Develop Reference

Print out the successful development of prime numbers from small to large

Given any natural number N> 1 (previously assigned). Print out the successful development of prime numbers from small to large.
Example:
9 --> 3 * 3
12 --> 2 * 2 * 3
My idea is find all GCD and add to list int, and write a function isPrimeNumber(int n), browse List< int > and check if isPrimeNumber().
But I can't solve problem print out the successful development of prime numbers from small to large
Here is what I tried
static void Main(string[] args)
{
Console.WriteLine("Enter n: ");
int n = Convert.ToInt32(Console.ReadLine());
List<int> arr = new List<int>();
for (int i = 1; i <= n; i++)
{
if (n % i == 0)
{
arr.Add(i);
}
}
/* I need Print out the successful development of prime numbers from small to large here */
}
static bool isPrimeNumber(int n)
{
if (n < 2)
{
return false;
}
for (int i = 2; i <= Math.Sqrt(n); i++)
{
if (n % i == 0)
{
return false;
}
}
return true;
}

As you posted your working solution for that, let me share a different implementation for that that is still simple to understand, but more efficient, because it only tests primes until it reaches the square root of n. After that, there will not be any other divisor, except the number n itself, if n is prime.
static IList<int> Factors(int num) {
var result = new List<int>();
// avoid scenarios like num=0, that would cause infinite loop
if (num <= 1) {
return result;
}
// testing with 2 outside the main loop, otherwise we would skip factor 3
// (2 * 2 > 3)
while (num % 2 == 0) {
result.Add(2);
num /= 2;
}
// only test primes until sqrt(num)
int i = 3;
while (i * i <= num) {
if (num % i == 0) {
result.Add(i);
num /= i;
} else {
i++;
}
}
// if not 1 here, num is prime
if (num > 1) {
result.Add(num);
}
return result;
}

I solved it
Here is code
static void lesson6()
{
Console.WriteLine("Enter n: ");
int n = Convert.ToInt32(Console.ReadLine());
int a = n;
List<int> arr = new List<int>();
for (int i = 2; i <= n; i++)
{
while (n % i == 0)
{
arr.Add(i);
n /= i;
}
}
Console.Write($"{a} = ");
int lastIndex = arr.Count - 1;
for (int i = 0; i < arr.Count; i++)
{
if (i == lastIndex)
{
Console.Write(arr[i]);
}
else
{
Console.Write(arr[i] + "*");
}
}
}

As pointed by derpirscher in the comment, there are several sources online with different approaches for integer factorization.
I recommend you to look for Trial Division algorithm, as it is the easier to understand, and is similar to your approach.
Based on the code you shared, there are some thing you should consider:
for (int i = 1; i <= n; i++)
{
if (n % i == 0)
{
arr.Add(i);
}
}
After finding that a prime is a divisor and appending to the list, you are going to the next number. However, a prime can figure many times in the factorization of a number. E.g: 12 -> { 2, 2, 3 }.
You need divide n by the prime and continue testing the until it is not a divisor anymore, then you can go test the next prime.
This way, your n is shrinking down each time you find a prime divisor, until it eventually become 1. Then you know you found all prime divisors.

Get all n-digit numbers whose sum of digits equals to given sum

How can I get all n-digit numbers whose sum of digits equals to given sum? I need the fastest solution because n can be equal with 9 and sum can be equal with 1000.
I have implemented the solution below but it's too slow...
List<int> l = new List<int>();
void findNDigitNumsUtil(int n, int sum, char[] ou, int index)
{
if (index > n || sum < 0)
return;
if (index == n)
{
if (sum == 0)
{
ou[index] = '\0';
string s = new string(ou);
l.Add(Int32.Parse(s));
}
return;
}
for (int i = 0; i <= 9; i++)
{
ou[index] = (char)(i + '0');
findNDigitNumsUtil(n, sum - i, ou,
index + 1);
}
}
void findNDigitNums(int n, int sum)
{
char[] ou = new char[n + 1];
for (int i = 1; i <= 9; i++)
{
ou[0] = (char)(i + '0');
findNDigitNumsUtil(n, sum - i, ou, 1);
}
}

I need the fastest solution
No, you need a fast-enough solution. You are probably unwilling to spend even a million dollars on custom hardware to get the fastest possible solution.
How can I get all n-digit numbers whose sum of digits equals to given sum?
Here, I'll give you the solution for a slightly different problem:
What are all the sequences of n digits drawn from 0-9 that sum to sum?
This is different because this counts 01 and 10 as sequences of length two that sum to 1, but 01 is not a two-digit number.
I'll give you a hint for how to solve this easier problem. You then take that solution and adapt it to your harder problem.
First, can you solve the problem for one-digit numbers? That's pretty easy. The one-digit numbers whose digits sum to n are the digit n if n is 0 through 9, and there is no solution otherwise.
Second: Suppose n > 1. Then the n-digit numbers that sum to sum are:
0 followed by all the n-1 digit numbers that sum to sum
1 followed by all the n-1 digit numbers that sum to sum-1
2 followed by all the n-1 digit numbers that sum to sum-2
...
9 followed by all the n-1 digit numbers that sum to sum-9
Write an implementation that solves that problem, and then adapt it to solve your problem.

You can treat n-digit number as an array of n digits. Then you can increment a particular number to the next number that also adds up to the sum. Stepping through all the next answers, you have generated all possible combinations.
Using a generator to yield each n-digit combination as an IEnumerable<int> (in fact, an int[]), you start with the "smallest" n-digit combination that yields the sum, and go through each one.
IEnumerable<IEnumerable<int>> DigitsToSum(int n, int sum) {
if (sum > 9 * n)
yield return Enumerable.Empty<int>();
else {
var ans = new int[n];
void distribute(int wsum, int downto) {
for (var j1 = n - 1; j1 > downto; --j1) {
if (wsum > 9) {
ans[j1] = 9;
wsum -= 9;
}
else {
ans[j1] = wsum;
wsum = 0;
}
}
}
ans[0] = Math.Max(1, sum-9*(n-1));
distribute(sum-ans[0], 0);
bool nextAns() {
var wsum = ans[n-1];
for (var j1 = n - 2; j1 >= 0; --j1) {
wsum += ans[j1];
if (ans[j1] < Math.Min(9, wsum)) {
++ans[j1];
distribute(wsum - ans[j1], j1);
return true;
}
}
return false;
}
do {
yield return ans;
} while (nextAns());
}
}
This is tremendously faster than my recursive double generator solution (somewhat like #EricLippert's suggestion) to iterate over all possibilities (e.g. using Count()).
You can put the digits back together to get a final numeric string for each number:
var ans = DigitsToSum(n, sum).Select(p => String.Join("", p));

Find N consecutive integers on a array, whose sum(of consecutive integers) S is a integer that i have

explanation
I have been staring at the problem for a few of minutes.
And i did some research before i ask this quest , but it were in different cases and they didn't included what i really need.
I found this piece of code in SO.
static int GetLargestSum(int[] array, int n, int sum)
{
int largestSum = 0;
int previousSum = 0;
for (int i = 0; i <= array.Length - n; i++)
{
if (i == 0)
{
for (int j = 0; j < n; j++)
{
largestSum += array[j];
}
previousSum = largestSum;
}
else
{
int currentSum = previousSum - array[i - 1] + array[i + n - 1];
if (currentSum > largestSum)
{
largestSum = currentSum;
}
previousSum = currentSum;
}
}
return largestSum;
}
And yes this works but if works for only the largest sum.
I tried to modify it to add the sum var into the code but that didn't actually went that well.
So i would really appreciate if someone helps me, bcs i am stuck in this algorithm.
Thank you!

The way to solve it would be to iterate over each segment of the array and evaluate its sum. A crude first draft would look something like this
public static int ConsecutiveSumArrangements(int[] vals, int count, int sum)
{
var number = 0;
for (int i = 0; i < (vals.Length - count); i++)
{
var segSum = vals.Skip(i).Take(count).Sum();
if (segSum == sum)
{
number++;
}
}
return number;
}

Maybe it is easier think in another way than try to correct this code. An idea is using slide window. Pseudo code look like
sum = 0
sol = 0
start = 0
end = 0
// sum of the first m elements
while end < m
sum = sum + s[end]
end = end + 1
If sum == d
sol = sol + 1
while end < n
sum = sum + s[end]
sum = sum - s[start]
end = end + 1
start = start + 1
if sum == d
sol = sol + 1
// in the loop we add the next element and subtract the first element
//so, we keep the length of m elements

Find the 10001st prime

I took a look at the following question from project euler:
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
I tried to take the square root of the number and than find all the prime numbers below the square root of the number and then divide the number by all the square roots and see if there is 0 left each time. If the number is not divisible by all the primes under its square root its a prime number. I did this to lower the itterations the programm has to make. Here is what I have now, I am not sure why it isn't working. Anybody knows what i did wrong?
List<int> primeNumbers = new List<int>();
bool prime = true;
bool MainPrime = true;
int check = 1;
for (long i = 3; i < long.MaxValue; i++)
{
if ((i % 2) != 0)
{
int root = Convert.ToInt32(Math.Sqrt(i));
for (int j = 1; j < root; j++)
{
for (int k = 2; k < j; k++)
{
if ((j% k) == 0)
{
prime = false;
}
}
if (prime)
{
primeNumbers.Add(j);
}
prime = true;
}
}
foreach (var item in primeNumbers)
{
if ((i%item) == 0)
{
MainPrime = false;
}
}
primeNumbers.Clear();
if (MainPrime)
{
check++;
}
if (check == 10001)
{
Console.WriteLine(i);
break;
}
}
Console.ReadKey();

Several points:
When finding possible prime divisors, you need to check all numbers up to the square root included, so your condition j < root is incorrect.
You don't have to recalculate the primes again for every number. Keep the list as you go and add new primes to it.
As soon as you find a divisor, you can break out of the foreach loop.
Improved code:
List<long> primeNumbers = new List<long>() { 2 };
for (long i = 3; i < long.MaxValue; i += 2)
{
if(!primeNumbers.Any(p => (i % p) == 0))
{
primeNumbers.Add(i);
if (primeNumbers.Count == 10001)
{
Console.WriteLine(i);
break;
}
}
}
Gives 104743 as the 10001st prime.

What we can do is we can use SieveOfEratosthenes to make an bool array in which all the prime numbers value are set to be true than after that;
1.As we found any prime number increment the count with 1;
2.And as count get equal to 10001 we print its value and break through the loop.
Have a Look at code in C++ (I recommend you to learn SieveOfEratosthenes first)
#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(long long unsigned n)
{
bool prime[n];
memset(prime, true, sizeof(prime)); //This is SieveOfEratosthenes
for (long long p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (long long i = p * p; i <= n; i += p)
prime[i] = false;
}
}
long long count=0; //initializing count as 0;
for (long long p = 2; p <= n; p++) //running the loop form 2 to n
{
if (prime[p]) //we have bool array in which all prime number set to true using sieve
count++; //increment the count because we found a prime number
if(count==10001) // and as count reaches to 10001 we found our number
{
cout<<p;break;} // print the answer and also break form the loop
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long unsigned n=999999;
SieveOfEratosthenes(n); //pass the value of n in sieve function
return 0;
}

Try this one out using python
sp=2
cnt = 1
while cnt <= 10001:
primeflag = 0
for j in range(2,sp):
if(sp%j == 0):
primeflag = 1
break;
if(primeflag == 1):
pass
else:
print(cnt ,sp)
cnt = cnt +1
sp =sp+1
#which Gives
#10001 104743

Help with Creating a Recursive Function C#

I am creating a forecasting application that will run simulations for various "modes" that a production plant is able to run. The plant can run in one mode per day, so I am writing a function that will add up the different modes chosen each day that best maximize the plant’s output and best aligns with the sales forecast numbers provided. This data will be loaded into an array of mode objects that will then be used to calculate the forecast output of the plant.
I have created the functions to do this, however, I need to make them recursive so that I am able to handle any number (within reason) of modes and work days (which varies based on production needs). Listed below is my code using for loops to simulate what I want to do. Can someone point me in the right direction in order to create a recursive function to replace the need for multiple for loops?
Where the method GetNumbers4 would be when there were four modes, and GetNumbers5 would be 5 modes. Int start would be the number of work days.
private static void GetNumber4(int start)
{
int count = 0;
int count1 = 0;
for (int i = 0; 0 <= start; i++)
{
for (int j = 0; j <= i; j++)
{
for (int k = 0; k <= j; k++)
{
count++;
for (int l = 0; l <= i; l++)
{
count1 = l;
}
Console.WriteLine(start + " " + (count1 - j) + " " + (j - k) + " " + k);
count1 = 0;
}
}
start--;
}
Console.WriteLine(count);
}
private static void GetNumber5(int start)
{
int count = 0;
int count1 = 0;
for (int i = 0; 0 <= start; i++)
{
for (int j = 0; j <= i; j++)
{
for (int k = 0; k <= j; k++)
{
for (int l = 0; l <= k; l++)
{
count++;
for (int m = 0; m <= i; m++)
{
count1 = m;
}
Console.WriteLine(start + " " + (count1 - j) + " " + (j - k) + " " + (k - l) + " " + l);
count1 = 0;
}
}
}
start--;
}
Console.WriteLine(count);
}
EDITED:
I think that it would be more helpful if I gave an example of what I was trying to do. For example, if a plant could run in three modes "A", "B", "C" and there were three work days, then the code will return the following results.
3 0 0
2 1 0
2 0 0
1 2 0
1 1 1
1 0 2
0 3 0
0 2 1
0 1 2
0 0 3
The series of numbers represent the three modes A B C. I will load these results into a Modes object that has the corresponding production rates. Doing it this way allows me to shortcut creating a list of every possible combination; it instead gives me a frequency of occurrence.
Building on one of the solutions already offered, I would like to do something like this.
//Where Modes is a custom classs
private static Modes GetNumberRecur(int start, int numberOfModes)
{
if (start < 0)
{
return Modes;
}
//Do work here
GetNumberRecur(start - 1);
}
Thanks to everyone who have already provided input.

Calling GetNumber(5, x) should yield the same result as GetNumber5(x):
static void GetNumber(int num, int max) {
Console.WriteLine(GetNumber(num, max, ""));
}
static int GetNumber(int num, int max, string prefix) {
if (num < 2) {
Console.WriteLine(prefix + max);
return 1;
}
else {
int count = 0;
for (int i = max; i >= 0; i--)
count += GetNumber(num - 1, max - i, prefix + i + " ");
return count;
}
}

A recursive function just needs a terminating condition. In your case, that seems to be when start is less than 0:
private static void GetNumberRec(int start)
{
if(start < 0)
return;
// Do stuff
// Recurse
GetNumberRec(start-1);
}

I've refactored your example into this:
private static void GetNumber5(int start)
{
var count = 0;
for (var i = 0; i <= start; i++)
{
for (var j = 0; j <= i; j++)
{
for (var k = 0; k <= j; k++)
{
for (var l = 0; l <= k; l++)
{
count++;
Console.WriteLine(
(start - i) + " " +
(i - j) + " " +
(j - k) + " " +
(k - l) + " " +
l);
}
}
}
}
Console.WriteLine(count);
}
Please verify this is correct.
A recursive version should then look like this:
public static void GetNumber(int start, int depth)
{
var count = GetNumber(start, depth, new Stack<int>());
Console.WriteLine(count);
}
private static int GetNumber(int start, int depth, Stack<int> counters)
{
if (depth == 0)
{
Console.WriteLine(FormatCounters(counters));
return 1;
}
else
{
var count = 0;
for (int i = 0; i <= start; i++)
{
counters.Push(i);
count += GetNumber(i, depth - 1, counters);
counters.Pop();
}
return count;
}
}
FormatCounters is left as an exercise to the reader ;)

I previously offered a simple C# recursive function here.
The top-most function ends up having a copy of every permutation, so it should be easily adapted for your needs..

I realize that everyone's beaten me to the punch at this point, but here's a dumb Java algorithm (pretty close to C# syntactically that you can try out).
import java.util.ArrayList;
import java.util.List;
/**
* The operational complexity of this is pretty poor and I'm sure you'll be able to optimize
* it, but here's something to get you started at least.
*/
public class Recurse
{
/**
* Base method to set up your recursion and get it started
*
* #param start The total number that digits from all the days will sum up to
* #param days The number of days to split the "start" value across (e.g. 5 days equals
* 5 columns of output)
*/
private static void getNumber(int start,int days)
{
//start recursing
printOrderings(start,days,new ArrayList<Integer>(start));
}
/**
* So this is a pretty dumb recursion. I stole code from a string permutation algorithm that I wrote awhile back. So the
* basic idea to begin with was if you had the string "abc", you wanted to print out all the possible permutations of doing that
* ("abc","acb","bac","bca","cab","cba"). So you could view your problem in a similar fashion...if "start" is equal to "5" and
* days is equal to "4" then that means you're looking for all the possible permutations of (0,1,2,3,4,5) that fit into 4 columns. You have
* the extra restriction that when you find a permutation that works, the digits in the permutation must add up to "start" (so for instance
* [0,0,3,2] is cool, but [0,1,3,3] is not). You can begin to see why this is a dumb algorithm because it currently just considers all
* available permutations and keeps the ones that add up to "start". If you want to optimize it more, you could keep a running "sum" of
* the current contents of the list and either break your loop when it's greater than "start".
*
* Essentially the way you get all the permutations is to have the recursion choose a new digit at each level until you have a full
* string (or a value for each "day" in your case). It's just like nesting for loops, but the for loop actually only gets written
* once because the nesting is done by each subsequent call to the recursive function.
*
* #param start The total number that digits from all the days will sum up to
* #param days The number of days to split the "start" value across (e.g. 5 days equals
* 5 columns of output)
* #param chosen The current permutation at any point in time, may contain between 0 and "days" numbers.
*/
private static void printOrderings(int start,int days,List<Integer> chosen)
{
if(chosen.size() == days)
{
int sum = 0;
for(Integer i : chosen)
{
sum += i.intValue();
}
if(sum == start)
{
System.out.println(chosen.toString());
}
return;
}
else if(chosen.size() < days)
{
for(int i=0; i < start; i++)
{
if(chosen.size() >= days)
{
break;
}
List<Integer> newChosen = new ArrayList<Integer>(chosen);
newChosen.add(i);
printOrderings(start,days,newChosen);
}
}
}
public static void main(final String[] args)
{
//your equivalent of GetNumber4(5)
getNumber(5,4);
//your equivalent of GetNumber5(5)
getNumber(5,5);
}
}