The code gives me the answer 43739 which is wrong. I don't know which part of the code I messed up, help would be much appreciated.
{
int primenumbers = 4;
int thenumber = 2;
int finalnumber = 0;
while (primenumbers <= 10001)
{
for (int x = 2; x < 10; x++)
{
if (thenumber % x == 0)
{
x = 10;
}
if (x == 9)
{
finalnumber = thenumber;
primenumbers += 1;
break;
}
}
thenumber += 1;
}
Console.WriteLine(finalnumber);
}
Let's split the initial problem into smaller ones: first, check for being prime:
private static bool IsPrime(int value) {
if (value <= 1)
return false;
else if (value % 2 == 0)
return 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;
}
Then enumerate prime numbers (2, 3, 5, 7, 11...):
// Naive; sieve algorithm will be better off
private static IEnumerable<int> Primes() {
for (int i = 1; ; ++i)
if (IsPrime(i))
yield return i;
}
Finally, query with a help of Linq:
using System.Linq;
...
int result = Primes().Skip(10000).First();
Console.Write(result);
And you'll get
104743
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();
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.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Prime Factors In C#
I am trying to get this coding to give me all the prime factors of the integer that is inputted, including its duplicates. I have this current code and it seems to be working sort of but it isn't showing all of it's prime factors and duplicates. Any help would be appreciated.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace _1_Numeric_Problem
{
class Program
{
static void Main(string[] args)
{
string myInput;
int myInt;
int p;
Console.WriteLine(("Please input an integer"));
myInput = Console.ReadLine();
myInt = Int32.Parse(myInput);
{
for (int i = 2; i > 1; i++)
{
if (i == 100000)
break;
if (myInt % i == 0)
{
if (i <= 3)
{
Console.Write("{0} ", i);
Console.ReadLine();
continue;
}
else
{
for (p = 2; p < i; p++)
if (i % p != 0)
{
Console.Write("{0} ", i);
Console.ReadLine();
return;
Console.ReadLine();
}
else
{
p++;
continue;
}
}
}
}
}
}
}
}
Try replacing the following code:
for (p = 2; p < i; p++) {
if (i % p != 0) {
Console.Write("{0} ", i);
Console.ReadLine();
return;
Console.ReadLine();
} else {
p++;
continue;
}
}
With this instead:
bool isPrime = true;
for (p = 2; p <= Math.Sqrt(i); p++) {
if (i % p == 0) {
isPrime = false;
break;
}
if (isPrime) {
Console.Write("{0} ", i);
Console.ReadLine();
}
}
Can't you just make a for loop like this?
for (int i = 2; i < myInt; i++)
{
if(myInt % i == 0)
//Do something with it.
}
The basic algorithm for integer factorization using trial division tries each potential factor starting from 2, and if it divides n, outputs the factor, reduces n, and searches for the next factor; note that f is not incremented if it divides n, since it might again divide the reduced n. The loop stops when f is greater than the square root of n, since at that point n must be prime. Here's the pseudocode:
function factors(n)
f := 2
while f * f <= n
if n % f == 0
output f
n := n / f
else
f := f + 1
output n
There are better ways to factor integers, but that should get you started. When you're ready for more, I modestly recommend this essay on my blog.
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();