.Binomial.Sample(P, n)
even if it might be using more then only one random number I'll guess it will be faster than O(n) even if its not O(1). I'll benchmark that. Big thanks to David and Rici.
UPDATE
The binomial sample above was O(n) so it did not help me. But thanks to a comment done by Fred I just switched to
MathNet.Numerics.Distributions.Normal.Sample(mean, stddev)
where
mean = n * P
stddev = Math.Sqrt(n * P * (1 - P));
and now it is O(1) !
Per #rici for small N you can use the CDF or PMF of the Binomial Distribution, and simply compare the random input with the probabibilities for 0,1,2..N successes.
Something like:
static void Main(string[] args)
{
var trials = 10;
var trialProbability = 0.25;
for (double p = 0; p <= 1; p += 0.01)
{
var i = GetSuccesses(trials, trialProbability, p);
Console.WriteLine($"{i} Successes out of {trials} with P={trialProbability} at {p}");
}
Console.ReadKey();
}
static int GetSuccesses(int N, double P, double rand)
{
for (int i = 0; i <= N; i++)
{
var p_of_i_successes = MathNet.Numerics.Distributions.Binomial.PMF(P, N, i);
if (p_of_i_successes >= rand)
return i;
rand -= p_of_i_successes;
}
return N;
}
I'm not going to write formula here, as it's already in wiki, and I don't really know good formatting here for such things.
Probability for each outcome can be determined by Bernulli formula
https://en.wikipedia.org/wiki/Bernoulli_trial
What you need to do is to calculate binominal coefficient, then probability calculation becomes quite easy - multiply binominal coefficient by p and q in appropriate powers. Fill in array P[0..n] that contains probability for each outcome - number of exactly i successes.
After set up go from 0 up to n and calculate rolling sum of probabilities.
Check lower/upper bounds against random value and once it's inside current interval, return the outcome.
So, deciding part will be like this:
sum=0;
for (int i = 0; i <= n; i++)
if (sum-eps < R && sum+P[i]+eps > R)
return i;
else
sum+=P[i];
Here eps is small floating point value to overcome floating point rounding issues, R is saved random value, P is an array of probabilities I mentioned before.
Unfortunately, this method is not practical for big N (20 or 100+):
you'll get quite big impact of rounding errors
random numbers generator can be not determinitive enough to cover every possible outcome with proper probabilities distribution
Based on how you've phrased the question, this is impossible to do.
Your are essentially asking how to ensure that a single coin flip (i.e. one randomized outcome) is exactly 50% heads and 50% tails, which is impossible.
Even if you were to use two random numbers, where you expect one heads and one tails; this test would fail in 50% of all cases (because you might get two heads or two tails).
Probablility is founded on the law of large numbers. This explicitly states that a small sampling cannot be expected to accurately reflect the expected outcome.
The LLN is important because it guarantees stable long-term results for the averages of some random events. For example, while a casino may lose money in a single spin of the roulette wheel, its earnings will tend towards a predictable percentage over a large number of spins. Any winning streak by a player will eventually be overcome by the parameters of the game. It is important to remember that the law only applies (as the name indicates) when a large number of observations is considered. There is no principle that a small number of observations will coincide with the expected value or that a streak of one value will immediately be "balanced" by the others (see the gambler's fallacy).
When I asked this as a comment; you replied:
#Flater No, I am making N actual draws but with only one random number.
But this doesn't make sense. If you only use one random value, and keep using that same value, then every draw is obviously going to give you the exact same outcome (that same number).
The closest I can interpret your question in a way that is not impossible would be that you were mistakenly referring to a single random seed as a single random number.
A random seed (or seed state, or just seed) is a number (or vector) used to initialize a pseudorandom number generator.
For a seed to be used in a pseudorandom number generator, it does not need to be random. Because of the nature of number generating algorithms, so long as the original seed is ignored, the rest of the values that the algorithm generates will follow probability distribution in a pseudorandom manner.
However, your explicitly mentioned expectations seem to disprove that assumption. You want to do something like:
GetSuccesses( n, P, Random.NextDouble())
and you're also expecting to get a O(1) operation, which flies in the face of the law of large numbers.
If you're actually talking about having a single random seed; then your expectation are not correct.
If you make N draws, the operation is still of O(N) complexity. Whether you randomize the seed after every draw or not is irrelevant, it's always O(N).
GetSuccesses( n, P, Random.NextDouble()) is going to give you one draw, not one seed. Regardless of terminology used, your expectation of the code is not related to using the same seed for several draws.
As the question is currently phrased; what you want is impossible. Repeated comments for clarification by several commenter have not yet yielded a clearer picture.
As a sidenote, I find it very weird that you've answered to every comment except when directly asked if you are talking about a seed instead of a number (twice now).
#David and #rici pointed me to the
MathNet.Numerics.Distributions.Binomial.Sample(P, n)
A benchmark told me that it also O(n) and in par with my original
int countSuccesses = 0;
while(N-- > 0)
{
if(Random.NextDouble()<P) countSuccesses++; // NextDouble is from 0.0 to 1.0
}
But thanks to a comment done by Fred:
You could turn that random number into a gaussian sample with mean N*P, which would have the same distribution as your initial function
I just switched to
MathNet.Numerics.Distributions.Normal.Sample(mean, stddev)
where
mean = n * P
stddev = Math.Sqrt(n * P * (1 - P));
and now it is O(1) !
and the function I wanted got to be:
private int GetSuccesses(double p, int n)
{
double mean = n * p;
double stddev = Math.Sqrt(n * p * (1 - p));
double hits = MathNet.Numerics.Distributions.Normal.Sample(Random, mean, stddev);
return (int)Math.Round(hits, 0);
}
As Paul pointed out,
this is an approximation, but one that I gladly accept.
I've been trying to make a factorial function in C++, and I just found that inputs which are greater than 10 are not calculated correctly. I tried C# but I faced the same problem.
using this recursive function :
int Factorial(int Number) {
if (Number == 0) return 1;
return Number * Factorial(Number - 1);
}
The program returns 0 for large numbers, and even small inputs such 15 or 16 are wrongly calculated, I mean the result differs from what I get in windows calculator.
That's because int is limited to 32 bits, and results of 10! much beyond 10 will give you a bigger result. For larger values, you can get an approximate result using double as the result. If you want more than about 16 digits precision, you need to use a multiprecision math library.
Using uint64_t would allow a larger number, but still fairly limited.
This is because factorials are big numbers and the don't fit in an int variable. It overflows. If You use unsigned long long instead of int, You can compute greater factorials.
If the result doesn't need to be precise, You can use double as well. In the other case, You can implement the multiplication on arrays and You can compute as big factorials as You want.
In C/C++, I recommend the GMP library
And because you asked for C family languages: Java has a BigInteger type which is also useful.
Variables cannot hold an infinite number of values. On a 32-bit machine, the maximum value an int can represent is 2^31-1, or 2147483647. You can use another type, such as unsigned int, long, unsigned long, long long, or the largest one, unsigned long long. Factorials are big numbers, and can easily overflow! If you want arbitrary precision, you should use a bignum library, such as GNU GMP.
Reason for You problem in simple.In C++ biggest in Long Long int. it has range of ~10^18. so you can't store numbers greater than that. and 100! have 158 digits in it so there is no way you can store that number in C++/C unless you can use vector/Array
You are a new Programmer and you are using Only C++/C .
Then i will not recommend for using GMP library for programming purpose(Algorithm or Programming contest Purpose) unless you are programming for some software.
I think you can implement your own and use it.. I use this one for my own purpose in Programming contest and Algorithm problem.
// Shashank Jain
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#define LL long long int
using namespace std;
vector<int> fact;
void do_it(int num)
{
int temp,carry;
vector<int>:: iterator it;
carry=0;
for(it=fact.begin();it!=fact.end();it++)
{
temp=(*it)*num;
temp+=carry;
*it=temp%10;
carry=temp/10;
}
if(carry!=0)
{
while(carry>0)
{
temp=carry%10;
fact.push_back(temp);
carry/=10;
}
}
}
int main()
{
int num,i,l;
cin>>num; // enter number for which you want to get factorial
fact.push_back(1);
for(i=2;i<=num;i++)
do_it(i);
l=fact.size();
cout<<"The Length of factorial is: "<<l<<endl;
for(i=l-1;i>=0;i--)
{
cout<<fact[i];
}
cout<<endl;
return 0;
}
Running Code link On Ideone
This can easily get the factorial of 2000 in less than 1 sec. Or else You can use GMP library. But Those are not allowed for Programming Contest like Google Code jam or Facebook Hacker Cup. or topcoder or anyother standard programming contest
I have the next function:
static bool isPowerOf(int num, int power)
{
double b = 1.0 / power;
double a = Math.Pow(num, b);
Console.WriteLine(a);
return a == (int)a;
}
I inserted the print function for analysis.
If I call the function:
isPowerOf(25, 2)
It return true since 5^2 equals 25.
But, if I call 16807, which is 7^5, the next way:
isPowerOf(16807, 5)
In this case, it prints '7' but a == (int)a return false.
Can you help? Thanks!
Try using a small epsilon for rounding errors:
return Math.Abs(a - (int)a) < 0.0001;
As harold suggested, it will be better to round in case a happens to be slightly smaller than the integer value, like 3.99999:
return Math.Abs(a - Math.Round(a)) < 0.0001;
Comparisons that fix the issue have been suggested, but what's actually the problem here is that floating point should not be involved at all. You want an exact answer to a question involving integers, not an approximation of calculations done on inherently inaccurate measurements.
So how else can this be done?
The first thing that comes to mind is a cheat:
double guess = Math.Pow(num, 1.0 / power);
return num == exponentiateBySquaring((int)guess, power) ||
num == exponentiateBySquaring((int)Math.Ceil(guess), power);
// do NOT replace exponentiateBySquaring with Math.Pow
It'll work as long as the guess is less than 1 off. But I can't guarantee that it will always work for your inputs, because that condition is not always met.
So here's the next thing that comes to mind: a binary search (the variant where you search for the upper boundary first) for the base in exponentiateBySquaring(base, power) for which the result is closest to num. If and only if the closest answer is equal to num (and they are both integers, so this comparison is clean), then num is a power-th power. Unless there is overflow (there shouldn't be), that should always work.
Math.Pow operates on doubles, so rounding errors come into play when taking roots. If you want to check that you've found an exact power:
perform the Math.Pow as currently, to extract the root
round the result to the nearest integer
raise this integer to the supplied power, and check you get the supplied target. Math.Pow will be exact for numbers in the range of int when raising to integer powers
If you debug the code and then you can see that in first comparison:
isPowerOf(25, 2)
a is holding 5.0
Here 5.0 == 5 => that is why you get true
and in 2nd isPowerOf(16807, 5)
a is holding 7.0000000000000009
and since 7.0000000000000009 != 7 => you are getting false. and Console.WriteLine(a) is truncating/rounding the double and only show 7
That is why you need to compare the nearest value like in Dani's solution
I have a large set of numbers, probably in the multiple gigabytes range. First issue is that I can't store all of these in memory. Second is that any attempt at addition of these will result in an overflow. I was thinking of using more of a rolling average, but it needs to be accurate. Any ideas?
These are all floating point numbers.
This is not read from a database, it is a CSV file collected from multiple sources. It has to be accurate as it is stored as parts of a second (e.g; 0.293482888929) and a rolling average can be the difference between .2 and .3
It is a set of #'s representing how long users took to respond to certain form actions. For example when showing a messagebox, how long did it take them to press OK or Cancel. The data was sent to me stored as seconds.portions of a second; 1.2347 seconds for example. Converting it to milliseconds and I overflow int, long, etc.. rather quickly. Even if I don't convert it, I still overflow it rather quickly. I guess the one answer below is correct, that maybe I don't have to be 100% accurate, just look within a certain range inside of a sepcific StdDev and I would be close enough.
You can sample randomly from your set ("population") to get an average ("mean"). The accuracy will be determined by how much your samples vary (as determined by "standard deviation" or variance).
The advantage is that you have billions of observations, and you only have to sample a fraction of them to get a decent accuracy or the "confidence range" of your choice. If the conditions are right, this cuts down the amount of work you will be doing.
Here's a numerical library for C# that includes a random sequence generator. Just make a random sequence of numbers that reference indices in your array of elements (from 1 to x, the number of elements in your array). Dereference to get the values, and then calculate your mean and standard deviation.
If you want to test the distribution of your data, consider using the Chi-Squared Fit test or the K-S test, which you'll find in many spreadsheet and statistical packages (e.g., R). That will help confirm whether this approach is usable or not.
Integers or floats?
If they're integers, you need to accumulate a frequency distribution by reading the numbers and recording how many of each value you see. That can be averaged easily.
For floating point, this is a bit of a problem. Given the overall range of the floats, and the actual distribution, you have to work out a bin-size that preserves the accuracy you want without preserving all of the numbers.
Edit
First, you need to sample your data to get a mean and a standard deviation. A few thousand points should be good enough.
Then, you need to determine a respectable range. Folks pick things like ±6σ (standard deviations) around the mean. You'll divide this range into as many buckets as you can stand.
In effect, the number of buckets determines the number of significant digits in your average. So, pick 10,000 or 100,000 buckets to get 4 or 5 digits of precision. Since it's a measurement, odds are good that your measurements only have two or three digits.
Edit
What you'll discover is that the mean of your initial sample is very close to the mean of any other sample. And any sample mean is close to the population mean. You'll note that most (but not all) of your means are with 1 standard deviation of each other.
You should find that your measurement errors and inaccuracies are larger than your standard deviation.
This means that a sample mean is as useful as a population mean.
Wouldn't a rolling average be as accurate as anything else (discounting rounding errors, I mean)? It might be kind of slow because of all the dividing.
You could group batches of numbers and average them recursively. Like average 100 numbers 100 times, then average the result. This would be less thrashing and mostly addition.
In fact, if you added 256 or 512 at once you might be able to bit-shift the result by either 8 or 9, (I believe you could do this in a double by simply changing the floating point mantissa)--this would make your program extremely quick and it could be written recursively in just a few lines of code (not counting the unsafe operation of the mantissa shift).
Perhaps dividing by 256 would already use this optimization? I may have to speed test dividing by 255 vs 256 and see if there is some massive improvement. I'm guessing not.
You mean of 32-bit and 64-bit numbers. But why not just use a proper Rational Big Num library? If you have so much data and you want an exact mean, then just code it.
class RationalBignum {
public Bignum Numerator { get; set; }
public Bignum Denominator { get; set; }
}
class BigMeanr {
public static int Main(string[] argv) {
var sum = new RationalBignum(0);
var n = new Bignum(0);
using (var s = new FileStream(argv[0])) {
using (var r = new BinaryReader(s)) {
try {
while (true) {
var flt = r.ReadSingle();
rat = new RationalBignum(flt);
sum += rat;
n++;
}
}
catch (EndOfStreamException) {
break;
}
}
}
Console.WriteLine("The mean is: {0}", sum / n);
}
}
Just remember, there are more numeric types out there than the ones your compiler offers you.
You could break the data into sets of, say, 1000 numbers, average these, and then average the averages.
This is a classic divide-and-conquer type problem.
The issue is that the average of a large set of numbers is the same
as the average of the first-half of the set, averaged with the average of the second-half of the set.
In other words:
AVG(A[1..N]) == AVG( AVG(A[1..N/2]), AVG(A[N/2..N]) )
Here is a simple, C#, recursive solution.
Its passed my tests, and should be completely correct.
public struct SubAverage
{
public float Average;
public int Count;
};
static SubAverage AverageMegaList(List<float> aList)
{
if (aList.Count <= 500) // Brute-force average 500 numbers or less.
{
SubAverage avg;
avg.Average = 0;
avg.Count = aList.Count;
foreach(float f in aList)
{
avg.Average += f;
}
avg.Average /= avg.Count;
return avg;
}
// For more than 500 numbers, break the list into two sub-lists.
SubAverage subAvg_A = AverageMegaList(aList.GetRange(0, aList.Count/2));
SubAverage subAvg_B = AverageMegaList(aList.GetRange(aList.Count/2, aList.Count-aList.Count/2));
SubAverage finalAnswer;
finalAnswer.Average = subAvg_A.Average * subAvg_A.Count/aList.Count +
subAvg_B.Average * subAvg_B.Count/aList.Count;
finalAnswer.Count = aList.Count;
Console.WriteLine("The average of {0} numbers is {1}",
finalAnswer.Count, finalAnswer.Average);
return finalAnswer;
}
The trick is that you're worried about an overflow. In that case, it all comes down to order of execution. The basic formula is like this:
Given:
A = current avg
C = count of items
V = next value in the sequence
The next average (A1) is:
(C * A) + V
A1 = ———————————
C + 1
The danger is over the course of evaulating the sequence, while A should stay relatively manageable C will become very large.
Eventually C * A will overflow the integer or double types.
One thing we can try is to re-write it like this, to reduce the chance of an overflow:
A1 = C/(C+1) * A/(C+1) + V/(C+1)
In this way, we never multiply C * A and only deal with smaller numbers. But the concern now is the result of the division operations. If C is very large, C/C+1 (for example) may not be meaningful when constrained to normal floating point representations. The best I can suggest is to use the largest type possible for C here.
Here's one way to do it in pseudocode:
average=first
count=1
while more:
count+=1
diff=next-average
average+=diff/count
return average
Sorry for the late comment, but isn't it the formula above provided by Joel Coehoorn rewritten wrongly?
I mean, the basic formula is right:
Given:
A = current avg
C = count of items
V = next value in the sequence
The next average (A1) is:
A1 = ( (C * A) + V ) / ( C + 1 )
But instead of:
A1 = C/(C+1) * A/(C+1) + V/(C+1)
shouldn't we have:
A1 = C/(C+1) * A + V/(C+1)
That would explain kastermester's post:
"My math ticks off here - You have C, which you say "go towards infinity" or at least, a really big number, then: C/(C+1) goes towards 1. A /(C+1) goes towards 0. V/(C+1) goes towards 0. All in all: A1 = 1 * 0 + 0 So put shortly A1 goes towards 0 - seems a bit off. – kastermester"
Because we would have A1 = 1 * A + 0, i.e., A1 goes towards A, which it's right.
I've been using such method for calculating averages for a long time and the aforementioned precision problems have never been an issue for me.
With floating point numbers the problem is not overflow, but loss of precision when the accumulated value gets large. Adding a small number to a huge accumulated value will result in losing most of the bits of the small number.
There is a clever solution by the author of the IEEE floating point standard himself, the Kahan summation algorithm, which deals exactly with this kind of problems by checking the error at each step and keeping a running compensation term that prevents losing the small values.
If the numbers are int's, accumulate the total in a long. If the numbers are long's ... what language are you using? In Java you could accumulate the total in a BigInteger, which is an integer which will grow as large as it needs to be. You could always write your own class to reproduce this functionality. The gist of it is just to make an array of integers to hold each "big number". When you add two numbers, loop through starting with the low-order value. If the result of the addition sets the high order bit, clear this bit and carry the one to the next column.
Another option would be to find the average of, say, 1000 numbers at a time. Hold these intermediate results, then when you're done average them all together.
Why is a sum of floating point numbers overflowing? In order for that to happen, you would need to have values near the max float value, which sounds odd.
If you were dealing with integers I'd suggest using a BigInteger, or breaking the set into multiple subsets, recursively averaging the subsets, then averaging the averages.
If you're dealing with floats, it gets a bit weird. A rolling average could become very inaccurate. I suggest using a rolling average which is only updated when you hit an overflow exception or the end of the set. So effectively dividing the set into non-overflowing sets.
Two ideas from me:
If the numbers are ints, use an arbitrary precision library like IntX - this could be too slow, though
If the numbers are floats and you know the total amount, you can divide each entry by that number and add up the result. If you use double, the precision should be sufficient.
Why not just scale the numbers (down) before computing the average?
If I were to find the mean of billions of doubles as accurately as possible, I would take the following approach (NOT TESTED):
Find out 'M', an upper bound for log2(nb_of_input_data). If there are billions of data, 50 may be a good candidate (> 1 000 000 billions capacity). Create an L1 array of M double elements. If you're not sure about M, creating an extensible list will solve the issue, but it is slower.
Also create an associated L2 boolean array (all cells set to false by default).
For each incoming data D:
int i = 0;
double localMean = D;
while (L2[i]) {
L2[i] = false;
localMean = (localMean + L1[i]) / 2;
i++;
}
L1[i] = localMean;
L2[i] = true;
And your final mean will be:
double sum = 0;
double totalWeight = 0;
for (int i = 0; i < 50) {
if (L2[i]) {
long weight = 1 << i;
sum += L1[i] * weight;
totalWeight += weight;
}
}
return sum / totalWeight;
Notes:
Many proposed solutions in this thread miss the point of lost precision.
Using binary instead of 100-group-or-whatever provides better precision, and doubles can be safely doubled or halved without losing precision!
Try this
Iterate through the numbers incrementing a counter, and adding each number to a total, until adding the next number would result in an overflow, or you run out of numbers.
( It makes no difference if the inputs are integers or floats - use the largest precision float you can and convert each input to that type)
Divide the total by the counter to get a mean ( a floating point), and add it to a temp array
If you had run out of numbers, and there is only one element in temp, that's your result.
Start over using the temp array as input, ie iteratively recurse until you reached the end condition described earlier.
depending on the range of numbers it might be a good idea to have an array where the subscript is your number and the value is the quantity of that number, you could then do your calculation from this