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static int Sss()
{
int k = int.Parse(Console.ReadLine());
int[] numbers = new int[k];
numbers = GenerateRandomNumbers(numbers);
for (int i = 0; i < numbers.Length; i++)
{
if (numbers[i] < 10 || numbers[i] > 99)
{
Console.WriteLine(numbers[i]);
I need to do a method, Ssk(k), which generates k random numbers and returns the product of these numbers which are double digits and ends with 5.
A non-negative integer x ends in 5 when written as a decimal if and only if x % 10 == 5 is true.
I note that you are returning an int, but ints can only go up to about two billion. The product of five two-digit numbers is almost certainly over that. You should use long, decimal, double or BigInteger instead, depending on your use case.
For any number you can check what the last digit is using by modulo operation, which in C# is a %, but works ONLY for integers. In your case you should check if numbers[i]%10==5.
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I need a method to return the firsts non zero numbers from a double in the following way: Any number >= 1 or == 0 will return the same; All the rest will return as per the following examples:
(Please note that I am using double because the potential imprecision is irrelevant in the use case whereas saving memory is relevant).
double NumberA = 123.2; // Returns 123.2
double NumberB = 1.2; // Returns 1.2
double NumberC = 0.000034; // Returns 3.4
double NumberD = 0.3; // Returns 3.0
double NumberE = -0.00000087; // Returns -8.7
One option would be to iteratively multiply by 10 until you get a number greater than 1:
public double RemoveLeadingZeros(double num)
{
if (num == 0) return 0;
while(Math.Abs(num) < 1) { num *= 10};
return num;
}
a more direct, but less intuitive, way using logarithms:
public double RemoveLeadingZeros(double num)
{
if (num == 0) return 0;
if (Math.Abs(num) < 1) {
double pow = Math.Floor(Math.Log10(num));
double scale = Math.Pow(10, -pow);
num = num * scale;
}
return num;
}
it's the same idea, but multiplying by a power of 10 rather then multiplying several times.
Note that double arithmetic is not always precise; you may end up with something like 3.40000000001 or 3.3999999999. If you want consistent decimal representation then you can use decimal instead, or string manipulation.
I would start with converting to a string. Something like this:
string doubleString = NumberC.ToString();
I don't know exactly what this will output, you will have to check. But if for example it is "0.000034" you can easily manipulate the string to meet your needs.
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I need someone to give me an idea on how to go on about this problem.Using a loop to calculate the fraction , There is no common value.I want to get the sum
Eg for fraction :
1 1/5 1/10 1/15 1/20 … 1/290 1/295 1/300
code snippet:-
int sum=0;
for(int i=1;i<=60 ;i++)
{
int sum=1
}
These sort of problems are actually surprisingly non-trivial due to issues with working with floating point, and decimal types for that matter.
Accepting that you want a loop solution for this (a closed form solution for n terms does exist), first note that your series can be written as
1 + 1/5(1 + 1/2 + 1/3 + ... + 1/60)
Then note that a good rule of thumb when working with floating point types is to add the small terms first.
So an algorithm would be of the form
double sum = 0.0;
for (int i = 60; i >= 1; --i){
sum += 1.0 / i;
}
sum = sum / 5 + 1;
Note the 1.0 in the numerator; that's there to defeat integer division.
Reference: Is floating point math broken?
͏Since you asked for a hint:
float sum = 1.0;
for (int i = 5; i <= ??; i += ??) {
sum += 1.0/i;
}
What goes in place of the ??s?
try this code:
double sum=1;
for(int i=5; i<=300; i+=5)
sum += (double) 1 / i;
The value of sum will be 1.93597408259035
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How can I find how many indivisible units of 1,000 are needed to overshadow a random number?
For example, for random number 5,123 I'm going to need 6 x 1,000 to overshadow it, so: MyAlgorithm(5123, 1000) = 6
Edit1: I am sorry if despite my endeavor to articulate my problem into a meaningful description my dyslexia took over, I hope this edit makes it a bit more comprehensible.
Well, if I understand your question, it sounds like you could simply convert the parameters to decimals, divide, then use Math.Ceiling:
int output = (int)Math.Ceiling((decimal)5123 / (decimal)1000); // 6
Alternatively, you could avoid the conversions and rely purely on integer division and the % operator (modulus), like this:
int output = (5123 / 1000) + (5123 % 1000 == 0 ? 0 : 1);
If you want this in a method simply wrap it up like this:
static int MyAlgorithm(int a, int b)
{
return (a / b) + (a % b == 0 ? 0 : 1);
}
if I've understood you correctly, this is really just a one-liner:
public static int MyAlgorithm(int input, int units)
{
return input%units == 0 ? input/units : input/units + 1;
}
the only case when it isn't simply the result of input/units + 1 is the case when there is no remainder
public int MyAlgorithm(int x, int y)
{
int result = x / y;
return (result < 0) ? (result - 1): (result + 1);
}
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I am trying to use the Math.Floor method on 2 user-input numbers but whenever I try to use the Math.Floor with my input3 and input3a it just doesn't work. I've seen examples for already set numbers such as in an array but not of numbers that the user inputs. Any help/clarification would be awesome.
static void Minimum()
{
Console.WriteLine("\n Enter two numbers and I shall determine the Minimum:\n");
Console.Write("> ");
Console.Write("\n> ");
// Read and parse input
string input3 = Console.ReadLine();
double d_input3 = Double.Parse(input3.Trim());
string input3a = Console.ReadLine();
double d_input3a = Double.Parse(input3a.Trim());
// Determine minimum of numbers
Console.WriteLine("\nThe Number {0} and {1}.\n", d_input3, d_input3a);
}
If think you don't need Math.Floor to determine a minimum :
Returns the largest integer less than or equal to the specified
double-precision floating-point number.
If you need to determine the minimum of both numbers, use Math.Min instead
Console.WriteLine("\nThe minimum number between {0} and {1} is {2}.\n", d_input3, d_input3a, Math.Min(d_input3, d_input3a));
The Math.Floor function takes one parameter and returns the value rounded down:
d_input3 = Math.Floor(d_input3);
d_input3a = Math.Floor(d_input3a);
The Math.Floor method is however not used to determine the lower value of two values. For that you would use the Math.Min function instead:
double lowest = Math.Min(d_input3, d_input3a);
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I have a quiz that have many questions and 5 possibilities of answers.
Lets take one question,it has follow answers:
148 - Good
5 - N/A
268 - Great
5 - Regular
11 - Bad
These are numbers collected directly from database.Now i need to show it as percentage.i.E:
Great - 45%
Good - 40
[..]
and so on
Any ideas?
int na = 5;
int good = 148;
int great = 268;
int regular = 5;
int bad = 11;
int sum = na + good + great + regular + bad;
naPercent = getPercent(na,sum);
float getPercent(int value, int sum)
{
return (value*100.0)/sum;
}
This is not a programming question, it is a math question. The percentage of each item is equal to the number of that item divided by the total number. In your example, the total number is 148+5+268+5+11 = 437. Great = 268 / 437 = 61.327%
total count for the answer / total count for all answers to this question combined * 100