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Problem with prime numbers not printed correctly [duplicate]

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Check if number is prime number
(31 answers)
Closed 2 years ago.
I have a problem with my code and i don't know how to solve it. Basically this program prints prime numbers based on the user input and at the end it prints their sum. This works perfectly until a certain amount, example: if i input 10, it shows ten correct prime numbers, but if i input 100, it also prints a number that is not prime, in this case 533. I don't know where i'm wrong.
Thanks for the support.
EDIT: I solved it on my own. Basically there was an error in the first "If" inside the for loop, i've simply added "c = n - 1;" after n++. Now it works perfectly.
Console.Write("How many prime numbers?: ");
int l = Convert.ToInt32(Console.ReadLine());
int n = 2;
int sum = 0;
sum += n;
Console.WriteLine(n);
n++;
int i = 1;
l++;
while (i < l)
{
for (int c = n - 1; c > 1; c--)
{
if (n % c == 0)
{
n++;
}
else if (n % c != 0 && c == 2)
{
sum += n;
Console.WriteLine(n);
n++;
i++;
}
}
}
Console.WriteLine("Sum: " + sum);

Let's start from extracting method:
public static bool IsPrime(int value) {
if (value <= 1)
return false;
if (value % 2 == 0)
retutn value == 2;
int n = (int) (Math.Sqrt(value) + 0.5);
for (int d = 3; d <= n; d += 2)
if (value % d == 0)
return false;
return true;
}
Having this method implemented you can easily compute the sum of the first N primes:
int N = 100;
long s = 0;
for (int p = 1; N > 0; ++p) {
if (IsPrime(p)) {
s += p;
N -= 1;
}
}
Console.Write(s);
Another (a bit more complex) possibility is prime numbers enumeration:
public static IEnumerable<long> Primes() {
yield return 2;
List<int> knownPrimes = new List<int>();
for (int p = 3; ; p += 2) {
int n = (int) (Math.Sqrt(p) = 0.5);
bool isPrime = true;
foreach (int d in knownPrimes) {
if (d > n)
break;
if (p % n == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
knownPrimes.Add(p);
yield return p;
}
}
}
Then you can query enumeration with a help of Linq:
using System.Linq;
...
long s = Primes()
.Take(N)
.Sum();

How to get integers to an array (from integer i to integer j). c#

I want to store integers(in an array or anything) that in range of int "i" and int "j".
eg:-Think, "int i = 1" and "int j = 10".I want to store integers from 1 and 10.
So that (1,2,3,4,5,6,7,8,9,10)
Because I want to answer to HackerRank "Beautiful Days at the Movies".
link below.
https://www.hackerrank.com/challenges/beautiful-days-at-the-movies/problem?isFullScreen=false
here is my code and it a garbage.
static int beautifulDays(int i, int j, int k) {
var total = 0;
for(var a = i; a <= j; a++ )
{
if (a != 0)
{
int ri = Reverse(i);
int rj = Reverse(j);
var ra = Reverse(a);
if((ra/k) % 1 == 0)
{
total++;
}
if((rj/k) % 1 == 0)
{
total++;
}
if((ri/k) % 1 == 0)
{
total++;
}
}
return total;
}
return total;
}
public static int Reverse(int inval)
{
int result = 0;
do
{
result = (result * 10) + (inval % 10);
inval = inval / 10;
}
while(inval > 0);
return result;
}
simply, can you give me the answer of HackerRank "Beautiful Days at the Movies".
link below.
https://www.hackerrank.com/challenges/beautiful-days-at-the-movies/problem?isFullScreen=false

Using Java you can easily stream a range of numbers with IntStream, then map the reverse function for each value, then filter those that fulfils the condition and count. With streams you don't need to store, you can get straight to the answer.
IntUnaryOperator reverse = (opperand) -> {
int reversed = 0;
int num = opperand;
while (num != 0) {
int digit = num % 10;
reversed = reversed * 10 + digit;
num /= 10;
}
return Math.abs(opperand - reversed);
};
return (int) IntStream.rangeClosed(i, j).map(reverse)
.filter(v -> v % k == 0).count();

Is a given number a prime or not, if not output all its divisors

The main part of my code is working, the only thing that doesn't work is the output of all its divisors. My result if it's not a prime should be like this:
Input -> 4
Output -> false 1 2 4
Console.WriteLine("Type your number: ");
int n = Convert.ToInt32(Console.ReadLine());
int a = 0, i;
for (i = 1; i <= n; i++)
{
if (n % i == 0)
{
a++;
}
}
if (a == 2)
{
Console.WriteLine("true");
}
else
{
Console.WriteLine("false" + i);
}
Console.ReadLine();

To print all the divisors, you'll need to gather them up in a collection of some sort – a list of integers, here.
In addition, all integers are divisible by 1, so you don't want to start there; neither do you want to end at n, since n % n == 0.
var divisors = new List<int>();
for (var i = 2; i < 2; i++)
{
if (n % i == 0)
{
divisors.Add(i);
}
}
if (divisors.Count == 0)
{
Console.WriteLine("true");
}
else
{
Console.WriteLine("false " + String.Join(" ", divisors));
}

Here is a working solution. You basically have to store your divisors somewhere or print them directly:
public static void Method(int n)
{
if (IsPrime(n))
{
Console.WriteLine($"{n} is a prime");
return;
}
var divisors = new List<int>();
for(var i = 1; i <= n; i++)
{
if (n % i == 0)
divisors.Add(i);
}
Console.WriteLine($"{n} isn't a prime");
Console.WriteLine($"The divisors are: {string.Join(", ", divisors)}");
}
public static bool IsPrime(int n)
{
for(var i = 2; i < n; i++)
{
if (n % i == 0)
return false;
}
return true;
}

From a brief inspection, there are two ways to generate the output. So far, you count the number of divisors, but neither store them nor write them to the output. You could replace
if (n % i == 0)
{
a++;
}
by
if (n % i == 0)
{
Console.WriteLine(i);
a++;
}
or store the divisors in a
List<int>
to generate the output afterwards.

Smallest Multiple

I have a code here written in C# that finds the smallest multiple by all numbers from 1 to 20. However, I find it very inefficient since the execution took awhile before producing the correct answer. I would like to know what are the different ways that I can do to improve the code. Thank You.
public static void SmallestMultiple()
{
const ushort ARRAY_SIZE = 21;
ushort[] array = new ushort[ARRAY_SIZE];
ushort check = 0;
for (uint value = 1; value < uint.MaxValue; value++)
{
for (ushort j = 1; j < ARRAY_SIZE; j++)
{
array[j] = j;
if (value % array[j] == 0)
{
check++;
}
}
if (check == 20)
{
Console.WriteLine("The value is {0}", value);
}
else
{
check = 0;
}
}
}

static void Main(string[] args)
{
int[] nums = Enumerable.Range(1, 20).ToArray();
int lcm = 1;
for (int i = 0; i < nums.Length; i++)
{
lcm = LCM(lcm, nums[i]);
}
Console.WriteLine("LCM = {0}", lcm);
}
public static int LCM(int value1, int value2)
{
int a = Math.Abs(value1);
int b = Math.Abs(value2);
// perform division first to avoid potential overflow
a = checked((a / GCD(a, b)));
return checked((a * b));
}
public static int GCD(int value1, int value2)
{
int gcd = 1; // Greates Common Divisor
// throw exception if any value=0
if (value1 == 0 || value2 == 0)
{
throw new ArgumentOutOfRangeException();
}
// assign absolute values to local vars
int a = Math.Abs(value1); // local var1
int b = Math.Abs(value2); // local var2
// if numbers are equal return the first
if (a == b) { return a; }
// if var "b" is GCD return "b"
if (a > b && a % b == 0) { return b; }
// if var "a" is GCD return "a"
if (b > a && b % a == 0) { return a; }
// Euclid algorithm to find GCD (a,b):
// estimated maximum iterations:
// 5* (number of dec digits in smallest number)
while (b != 0)
{
gcd = b;
b = a % b;
a = gcd;
}
return gcd;
}
}
Source : Fast Integer Algorithms: Greatest Common Divisor and Least Common Multiple, .NET solution

Since the result must also be divisible by 19 (which is the greatest prime number) up to 20, you might only cycle through multiples of 19.
This should get to to the result about 19 times faster.
Here's the code that does this:
public static void SmallestMultiple()
{
const ushort ARRAY_SIZE = 21;
ushort[] array = new ushort[ARRAY_SIZE];
ushort check = 0;
for (uint value = 19; value < uint.MaxValue; value += 19)
{
for (ushort j = 1; j < ARRAY_SIZE; j++)
{
array[j] = j;
if (value % array[j] == 0)
{
check++;
}
}
if (check == 20)
{
Console.WriteLine("The value is {0}", value);
return;
}
else
{
check = 0;
}
}
}
On my machine, this finds the result 232792560 in a little over 2 seconds.
Update
Also, please note that the initial program did not stop when reaching a solution; I have added a return statement to make it stop.

You're just looking for the LCM of the numbers from 1 to 20:
Where the GCD can be efficiently calculated with the Euclidean algorithm.
I don't know C#, but this Python solution shouldn't be hard to translate:
def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
def lcm(a, b):
return (a * b) / gcd(a, b)
numbers = range(1, 20 + 1)
print reduce(numbers, lcm)
It's pretty fast too:
>>> %timeit reduce(lcm, range(1, 20000))
1 loops, best of 3: 258 ms per loop

EDIT: v2.0 - Major speed improvement
Building on w0lf's solution. A faster solution:
public static void SmallestMultiple()
{
// this is a bit quick and dirty
// (not too difficult to change to generate primeProduct dynamically for any range)
int primeProduct = 2*3*5*7*11*13*17*19;
for (int value = primeProduct; ; value += primeProduct)
{
bool success = true;
for (int j = 11; j < 21; j++)
{
if (value % j != 0)
{
success = false;
break;
}
}
if (success)
{
Console.WriteLine("The value is {0}", value);
break;
}
}
}
You needn't check 1-10 since if something is divisible by x (e.g. 12), it is divisible by x/n (e.g. 12/2 = 6). The smallest multiple will always be a multiple of a product of all the primes involved.
Didn't benchmark C# solution, but equivalent Java solution runs in about 0.0000006 seconds.

Well I'm not sure what exactly you are trying to accomplish here but your out side for loop will run approximately 4,294,967,295 time (uint.MaxValue). So that will take some time...
If you have a way to keep from going to uint.MaxValue - like breaking your loop when you have accomplished what you need to - that will speed it up.
Also, since you are setting array[j] equal to j and then apparently never using the array again why not just do:
value % j
instead of
value % array[j]

Using also code written by W0lf (sorry but i cannot comment on your post) I would improve it (only a little) deleting the array that I think is useless..
public static void SmallestMultiple()
{
const ushort ARRAY_SIZE = 21;
ushort check = 0;
for (uint value = 1; value < uint.MaxValue; value++)
{
for (ushort j = 1; j < ARRAY_SIZE; j++)
{
if (value % j == 0)
{
check++;
}
}
if (check == 20)
{
Console.WriteLine("The value is {0}", value);
}
else
{
check = 0;
}
}
}

Sum of digits in C#

What's the fastest and easiest to read implementation of calculating the sum of digits?
I.e. Given the number: 17463 = 1 + 7 + 4 + 6 + 3 = 21

You could do it arithmetically, without using a string:
sum = 0;
while (n != 0) {
sum += n % 10;
n /= 10;
}

I use
int result = 17463.ToString().Sum(c => c - '0');
It uses only 1 line of code.

For integer numbers, Greg Hewgill has most of the answer, but forgets to account for the n < 0. The sum of the digits of -1234 should still be 10, not -10.
n = Math.Abs(n);
sum = 0;
while (n != 0) {
sum += n % 10;
n /= 10;
}
It the number is a floating point number, a different approach should be taken, and chaowman's solution will completely fail when it hits the decimal point.

public static int SumDigits(int value)
{
int sum = 0;
while (value != 0)
{
int rem;
value = Math.DivRem(value, 10, out rem);
sum += rem;
}
return sum;
}

int num = 12346;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;

I like the chaowman's response, but would do one change
int result = 17463.ToString().Sum(c => Convert.ToInt32(c));
I'm not even sure the c - '0', syntax would work? (substracting two characters should give a character as a result I think?)
I think it's the most readable version (using of the word sum in combination with the lambda expression showing that you'll do it for every char). But indeed, I don't think it will be the fastest.

I thought I'd just post this for completion's sake:
If you need a recursive sum of digits, e.g: 17463 -> 1 + 7 + 4 + 6 + 3 = 21 -> 2 + 1 = 3
then the best solution would be
int result = input % 9;
return (result == 0 && input > 0) ? 9 : result;

int n = 17463; int sum = 0;
for (int i = n; i > 0; i = i / 10)
{
sum = sum + i % 10;
}
Console.WriteLine(sum);
Console.ReadLine();

I would suggest that the easiest to read implementation would be something like:
public int sum(int number)
{
int ret = 0;
foreach (char c in Math.Abs(number).ToString())
ret += c - '0';
return ret;
}
This works, and is quite easy to read. BTW: Convert.ToInt32('3') gives 51, not 3. Convert.ToInt32('3' - '0') gives 3.
I would assume that the fastest implementation is Greg Hewgill's arithmetric solution.

private static int getDigitSum(int ds)
{
int dssum = 0;
while (ds > 0)
{
dssum += ds % 10;
ds /= 10;
if (dssum > 9)
{
dssum -= 9;
}
}
return dssum;
}
This is to provide the sum of digits between 0-9

public static int SumDigits1(int n)
{
int sum = 0;
int rem;
while (n != 0)
{
n = Math.DivRem(n, 10, out rem);
sum += rem;
}
return sum;
}
public static int SumDigits2(int n)
{
int sum = 0;
int rem;
for (sum = 0; n != 0; sum += rem)
n = Math.DivRem(n, 10, out rem);
return sum;
}
public static int SumDigits3(int n)
{
int sum = 0;
while (n != 0)
{
sum += n % 10;
n /= 10;
}
return sum;
}
Complete code in: https://dotnetfiddle.net/lwKHyA

int j, k = 1234;
for(j=0;j+=k%10,k/=10;);

A while back, I had to find the digit sum of something. I used Muhammad Hasan Khan's code, however it kept returning the right number as a recurring decimal, i.e. when the digit sum was 4, i'd get 4.44444444444444 etc.
Hence I edited it, getting the digit sum correct each time with this code:
double a, n, sumD;
for (n = a; n > 0; sumD += n % 10, n /= 10);
int sumI = (int)Math.Floor(sumD);
where a is the number whose digit sum you want, n is a double used for this process, sumD is the digit sum in double and sumI is the digit sum in integer, so the correct digit sum.

static int SumOfDigits(int num)
{
string stringNum = num.ToString();
int sum = 0;
for (int i = 0; i < stringNum.Length; i++)
{
sum+= int.Parse(Convert.ToString(stringNum[i]));
}
return sum;
}

If one wants to perform specific operations like add odd numbers/even numbers only, add numbers with odd index/even index only, then following code suits best. In this example, I have added odd numbers from the input number.
using System;
public class Program
{
public static void Main()
{
Console.WriteLine("Please Input number");
Console.WriteLine(GetSum(Console.ReadLine()));
}
public static int GetSum(string num){
int summ = 0;
for(int i=0; i < num.Length; i++){
int currentNum;
if(int.TryParse(num[i].ToString(),out currentNum)){
if(currentNum % 2 == 1){
summ += currentNum;
}
}
}
return summ;
}
}

The simplest and easiest way would be using loops to find sum of digits.
int sum = 0;
int n = 1234;
while(n > 0)
{
sum += n%10;
n /= 10;
}

#include <stdio.h>
int main (void) {
int sum = 0;
int n;
printf("Enter ir num ");
scanf("%i", &n);
while (n > 0) {
sum += n % 10;
n /= 10;
}
printf("Sum of digits is %i\n", sum);
return 0;
}

Surprised nobody considered the Substring method. Don't know whether its more efficient or not. For anyone who knows how to use this method, its quite intuitive for cases like this.
string number = "17463";
int sum = 0;
String singleDigit = "";
for (int i = 0; i < number.Length; i++)
{
singleDigit = number.Substring(i, 1);
sum = sum + int.Parse(singleDigit);
}
Console.WriteLine(sum);
Console.ReadLine();