We Keep Coding

Adding int numbers?

public static int addIntNumbers()
{
int input = int.Parse(Console.ReadLine());
int sum = 0;
while (input !=0)
{
sum += input % 10;
input /= 10;
Console.WriteLine(sum);
}
return sum;
}
I don't understand this syntax: After the while condition, sum += input % 10, which basically means sum = sum(which is 0) + input % 10, so lets say I input 24 so the sum of this should be 4, I think ?
And then the second line which I have no idea what it is even doing.
Any suggestions ?

The best way might be to add comments. However I can already tell that whoever wrote this, did not know what he as doing. For starters, there were not comments, the naming is abysmal and the I/O is actually handeled inside the function.
//The name is not right. This is not a proper sum function
//I think it is getting a sum of all digits in the input
public static int addIntNumbers()
{
//Get a input from the user, parse it to int
//That really belons outside. Just the int in per argument
int input = int.Parse(Console.ReadLine());
//Initialize sum to 0
int sum = 0;
//Input is also used as sort of "running variable".
//The loop will break if input reaches 0
while (input !=0)
{
//sum = sum + input % 10
//It tries to divide input by 10, get's the rest, then adds that rest to sum
sum += input % 10;
//Divide input by 10. Note that all decimal parts will be dropped
//That means it will reach 0 invariably
input /= 10;
//Output the current sum for debugging
Console.WriteLine(sum);
}
//The function returns
return sum;
}

Your code calculates digit-by-digit sum of an integer (the sum is positive if input is positive, negative if input is negative).
If you are a C# beginner, this might help you:
while (input !=0)
{
sum = sum + (input % 10); //sum = sum + remainder of division by ten (separation of least significant digit)
input = input / 10; //input is integer-divided by ten, which results in discarding of the least significant digit
Console.WriteLine(sum);
}
If you don't understand, get familiar with a difference between
4/6
and 4.0/6.
The first one is integer division, the other is floating point division.

Some things to help you understand what's going on here:
First, assuming you're in Visual Studio, you can set a break point in your code by clicking to the left of the line number, in the margin. A red dot will show up and when your code hits that point, it will pause. While paused, you can look at the "Locals" tab or hover over variable names in your code to see what values are at that point in time. You can then use F10 to step forward one line at a time and see how things change.
Second, the /= operator is similar to the += operator, except with division. So, "x /= 10" is exactly the same as "x = x / 10".
This program is adding up each digit of the number you type in by getting the ones digit, adding it to sum, then dividing the number by 10 to get rid of the old ones digit.

Finding sum of consecutive primes

I am working on improving my C# skills, and in the process I am trying to solve some of the problems on Project Euler, in this case problem 50. The problem states:
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-
hundred.
The longest sum of consecutive primes below one-thousand that adds to
a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most
consecutive primes?
Seems simple enough. I wrote a method to tell if something is prime, made a list of the primes below 1 million (which is easily more than I need, but I don't know how many I actually need), and iterated through that list to find the sums of the primes. Here is my code:
public static void Main()
{
IEnumerable<int> primes = Enumerable.Range(0, 1000000)
.Where(i => isPrime(i));
int sum = 0;
List<int> history = new List<int>();
foreach (int bar in primes)
{
if (sum + bar < 1000000)
{
sum += bar;
Console.WriteLine(sum);
history.Add(bar);
}
}
while (!isPrime(sum))
{
sum -= history[history.Count - 1];
history.Remove(history[history.Count - 1]);
}
Console.WriteLine(sum);
Console.ReadLine();
}
public static bool isPrime(int num)
{
if (num <= 1)
{
return false;
}
else if (num == 2)
{
return true;
}
else if (num % 2 == 0)
{
return false;
}
else
{
var boundary = (int)Math.Floor(Math.Sqrt(num));
for (int i = boundary; i > 1; i--)
{
if (num % i == 0)
{
return false;
}
}
return true;
}
}
If I am correct, this should find the sum of my primes up to a million, then subtract primes until the sum is a prime number itself. When I run this, the code sums up to 997661, but that is not prime. I subtract the recently added primes until I get a result of 958577, which is prime, but this is not the correct answer. I am fairly certain my method to find primes is correct, but I cannot figure out what is causing my answer to be wrong. What's worse, I don't know the correct answer, so I can't work backwards to see what is causing the issue.
I suspect something may be broken inside of my while loop, like maybe I am removing the wrong values from the list. If anyone can offer some insight into why my program is not working, I would very much appreciate it.

Find the longest list of primes with a sum less than 1000000. That's the list that starts at 2 and goes as high as possible. Let the length of this list be L.
Now, iterate through all lists with sums less than 1000000, starting with the list of length L, then all lists of length L-1, then L-2, etc.
Stop when you get a prime sum.
About 1 in every 15 integers near 1000000 is prime, so you wont have to check very many lists, and of course you should make subsequent lists by adding and removing primes from the ends instead of recalculating the whole sum.

well i am improving my python skills and solved the question using python.I think the question might be wrong and i did the calculation manually.well here is my answer
The prime 41, can be written as the sum of six consecutive primes: 41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred. The longest sum of
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
solution:-
i tried to solve it in step by step process and here is my justification towards my assumption.
import sympy
sum=0
lst1=[]
for num in range(1,100):
#isprime(n):return True when the num is prime and false when the num is composite
if sympy.isprime(num) is True:
sum+=num
lst1.append(sum)
print("The sum list 1 is: ",lst1)
lst2=[]
for sum in lst1:
if sum<100:
if sympy.isprime(sum)==True:
lst2.append(sum)
print("The list 2 is: ",lst2)
print("The required answer is :",max(lst2))
the longest sum of consecutive primes that adds to a prime below one-hundred is 41
enter image description here
so similarly i changed the limits from 100 to 1000 as below..
import sympy
sum=0
lst1=[]
for num in range(1,1000):
#isprime(n):return True when the num is prime and false when the num is composite
if sympy.isprime(num) is True:
sum+=num
lst1.append(sum)
print("The sum list 1 is: ",lst1)
lst2=[]
for sum in lst1:
if sum<1000:
if sympy.isprime(sum)==True:
lst2.append(sum)
print("The list 2 is: ",lst2)
print("The required answer is :",max(lst2))
enter image description here
here the answer is different which gave me 281 is the highest whereas in the question it was mentioned 953 to be actual answer.
just by changing your upper limits from 1000 to 1000000 we get answer as 958577
enter image description here
EXPLANATION:when you add up manually u will get 963 instead of 953 which makes 963 a composite number.Therefore,i think there is a mistake in this question and maybe the developer gave a wrong answer or so.

class Example {
public static void main(String[] args) {
int count = 0;
int sum = 0;
for (int j = 2; j < 1000; j++) {
count = 0;
for (int i = 1; i <= j; i++) {
if (j % i == 0) {
count++;
}
}
if (count == 2) {
sum += j;
if (sum >= 1000) {
sum -= j;
break;
}
}
}
System.out.println("longest sum of consecutive primes that adds to a prime below 1000: " + sum);
}
}

Enter a code branch every ten percent till 100 percent

I can think of some very convoluted methods with loops and nested loops to solve this problem but I'm trying to be more professional than that.
My scenario is that I need to enter a section of code every ten percent but it isn't quite working as expected. It is entering the code about every percent which is due to my code but I lack the knowledge to know how to change it.
int currentPercent = Math.Truncate((current * 100M) / total);
//avoid divide by zero error
if (currentPercent > 0)
{
if (IsDivisible(100, currentPercent))
{
....my code that works fine other than coming in too many times
}
}
Helper referenced above where the trouble is:
private bool IsDivisible(int x, int y)
{
return (x % y) == 0;
}
So obviously it works as it should. Mod eliminates currentPercent of 3 but 1 & 2 pass when really I don't want a true value until currentPercent = 10 and then not again till 20...etc.
Thank you and my apologies for the elementary question

Mod will only catch exact occurrences of your interval. Try keeping track of your next milestone, you'll be less likely to miss them.
const int cycles = 100;
const int interval = 10;
int nextPercent = interval;
for (int index = 0; index <= cycles; index++)
{
int currentPercent = (index * 100) / cycles;
if (currentPercent >= nextPercent)
{
nextPercent = currentPercent - (currentPercent % interval) + interval;
}
}

I might misunderstand you, but it seems like you're trying to do something extremely simple more complex than it needs to be. What about this?
for (int i = 1; i <= 100; i++)
{
if (i % 10 == 0)
{
// Here, you can do what you want - this will happen
// every ten iterations ("percent")
}
}
Or, if your entire code enters from somewhere else (so no loop in this scope), the important part is the i % 10 == 0.

if (IsDivisible(100, currentPercent))
{
....my code that works fine other than coming in too many times
}
try changing that 100 to a 10. And I think your x and y are also backwards.
You can try a few sample operations using google calculator.
(20 mod 10) = 0

Not sure if I fully understand, but I think this is what you want? You also reversed the order of modulo in your code (100 mod percent, rather than the other way around):
int currentPercent = current * 100 / total;
if (currentPercent % 10 == 0)
{
// your code here, every 10%, starting at 0%
}
Note that code this way only works properly if you are guaranteed to hit every percentage-mark. If you could, say, skip from 19% to 21% then you'll need to keep track of which percentage the previous time was to see if you went over a 10% mark.

try this:
for (int percent = 1; percent <= 100; percent++)
{
if (percent % 10 == 0)
{
//code goes here
}
}

Depending on how you increment your % value, this may or may not work % 10 == 0. For example jumping from 89 to 91 % would effectively skip the code execution. You should store last executed value, 80 in this case. Then check if interval is >= 10, so 90 would work, as well as 91.

Getting Factors of a Number

I'm trying to refactor this algorithm to make it faster. What would be the first refactoring here for speed?
public int GetHowManyFactors(int numberToCheck)
{
// we know 1 is a factor and the numberToCheck
int factorCount = 2;
// start from 2 as we know 1 is a factor, and less than as numberToCheck is a factor
for (int i = 2; i < numberToCheck; i++)
{
if (numberToCheck % i == 0)
factorCount++;
}
return factorCount;
}

The first optimization you could make is that you only need to check up to the square root of the number. This is because factors come in pairs where one is less than the square root and the other is greater.
One exception to this is if n is an exact square then its square root is a factor of n but not part of a pair.
For example if your number is 30 the factors are in these pairs:
1 x 30
2 x 15
3 x 10
5 x 6
So you don't need to check any numbers higher than 5 because all the other factors can already be deduced to exist once you find the corresponding small factor in the pair.
Here is one way to do it in C#:
public int GetFactorCount(int numberToCheck)
{
int factorCount = 0;
int sqrt = (int)Math.Ceiling(Math.Sqrt(numberToCheck));
// Start from 1 as we want our method to also work when numberToCheck is 0 or 1.
for (int i = 1; i < sqrt; i++)
{
if (numberToCheck % i == 0)
{
factorCount += 2; // We found a pair of factors.
}
}
// Check if our number is an exact square.
if (sqrt * sqrt == numberToCheck)
{
factorCount++;
}
return factorCount;
}
There are other approaches you could use that are faster but you might find that this is already fast enough for your needs, especially if you only need it to work with 32-bit integers.

Reducing the bound of how high you have to go as you could knowingly stop at the square root of the number, though this does carry the caution of picking out squares that would have the odd number of factors, but it does help reduce how often the loop has to be executed.

Looks like there is a lengthy discussion about this exact topic here: Algorithm to calculate the number of divisors of a given number
Hope this helps

The first thing to notice is that it suffices to find all of the prime factors. Once you have these it's easy to find the number of total divisors: for each prime, add 1 to the number of times it appears and multiply these together. So for 12 = 2 * 2 * 3 you have (2 + 1) * (1 + 1) = 3 * 2 = 6 factors.
The next thing follows from the first: when you find a factor, divide it out so that the resulting number is smaller. When you combine this with the fact that you need only check to the square root of the current number this is a huge improvement. For example, consider N = 10714293844487412. Naively it would take N steps. Checking up to its square root takes sqrt(N) or about 100 million steps. But since the factors 2, 2, 3, and 953 are discovered early on you actually only need to check to one million -- a 100x improvement!
Another improvement: you don't need to check every number to see if it divides your number, just the primes. If it's more convenient you can use 2 and the odd numbers, or 2, 3, and the numbers 6n-1 and 6n+1 (a basic wheel sieve).
Here's another nice improvement. If you can quickly determine whether a number is prime, you can reduce the need for division even further. Suppose, after removing small factors, you have 120528291333090808192969. Even checking up to its square root will take a long time -- 300 billion steps. But a Miller-Rabin test (very fast -- maybe 10 to 20 nanoseconds) will show that this number is composite. How does this help? It means that if you check up to its cube root and find no factors, then there are exactly two primes left. If the number is a square, its factors are prime; if the number is not a square, the numbers are distinct primes. This means you can multiply your 'running total' by 3 or 4, respectively, to get the final answer -- even without knowing the factors! This can make more of a difference than you'd guess: the number of steps needed drops from 300 billion to just 50 million, a 6000-fold improvement!
The only trouble with the above is that Miller-Rabin can only prove that numbers are composite; if it's given a prime it can't prove that the number is prime. In that case you may wish to write a primality-proving function to spare yourself the effort of factoring to the square root of the number. (Alternately, you could just do a few more Miller-Rabin tests, if you would be satisfied with high confidence that your answer is correct rather than a proof that it is. If a number passes 15 tests then it's composite with probability less than 1 in a billion.)

You can limit the upper limit of your FOR loop to numberToCheck / 2
Start your loop counter at 2 (if your number is even) or 3 (for odd values). This should allow you to check every other number dropping your loop count by another 50%.
public int GetHowManyFactors(int numberToCheck)
{
// we know 1 is a factor and the numberToCheck
int factorCount = 2;
int i = 2 + ( numberToCheck % 2 ); //start at 2 (or 3 if numberToCheck is odd)
for( ; i < numberToCheck / 2; i+=2)
{
if (numberToCheck % i == 0)
factorCount++;
}
return factorCount;
}

Well if you are going to use this function a lot you can use modified algorithm of Eratosthenes http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes and store answars for a interval 1 to Max in array. It will run IntializeArray() once and after it will return answers in 0(1).
const int Max =1000000;
int arr [] = new int [Max+1];
public void InitializeArray()
{
for(int i=1;i<=Max;++i)
arr[i]=1;//1 is factor for everyone
for(int i=2;i<=Max;++i)
for(int j=i;i<=Max;i+=j)
++arr[j];
}
public int GetHowManyFactors(int numberToCheck)
{
return arr[numberToCheck];
}
But if you are not going to use this function a lot I think best solution is to check unitll square root.
Note: I have corrected my code!

An easy to implement algorithm that will bring you much farther than trial division is Pollard Rho
Here is a Java implementation, that should be easy to adapt to C#: http://www.cs.princeton.edu/introcs/78crypto/PollardRho.java.html

https://codility.com/demo/results/demoAAW2WH-MGF/
public int solution(int n) {
var counter = 0;
if (n == 1) return 1;
counter = 2; //1 and itself
int sqrtPoint = (Int32)(Math.Truncate(Math.Sqrt(n)));
for (int i = 2; i <= sqrtPoint; i++)
{
if (n % i == 0)
{
counter += 2; // We found a pair of factors.
}
}
// Check if our number is an exact square.
if (sqrtPoint * sqrtPoint == n)
{
counter -=1;
}
return counter;
}

Codility Python 100 %
Here is solution in python with little explanation-
def solution(N):
"""
Problem Statement can be found here-
https://app.codility.com/demo/results/trainingJNNRF6-VG4/
Codility 100%
Idea is count decedent factor in single travers. ie. if 24 is divisible by 4 then it is also divisible by 8
Traverse only up to square root of number ie. in case of 24, 4*4 < 24 but 5*5!<24 so loop through only i*i<N
"""
print(N)
count = 0
i = 1
while i * i <= N:
if N % i == 0:
print()
print("Divisible by " + str(i))
if i * i == N:
count += 1
print("Count increase by one " + str(count))
else:
count += 2
print("Also divisible by " + str(int(N / i)))
print("Count increase by two count " + str(count))
i += 1
return count
Example by run-
if __name__ == '__main__':
# result = solution(24)
# result = solution(35)
result = solution(1)
print("")
print("Solution " + str(result))
"""
Example1-
24
Divisible by 1
Also divisible by 24
Count increase by two count 2
Divisible by 2
Also divisible by 12
Count increase by two count 4
Divisible by 3
Also divisible by 8
Count increase by two count 6
Divisible by 4
Also divisible by 6
Count increase by two count 8
Solution 8
Example2-
35
Divisible by 1
Also divisible by 35
Count increase by two count 2
Divisible by 5
Also divisible by 7
Count increase by two count 4
Solution 4
Example3-
1
Divisible by 1
Count increase by one 1
Solution 1
"""
Github link

I got pretty good results with complexity of O(sqrt(N)).
if (N == 1) return 1;
int divisors = 0;
int max = N;
for (int div = 1; div < max; div++) {
if (N % div == 0) {
divisors++;
if (div != N/div) {
divisors++;
}
}
if (N/div < max) {
max = N/div;
}
}
return divisors;

Python Implementation
Score 100% https://app.codility.com/demo/results/trainingJ78AK2-DZ5/
import math;
def solution(N):
# write your code in Python 3.6
NumberFactor=2; #one and the number itself
if(N==1):
return 1;
if(N==2):
return 2;
squareN=int(math.sqrt(N)) +1;
#print(squareN)
for elem in range (2,squareN):
if(N%elem==0):
NumberFactor+=2;
if( (squareN-1) * (squareN-1) ==N):
NumberFactor-=1;
return NumberFactor

Append a digit to an integer and make sure sum of each digits ends with 1

What is the algorithm in c# to do this?
Example 1:
Given n = 972, function will then append 3 to make 9723, because 9 + 7 + 2 + 3 = 21 (ends with 1). Function should return 3.
Example 2:
Given n = 33, function will then append 5 to make 335, because 3 + 3 + 5 = 11 (ends with 1). Function should return 5.

Algorithms are language independent. Asking for "an algorithm in C#" doesn't make much sense.
Asking for the algorithm (as though there is only one) is similarly misguided.
So, let's do this step by step.
First, we note that only the last digit of the result is meaningful. So, we'll sum up our existing digits, and then ignore all but the last one. A good way to do this is to take the sum modulo 10.
So, we have the sum of the existing digits, and we want to add another digit to that, so that the sum of the two ends in 1.
For the vast majority of cases, that will mean sum + newDigit = 11. Rearranging gives newDigit = 11 - sum
We can then take this modulo 10 (again) in order to reduce it to a single digit.
Finally, we multiply the original number by 10, and add our new digit to it.

The algorithm in general:
(10 - (sum of digits mod 10) + 1) mod 10
The answer of the above expression is your needed digit.
sum of digits mod 10 gives you the current remainder, when you subtract this from 10 you get the needed value for a remainder of 0. When you add 1 you get the needed value to get a remainder of 1. The last mod 10 gives you the answer as a 1 digit number.
So in C# something like this:
static int getNewValue(string s)
{
int sum = 0;
foreach (char c in s)
{
sum += Convert.ToInt32(c.ToString());
}
int newDigit = (10 - (sum % 10) + 1) % 10;
return newDigit;
}

Another alternative using mod once only
int sum = 0;
foreach (char c in s)
sum += Convert.ToInt32(c.ToString());
int diff = 0;
while (sum % 10 != 1)
{
sum++;
diff++;
}
if (diff > 0)
s += diff.ToString();

Well, it's easier in C++.
std::string s = boost::lexical_cast<string>( i );
i = i * 10 + 9 - std::accumulate( s.begin(), s.end(), 8 - '0' * s.size() ) % 10;
Addicted to code golf…